5 - Nucleic Acid Structure

In Chapter 5, we will first learn about the major experiments that established DNA as the genetic material of nearly all organisms; recall that we learned in Chapter 1 that some viruses use RNA as the genetic material.  Then we will learn about the structures of both DNA and RNA.

A. DNA is the Genetic Material

Transformation

The story of DNA starts with the work of the British physician Frederick Griffith in the late 1920s.  Griffith didn't originally set out to study DNA; instead, Griffith was interested in developing a vaccine against the bacterium Streptococcus pneumoniae, one of the major causes of pneumonia, ear infections, and meningitis in children and elderly adults.  Some strains (varieties) of S. pneumoniae produce a polysaccharide capsule that surrounds the bacterial cell wall. 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 strains without capsules form rough colonies (called type R) on culture media.

Griffith showed the following (see figure 5.1):

  • When smooth (type S) strains of S. pneumoniae were injected into mice, the mice died. The smooth strain of S. pneumoniae could be isolated from the blood of the dead mouse.
  • When rough (type R) strains of S. pneumoniae were injected into mice, the mice survived. No bacteria were isolated from the blood of the living mouse.
  • When smooth (type S) S. pneumoniae were heat-killed to lyse the bacterial cells, and the bacterial extract was then injected into mice, the mice survive. No bacteria were isolated from the blood of the living mouse.
  • When live rough (type R) bacteria were mixed with heat-killed smooth (type S) bacteria and injected into mice, the mice died. Note that although neither the rough (type R) bacteria nor the heat-killed smooth (type S) bacteria killed mice on their own, the mixture of the two killed the mice. The living bacteria isolated from the blood of the dead mice were smooth (type S) bacteria.  

In this final experiment, Griffith reasoned that some chemical released from the heat-killed smooth (type S) bacteria was internalized by the living rough (type R) bacteria.  This chemical could change phenotype, converting the rough (type R) bacteria into smooth (type S) bacteria.  This change in phenotype is called transformation. The transformed bacteria then passed the type S trait to their progeny when the bacteria divided. As a result, the chemical responsible for transformation (i.e., the transforming principle) had properties of the genetic material.  The transforming principle changed the phenotype of the bacterial cells and was inherited when the bacterial cells divided.  Unfortunately, Griffith did not identify the chemical responsible for transformation.

5.1_The_Griffith_Experiment.jpg
Figure 5.1 The Griffith Experiment --- image used from OpenStax (access for free at https://openstax.org/books/biology-2e/pages/1-introduction)

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 Transforming Principle

    Oswald Avery, Maclyn McCarty, and Colin MacLeod sought to identify the chemical responsible for transforming rough (type R) into smooth (type S) bacteria in Griffith's experiment. Avery and his colleagues focused on three candidate chemicals: DNA, RNA, and protein (see figure 5.2).  To distinguish these three classes of chemicals, Avery, McCarty, and MacLeod performed three experiments:

    • In one experiment, rough (type R) bacterial cells were mixed with bacterial extracts from heat-killed smooth (type S) cells in the presence of ribonuclease (RNase) to digest RNA.  When RNA was eliminated from the type S bacterial extract, the rough bacteria were still transformed into smooth bacteria. Thus, digestion of RNA had no effect on transformation.
    • In a second experiment, rough (type R) bacterial cells were mixed with bacterial extracts from heat-killed smooth (type S) cells in the presence of protease to digest proteins.  When protein was eliminated from the type S bacterial extract, the rough bacteria were still transformed into smooth bacteria. Thus, digestion of protein had no effect on transformation.
    • In a final experiment, rough (type R) bacterial cells were mixed with bacterial extracts from heat-killed smooth (type S) cells in the presence of deoxyribonuclease (DNase) to digest DNA.  Interestingly, the rough bacteria were not transformed into smooth bacteria when DNA was eliminated from the bacterial extract.  Thus, the digestion of DNA prevented transformation.

    Avery and colleagues concluded from these three experiments that DNA 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://openstax.org/books/biology-2e/pages/1-introduction)

    Key Questions

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

      Bacteriophage T2

      To confirm the results from the Avery, McCarty, and MacLeod experiment, Alfred Hershey and Martha Chase examined the genetic material of the T2 bacteriophage (see figure 5.3)T2 bacteriophage is a type of virus that infects the bacterium Escherichia coli. T2 bacteriophage contains two molecular components: DNA and protein. The DNA of the bacteriophage is encased within a bacteriophage head structure made of protein. T2 bacteriophage also contains other protein structures, including a tube-like sheath, tail fibers, and a base plate. 

      During the bacteriophage life cycle, T2 bacteriophage acts like a hypodermic needle, injecting the bacteriophage genetic material into a host E. coli cell. 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. Hershey and Chase were interested in determining which of the two bacteriophage T2 components, DNA or protein, was injected into E. coli, leading to the production of progeny T2 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 Bacteriophages.  (Top) A bacteriophage as visualized in a scanning electron microscope.--- licensed under CC BY 4.0 (Bottom)  Bacteriophage DNA is encased within a head structure made of protein. --- Tevenphage by Adenosine licensed under CC BY-SA 2.5

      The Hershey-Chase Experiment

      The Hershey and Chase experiment relied on two important experimental details:

      • T2 bacteriophage proteins can be distinguished from bacteriophage DNA using radioactive labeling. Bacteriophage proteins were radiolabeled with 35S, a radioactive isotope of sulfur, and the bacteriophage DNA was radiolabeled with 32P, a radioactive isotope of phosphorus. Note that sulfur is found in proteins and not in DNA; phosphorus is a component of DNA and is not found in proteins. Thus, Hershey and Chase could determine whether DNA or protein is the bacteriophage T2 genetic material by determining whether 35S or 32P is injected into the host E. coli cell during a T2 bacteriophage infection.
      • A kitchen blender can separate the empty T2 bacteriophage components that remain attached to the E. coli surface after infection has occurred from the bacteriophage components that are injected into the cytoplasm of the E. coli cell.

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

      1. In one experiment, bacteriophage T2 proteins were radiolabeled with 35S. In another experiment, bacteriophage T2 DNA was radiolabeled with 32P.
      2. The radiolabeled bacteriophages were mixed in two separate reactions with E. coli cells to allow T2 bacteriophage infections to occur.
      3. After bacteriophage T2 injected its genetic material, the reactions were subjected to blending. During blending, the empty bacteriophage components that remained on the surfaces of the E. coli cells were released.
      4. The infected E. coli were collected in a centrifuge. Since bacteriophage components (phage head, tail, tail fibers) are considerably smaller than an E. coli cell, the empty bacteriophage subunits remain in the supernatant (liquid) after centrifugation. The larger E. coli cells containing the bacteriophage genetic material settled into the 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://openstax.org/books/biology-2e/pages/1-introduction)

      Blending removed most of the 35S from the E. coli cells in the pellet. Thus, proteins are not injected into E. coli to direct the formation of progeny T2 bacteriophages. In contrast, most of the 32P was found in the host E. coli cells in the pellet after blending and centrifugation, indicating that DNA is injected into host E. coli cells. Note that after the completion of the bacteriophage life cycle, the progeny bacteriophages also contained 32P; the progeny bacteriophages contained little 35S.  Thus, the 32P labeled DNA is heritable. To summarize, the Hershey-Chase experiment showed that DNA serves as the genetic material for bacteriophage T2. The work of Avery, McCarty, and MacLeod combined with the results of the Hershey-Chase experiment provided compelling evidence that DNA is the genetic material of many viruses and bacteria.

      Key Questions

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

      DNA is the Genetic Material of Eukaryotes

      Eukaryotic cells are not as easy to work with in the lab as bacteria and bacteriophages. As a result, it was 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.

      Several lines of indirect evidence suggest 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 (i.e., protein is found throughout the cell). 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. In contrast, the amount of protein is the same in both haploid and diploid cells. Finally, ultraviolet light (UV light) causes mutations that negatively affect the phenotype of a cell. The wavelength of UV light that produces the highest frequency of mutations 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 does not alter phenotype.

      Recombinant DNA technology (see Chapter 12) provided direct evidence that DNA is the genetic material of eukaryotic cells. In this technique, a DNA sequence from a eukaryotic cell is isolated and then introduced into a bacterial cell. This eukaryotic DNA sequence can then be transcribed by the bacterial cell to make a messenger RNA (mRNA); the mRNA is then translated by bacterial ribosomes to make a protein. The resulting protein often changes the phenotype of the bacterial cell. Moreover, the introduced eukaryotic DNA sequence is passed on to the progeny bacterial cells during bacterial cell division.  As an example, recombinant DNA technology allowed scientists to insert the human insulin gene into bacteria.  These bacterial cells then produce the human insulin protein (change in phenotype) and transfer the human insulin gene to daughter bacterial cells after cell division (inherited). The fact that an introduced eukaryotic gene results in protein production, alters the phenotype of a bacterial cell, and is inherited by progeny bacterial cells provided 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 shows that DNA is the genetic material of eukaryotes?

        B. The Structure of DNA and RNA

        Overview of Nucleic Acid Structure

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

        • Nucleotide. Nucleotides are the basic subunits of nucleic acid molecules.
        • Nucleic acid strand. A nucleic acid strand is a chain of nucleotides covalently linked together via phosphodiester bonds.
        • Double helix. Two nucleic acid strands of either DNA or RNA can hydrogen bond together to form a double helix structure.
        • Chromosomes. DNA molecules associate with NAP proteins or histone proteins to form prokaryotic and eukaryotic chromosomes. We discussed the structures of prokaryotic and eukaryotic chromosomes in Chapter 2.
        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 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://openstax.org/books/biology-2e/pages/1-introduction)

        Nucleotides

        Nucleotides are the building blocks of DNA and RNA molecules.  A nucleotide is composed of three molecular components (see figure 5.6):

        • Phosphate groups. When the cell synthesizes DNA and RNA strands (see Chapters 6 and 9), the nucleotides used as substrates contain three phosphate groups.  However, when a nucleotide is incorporated into a DNA or RNA strand, two of the phosphate groups are released.  The phosphate groups are negatively charged, giving the DNA double-helix a negative charge.
        • A pentose sugar. In DNA, the pentose sugar is deoxyribose; in RNA, the pentose sugar is ribose. Deoxyribose and ribose are distinguished by the chemical groups attached to the five carbon atoms within the pentose sugar.  Specifically, the five carbon atoms within both deoxyribose and ribose are numbered 1’ (one-prime) to 5’ (five prime) (see figure 5.6). The nitrogenous base (see below) is attached to 1’ carbon of both deoxyribose and ribose. The 2’ carbon in deoxyribose is attached to two hydrogen atoms; the 2’ carbon in ribose is attached to a hydroxyl group and a hydrogen atom. As a result, one of the fundamental differences between a DNA nucleotide and an RNA nucleotide is the chemical group attached to the 2’ carbon. The 3’ carbon in both deoxyribose and ribose is attached to a hydroxyl group. It is important to note that the hydroxyl group attached to the 3’ carbon plays a critical role in DNA replication and transcription (see Chapters 6 and 9). The 4’ carbon in both deoxyribose and ribose is involved in maintaining the integrity of the pentose ring structure. The 5’ carbon in both deoxyribose and ribose is attached to a phosphate group. This phosphate group also plays an important role in DNA replication and transcription.
        • A nitrogenous base. There are two types of nitrogenous bases found in DNA and RNA nucleotides (see figure 5.6):
          • Pyrimidines. The pyrimidine nitrogenous bases consist of a single carbon/nitrogen ring structure. The pyrimidine bases are cytosine (C), thymine (T), and uracil (U). Thymine (T) is found exclusively in DNA; Uracil (U) is found only in RNA. Cytosine (C) is found in both DNA and RNA.
          • Purines. The purine nitrogenous bases consist of a double carbon/nitrogen ring structure. The purines bases include adenine (A) and guanine (G). Adenine (A) and guanine (G) are found in both DNA and RNA molecules.

        The nucleotide substrates used to build DNA and RNA are called deoxynucleoside triphosphates (dNTPs) and nucleoside triphosphates (NTPs), respectively.  In the dNTP/NTP nomenclature "N" refers to any of the nitrogenous bases (A,U,C,G, or T); TP refers to triphosphate. For example, dTTP contains deoxyribose as the pentose sugar; deoxyribose is attached to thymine at the 1' carbon with three phosphate groups attached to the 5' carbon.  dTTP is a substrate for DNA replication.  GTP contains ribose as the pentose sugar; ribose is attached to guanine at the 1' carbon with three phosphate groups attached to the 5' carbon.  GTP is a substrate for transcription.

        Key Questions

        • What are the functions of the five carbon atoms 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?
        • What is meant by "dNTP" and "NTP"?
        EDI_5.6_Nucleotide_Structure-01_1.jpg
        Figure 5.6 Nucleotide Structure --- image used from OpenStax (access for free at https://openstax.org/books/biology-2e/pages/1-introduction)

        Nucleic Acid Strands

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

        A DNA or RNA strand is said to have 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 the nucleotide at the other end of the nucleic acid strand (the 3’ end of the strand) contains a free hydroxyl group.

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

        EDI_5.7_Structure_of_a_DNA_Strand-01_1.jpg
        Figure 5.7 Structure of a DNA Strand.  The 5'  end of the DNA strand is at the top of the image; the 3' end of the DNA strand is at the bottom of the image.  The DNA sequence of this nucleic acid strand is 5'-CGAT-3'. --- Image created by SL

        Key Questions

        • How are adjacent nucleotides linked within a nucleic acid strand?
        • What is meant by directionality (polarity)?
        • How do you know which end of the nucleic acid 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 structure of the DNA double helix; however, their work was built upon the contributions of other noteworthy scientists (see figure 5.8).  For example, in the early 1950s, Linus Pauling, who won two Nobel Prizes himself, demonstrated that ball-and-stick models could be built to show the locations of individual atoms within biological molecules. Pauling is known for using ball-and-stick models to describe the secondary structures within proteins. Watson and Crick mimicked Pauling’s approach by building a large ball-and-stick model of the DNA double helix.

          The structure of the DNA double helix was partly determined by X-ray diffraction. The X-ray diffraction technique involves subjecting a substance, such as DNA, to X-rays. When the X-rays pass through the DNA, the atoms within the DNA diffract the X-rays to produce a unique pattern that can be captured on photographic film. This pattern is then interpreted using 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 and a diameter of two nanometers (2 nm wide), suggestive of two nucleic acid strands. The X-ray diffraction results were critical for solving the structure of DNA; Wilkins shared the Nobel Prize with Watson and Crick.

          Moreover, Erwin Chargaff's work suggested a relationship between the nitrogenous bases in DNA.  Chargaff isolated DNA from many different organisms, including 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 the AT/GC rule or 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:

          • The phosphate-sugar backbones of the two nucleic acid strands are on the outside of the DNA molecule.
          • The nitrogenous bases are on the inside of the DNA molecule.
          • The two nucleic acid strands are antiparallel (see below), meaning that the two DNA strands have opposite polarity (i.e., run in the opposite directions).
          • Adenine forms two hydrogen bonds with thymine within the center of the DNA molecule; guanine forms three hydrogen bonds with cytosine. These hydrogen bonding interactions between nitrogenous bases are called base pairing.

          Watson and Crick also proposed a mechanism by which the DNA double helix is copied prior to cell division (semi-conservative replication). We will describe the details of semi-conservative replication in Chapter 6.

          Key Questions

          • What were the contributions of Pauling, Franklin, and Chargaff to the DNA story?
          • Describe Watson and Crick's model of the DNA double helix.

          The DNA Double Helix

          Watson and Crick proposed that the two phosphate-sugar backbones within DNA are found on the outside of the DNA molecule, directly exposed to water within the cell (see figure 5.9).  Also, hydrogen bonds are formed between pairs of nitrogenous bases, called base pairs (bp), located in the interior of the double-helix. Adenine (A) always forms two hydrogen bonds with thymine (T), and guanine (G) always forms three hydrogen bonds with cytosine (C). This relationship between nitrogenous bases is called the AT/GC rule or Chargaff’s rule.  Because of these hydrogen bonding interactions, DNA molecules with a higher proportion of GC base pairs are more stable than DNA molecules that are rich in AT base pairs.  

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

          The two nucleic acid strands of DNA are antiparallel (see figure 5.10). One DNA strand starts with the free 5’ phosphate group at the top of the DNA strand and ends with the free 3’ hydroxyl group at the bottom. The other DNA strand runs in the opposite direction; the free 5’ phosphate group is at the bottom end of the DNA strand, the free 3’ hydroxyl group is at the top. Moreover, you can predict the nitrogenous base sequence of one strand of DNA if you know the sequence of the other DNA strand. For example, if one DNA strand is 5’-GCCATG-3’, then the opposite DNA strand is 3’-CGGTAC-5’. As a result, the two DNA strands are said to be complementary (see figure 5.10).

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

          5.10_Complementary_Base_Pairs.jpg
          Figure 5.10 Complementary Base Pairs --- Image used from OpenStax (access for free at https://openstax.org/books/biology-2e/pages/1-introduction)

          Key Questions

          • What is meant by Chargaff's rule?
          • What is the distance between adjacent base pairs within DNA?
          • How many base pairs are found per helical turn in DNA?
          • How wide is the DNA double helix?
          • How does DNA 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 the clockwise direction as you look down the axis of the DNA molecule. The base pairs themselves are flat (planar) and stack on top of each other, much like the stairs of a spiral staircase. On the outside of the DNA double helix, there are major and minor grooves where the nitrogenous bases are exposed (see figure 5.9 above). Proteins that bind to specific base pair sequences within the DNA double-helix interact mainly with the major groove and to a lesser extent with the minor groove. These proteins bind to the DNA double helix to control DNA replication and transcription.

            Key Questions

            • How do DNA-binding proteins recognize the nitrogenous bases within the DNA double helix?

            Alternative Forms of DNA

            The structure of the DNA double helix detailed above is the standard form of DNA, called B DNA. B DNA is the predominant form of DNA that is found in aqueous environments, including living cells. Interestingly, there are at least two alternative forms of DNA, called A DNA and Z DNA (see figure 5.11). The A form of DNA is produced in the laboratory under high salt (low water) conditions. The A DNA structure is more compact than B DNA, has 11 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 double helix axis in B DNA, the base pairs in A DNA are tilted relative to the axis of the molecule.  The A form of DNA was studied by Rosalind Franklin using X-ray diffraction (see above). The Z form of DNA is a left-handed double helix, having a more extended structure.  Z DNA has 12 base pairs per turn of the helix and is only 1.8 nm wide. The Z form of DNA is favored over the B DNA form 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 in certain circumstances within living cells.  For example, scientists speculate that Z DNA formation near a particular gene may influence whether the gene can be activated by transcription to produce an RNA molecule.

             

            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?
            • Which form of DNA is the major form found within living cells?

            RNA Structure

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

            Since many 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. These base pairing interactions within a single RNA molecule form regions of double-stranded RNA. One important RNA structure formed in this way is a stem-loop (hairpin loop) which plays a role in transcription termination (see Chapter 9).  Additionally, transfer RNA (tRNA) molecules, which play a critical role in the translation process (see Chapter 11), are noteworthy because each tRNA molecule 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://openstax.org/books/biology-2e/pages/1-introduction)

            Key Questions

            • How are DNA and RNA strands similar?
            • What is a stem-loop?

            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 bind to DNA primarily at 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”.