5 - Nucleic Acid Structure
In Part 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 Part 1 that some viruses use RNA as the genetic material). In the second portion of Part 5, we will learn about the structure of both DNA and RNA.
A. Experiments that Established DNA as the Genetic Material
The physician Frederick 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.
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):
- When smooth (type S) strains of S. pneumoniae are injected into mice, the mice die. The smooth strain of S. pneumoniae can also be cultured from the blood of the dead mouse.
- When rough (type R) strains of S. pneumoniae are injected into mice, the mice survive. No bacteria can be isolated from the blood of the living mouse.
- When smooth (type S) S. pneumoniae are heat-killed and injected into mice, the mice survive. No bacteria can be isolated from the blood of the living mouse.
- When live rough (type R) bacteria are mixed with heat-killed smooth (type S) bacteria and injected into mice, the mice die. The living bacteria isolated from the blood of the dead mice are smooth (type S) bacteria.
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. This chemical could change phenotype, converting the rough (type R) bacteria into smooth (type S) bacteria (transformation). 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.
- 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 (i.e., the chemical) 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).
- When rough (type R) bacterial cells were mixed with material from heat-killed smooth (type S) cells in the presence of ribonuclease (RNase) to digest RNA, the rough bacteria were still transformed into smooth bacteria. Thus, digestion of RNA had no effect on transformation.
- When rough (type R) bacterial cells were mixed with material from heat-killed smooth (type S) cells in the presence of protease to digest proteins, the rough bacteria were still transformed into smooth bacteria. Thus, digestion of protein had no effect on transformation.
- When rough (type R) bacterial cells were mixed with material from heat-killed smooth (type S) cells in the presence of deoxyribonuclease (DNase) to digest DNA, the rough bacteria were not transformed into smooth bacteria. Thus, the digestion of DNA prevents transformation.
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.
- 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) called 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 progeny bacteriophage particles. In essence, Hershey and Chase were asking whether DNA or protein is the genetic material of bacteriophage T2.
The Hershey and Chase experiment relied on two important ideas:
- Bacteriophage components, which remain attached to the surface of the bacterium during an infection, can be separated from bacteriophage components that are injected into the cytoplasm of the E. coli cell using a kitchen blender.
- 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 proteins is the T2 genetic material by determining if either 35S or 32P is injected into a host E. coli cell during a bacteriophage infection.
The Hershey and Chase experiment was done as follows (see figure 5.4):
- In one experiment, bacteriophage T2 proteins were labeled with 35S. In another experiment, bacteriophage T2 DNA was labeled with 32P.
- The radiolabeled bacteriophages were mixed in two separate reactions with E. coli cells to allow bacteriophage infections to occur.
- The reactions were subjected to blending. During blending, the bacteriophage components attached to the surfaces of the E. coli cells were released.
- 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.
- The amount of radioactivity in the supernatant and pellet was calculated.
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, indicating that DNA is injected into host E. coli cells. 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.
- 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.
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 (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.
- 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):
- Nucleotide. Nucleotides are the basic subunits (the building blocks) of nucleic acid molecules.
- Nucleic acid strand. A nucleic acid strand is a chain of nucleotides covalently linked together.
- 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 proteins to form prokaryotic and eukaryotic chromosomes. We discussed the structures of prokaryotic and eukaryotic chromosomes in Part 2.
A nucleotide is composed of the following (see figure 5.6):
- One, two, or three phosphate groups. When a nucleotide is incorporated into a DNA or RNA molecule, the nucleotide contains a single phosphate group. When the cell synthesizes DNA and RNA strands, however, the nucleotides used as substrates contain three phosphate groups.
- A pentose sugar. In DNA, the pentose sugar is deoxyribose; in RNA, the pentose sugar is ribose. The carbon atoms within both deoxyribose and ribose are numbered 1’ (one-prime) to 5’ (five prime). The nitrogenous base (see below) in both deoxyribose and ribose is attached to 1’ carbon. The 2’ carbon in a DNA nucleotide is attached to two hydrogen atoms, the 2’ carbon in an RNA nucleotide 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 groups 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 an important role in DNA and RNA synthesis. 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 and RNA synthesis.
- A nitrogenous base. There are two groups of nitrogenous bases:
- Pyrimidines. The pyrimidine bases consist of a single-ring structure. The pyrimidine bases are cytosine (C), thymine (T), and uracil (U). T is found only in DNA; U is found only in RNA. C is found in both DNA and RNA.
- Purines. The purine bases consist of a double-ring structure. The purines bases include adenine (A) and guanine (G). A and G are found in both DNA and RNA.
The nucleotide substrates that are used to build DNA and RNA are called deoxynucleoside triphosphates and nucleoside triphosphates, respectively. These substrate molecules are labeled dNTPs and NTPs. 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.
- 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?
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).
- 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 atoms within the DNA diffract the X-rays to produce a unique pattern. This pattern can be 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, a diameter of 2 nanometers (2 nm) suggestive of two nucleic acid strands, and 10 base pairs (bp) per turn of the helix.
Erwin Chargaff isolated DNA from many different organisms (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:
- The 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).
- Adenine forms two hydrogen bonds with thymine; guanine forms three hydrogen bonds with cytosine. These interactions are called base pairing.
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 the process of semi-conservative replication in Part 6.
- 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-sugar backbone within DNA is found on the outside of the DNA molecule; the nitrogenous bases are found on the inside of the DNA molecule. The backbones of the DNA molecule are directly exposed to water within the cell (see figure 5.9).
DNA is 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 relationship between nitrogenous bases 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.
The two nucleic acid strands of DNA are antiparallel. One strand starts with the free 5’ phosphate group at the top of the strand and ends with a free 3’ hydroxyl group at the bottom of the strand. The other DNA strand runs in the opposite direction; the free 5’ phosphate group is at the bottom end of the strand, the free 3’ hydroxyl group is at the top of the strand.
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).
- 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 major and minor grooves where the nitrogenous bases are directly exposed to the environment (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 double helix to control DNA replication and transcription.
- 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 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 (dehydrating) 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 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. 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 blocks the major groove, transcription of nearby genes is likely inhibited. Thus, the formation of triplex DNA may provide scientists a means to inhibit overactive genes, for example, in cancerous cells.
- How do A-DNA, B-DNA, and Z-DNA differ from one another?
- What effect does the triple helix have on gene transcription?
Nucleic acid strands composed of RNA nucleotides are very similar to DNA strands. RNA strands have a backbone composed of negatively charged 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 contain ribose, 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).
- What are the structural differences between DNA and RNA?
Fill in the Blank:
- In the Griffith experiment, the ______________ strain of bacteria kills the mouse.
- The bacterium ___________________ was used in the Hershey-Chase experiment, while the bacterium ______________________ was examined in the Frederick Griffith experiment.
- Avery, McCarty, and MacLeod found that the enzyme _______________ destroyed the transforming principle in bacteria, whereas the enzymes _____________ and _______________ did not.
- The _______________________________ experiment showed that DNA is the genetic material of bacteriophage T2.
- In eukaryotes, a __________ cell has twice the amount of DNA as a __________ cell, but both cells have similar amounts of protein.
- Each base pair within the DNA double helix is separated by ________ nanometers (nm).
- The two purines found in RNA are ________________ and ___________________.
- The pentose sugar in dCTP is ______________________, while the pentose sugar in ATP is ______________________.
- Proteins that interact directly with DNA primarily bind to the _____________ groove.
- The _____ form of DNA exists in aqueous environments while the _____ form of DNA exists when cytosines are methylated.
- The ______________________ experiment is sometimes called the “blender experiment”.
End-of-Chapter Survey: How would you rate the overall quality of this chapter?
- Very Low Quality
- Low Quality
- Moderate Quality
- High Quality
- Very High Quality