Modified from Peller S, Rotter V: TP53 in hematological cancer: low incidence of mutations with significant clinical relevance, Hum Mutat 21:277-284, 2003.

BOX 32-2

Inherited Hematologic Disorders Detected by Molecular Diagnostic Methods

Hemoglobinopathies

Sickle cell anemia

Hemoglobin C disease

Hemoglobin SC disease

Hereditary persistence of fetal hemoglobin

Coagulation factor V Leiden mutation

Prothrombin G20210A mutation

Erythrocytic disorders

Lipid storage diseases

Neutrophil disorders

Modified from Crisan D: Pioneering advances in molecular hematology: use of nucleic-acid technology in hematologic disease diagnosis and monitoring—molecular pathology, part 4, MLO Med Lab Obs 27(10):48-56, 1995.

BOX 32-3

Hematologically Important Pathogens Detected by Molecular Diagnostic Methods

Viral pathogens associated with hemolytic anemia

Parvovirus B19

Cytomegalovirus

Epstein-Barr virus

Human immunodeficiency virus types 1 and 2

Human T-cell lymphoma virus type 1

Modified from Paessler M, Bagg A: Use of molecular techniques in the analysis of hematologic diseases. In Hoffman R, Benz EJ Jr, Shattil SJ, et al, editors: Hematology: basic principles and practice, ed 4, Philadelphia, 2005, Churchill Livingstone, pp 2713-2726.

Structure and Function of Dna

The Central Dogma: DNA to RNA to Protein

Much of the stored information needed to carry out cell processes resides in deoxyribonucleic acid (DNA); therefore, proper cellular storage, maintenance, and replication of DNA are necessary to ensure homeostasis. Since molecular testing takes advantage of DNA structure and replication, a review of molecular biology is helpful.

The central dogma in genetics is that information stored in the DNA is replicated to daughter DNA, transcribed to messenger ribonucleic acid (mRNA), and translated into a functional protein (Figure 32-3). This process is essential to carry out cellular functions while preserving a record of the stored information. In eukaryotes, the initial DNA sequence is composed of exons separated by untranslated introns. The introns are enzymatically excised during transcription from DNA to RNA, and the mature mRNA sequence is then translated. Translation is an enzymatic process wherein mRNA three-member base sequences called codons drive the addition of individual amino acids to the growing peptide. The mature protein then carries out its cellular function, which may be structural or may involve recognition, regulation, or enzymatic activity.

Figure 32-3 RNA polymerase transcribes DNA into a primary RNA transcript. The introns in the primary RNA transcript are excised and the exons are spliced. The spliced messenger RNA (mRNA) enters the cytoplasm of the cell. The ribosomes translate the mRNA into protein.

The structural units that carry DNA’s message are called genes. The human β-globin gene, part of the hemoglobin molecule, provides a good example of replication and transcription, because it was one of the first sequenced and demonstrates the result of aberrant sequence maintenance. A normal (or wild-type) β-globin gene contains a sequence of bases that code for a β-globin peptide of 146 amino acids. One inherited mutation changes a single DNA base. This is called a point mutation. The mutation occurs in the portion of the sequence that codes for the sixth amino acid of β-globin. The mutation substitutes the amino acid valine for glutamine in the growing peptide. Valine modifies the overall charge, producing a protein that polymerizes in a low-oxygen environment. This leads to sickled erythrocytes, circulatory ischemia, and poor oxygen exchange between blood and tissues.1,2 A mutation in one of the two copies (alleles) of this gene inherited from the parents results in a heterozygous condition, or a sickle cell trait. In a heterozygote, the symptoms of the disease are often unseen or are present only during times of physical stress. If both alleles are mutated, there is overt homozygous sickle cell disease, and the symptoms are severe.

Every active gene is translated. Human somatic cells contain 20,000 to 25,000 genes in 2 meters of DNA.3,4 Significant packing (see Chapter 31) takes place to reduce the volume of the nucleic acid to the size of chromosomes.

DNA at the Molecular Level

DNA is a duplex molecule composed of two complementary hydrogen-bonded nucleotide strands (Figure 32-4). Deoxyribonucleotides and ribonucleotides are the building blocks of DNA and RNA, respectively. Each nucleotide is composed of a 5-carbon sugar (pentose), a nitrogenous base, and a phosphate group (Figure 32-5). The numbers one prime (1′) to five prime (5′) designate the pentose’s carbons. In DNA, the pentose is a ribose in which the hydroxyl group on the 2′ carbon is replaced by a hydrogen molecule, hence 2′-deoxyribose. In RNA, the 2′ ribose retains the 2′ hydroxyl group. The hydroxyl group present on the 3′ carbon of the sugar is crucial for polymerization of the nucleotide monomers to form the nucleic acid strand.

Figure 32-4 DNA is a double-stranded helical macromolecule consisting of nucleotide subunits joined in sequence by deoxyribose molecules (pentagons) and phosphate radicals (circles). The bases thymine (T), adenine (A), cytosine (C), and guanine (G) are illustrated in their standard pairs: thymine to adenine, cytosine to guanine.

Figure 32-5 A, The pentose sugar deoxyribose, a phosphate group, and a nitrogenous base compose a DNA nucleotide. The carbons of the deoxyribose molecule are numbered 1′ through 5′. B, A nucleotide results from the formation of a phosphodiester bond between the nitrogenous base and the hydroxyl group on the 1′ carbon of deoxyribose and a glycosidic bond between the phosphate group and the hydroxyl group on the 5′ carbon of deoxyribose. C, A nucleotide illustrating the glycosidic and phosphodiester bonds.

The nitrogenous base is linked to the sugar by a glycosidic bond at the 1′ carbon. Four different bases form DNA, but the linkage to the sugar is the same for each. The phosphate group is linked to sugar at the 5′ carbon by a phosphodiester bond. The phosphate group is also crucial for addition of nucleotides to the growing polymer. A sugar, whether ribose or deoxyribose, linked to a nitrogenous base, but without a phosphate group, is called a nucleoside. A nucleoside cannot be incorporated into DNA, and neither can a nucleotide consisting of only one phosphate group (deoxynucleotide monophosphate, or dNMP). To be incorporated into a growing strand of DNA, the nucleotide must have three phosphate groups linked to one another, referred to as the α-, β-, and γ-phosphates. The α-phosphate is linked to the sugar.

Creation of a phosphodiester bond between the 3′ hydroxyl group of the existing strand and the 5′ α-phosphate of the nucleotide monomer requires the protein enzyme DNA polymerase. This enzyme recognizes the hydroxyl group on the 3′ carbon of the sugar and bonds the 3′ hydroxyl group of one nucleotide with the α-phosphate group of another (Figure 32-6). Polymerization of subsequent nucleotides forms a DNA strand.

Figure 32-6 The enzyme DNA polymerase catalyzes the reaction between the hydroxyl group on the 3′ carbon of one nucleotide with the phosphate group bound to the 5′ carbon of the downstream nucleotide. The α-phosphate group is split by the 3′-OH, with release of the β- and γ-phosphates.

DNA consists of two strands that are antiparallel and complementary (Figure 32-7). One strand begins with a phosphate group attached to the 5′ carbon of the first nucleotide oriented to the left and ends with the hydroxyl group on the 3′ carbon of the last nucleotide oriented to the right. This strand is in the 5′-to-3′ direction. The other strand runs in the 3′-to-5′ direction, or antiparallel. The nucleotide sequences composing these strands provide the encoded messages of our genes. Therefore, the addition of nucleotides is highly regulated.

Figure 32-7 DNA consists of two antiparallel and complementary strands. One strand begins with a 5′ phosphate group and ends with a 3′ hydroxyl group. This strand is read in the 5′-to-3′ direction. The other strand begins with a 3′ hydroxyl group and ends with a 5′ phosphate group. This strand is shown in the 3′-to-5′ orientation.

One regulation mechanism arises from the complementary characteristic of the nucleotides. A nucleotide’s identity depends on the type of nitrogenous base present on the template. There are two categories of nitrogenous bases in nucleic acids, purines and pyrimidines (Figure 32-8). The bases adenine (A) and guanine (G) are double-ringed purines, whereas thymine (T) and cytosine (C) are single-ringed pyrimidines. Adenine forms hydrogen bonds at two points with thymine (A:T), whereas guanine forms hydrogen bonds at three points with cytosine (G:C). If a strand has a 5′-CTAG-3′ sequence, the complementary nucleotides on the 3′-to-5′ strand are 3′-GATC-5′. In RNA, the pyrimidine uracil (U) takes the place of thymine and forms hydrogen bonds with adenine. Hydrogen bonds between A:T and G:C hold the strands together (Figure 32-9). RNA is most often single-stranded.

Figure 32-8 The single-ringed pyrimidines thymine and cytosine and the double-ringed purines adenine and guanine are the code-carrying nitrogenous bases of DNA.

Figure 32-9 A, The purine adenine forms two hydrogen bonds with the pyrimidine thymine. The purine guanine forms three hydrogen bonds with the pyrimidine cytosine. B, The two strands maintain a consistent distance from each other, which allows DNA to twist into a helix.

In addition to conferring identity to the nucleotide, the nitrogenous bases assist in maintaining a constant width between the strands of a DNA molecule. DNA resembles a ladder, with the repeating sugar and phosphate groups forming the sides of the ladder and the bases forming the rungs. The pairing of a double-ringed purine on one strand with a single-ringed pyrimidine on the other maintains a consistent distance between the DNA strands. This makes DNA flexible, which allows the molecule to twist into a helix. Twisting stabilizes the molecule and protects the bases from their environment.

Transcription and Translation

DNA provides a permanent set of instructions. The cellular enzyme RNA polymerase transcribes the code. RNA polymerase recognizes starter sequences called promoters. Promoters lie upstream of coding sequences and bind RNA polymerase to separating DNA strands. The enzyme then slides along the DNA strand 3′ to 5′, “reading” the code and polymerizing (assembling) the complementary ribonucleotides. As the complementary ribonucleotides form hydrogen bonds with the bases of the exposed DNA strand, the RNA polymerase creates phosphodiester bonds to extend the single-stranded primary RNA transcript (Figure 32-10). If the nucleotide sequence of the DNA strand is 3′-CTAG-5′, the primary RNA transcript is 5′-GAUC-3′.

Figure 32-10 RNA polymerase binds to a sequence of DNA called the *promoter region,* which causes the DNA strands to separate. Using one of the DNA strands as a template, RNA polymerase moves along and simultaneously reads the DNA strand, forming the primary messenger RNA (mRNA) transcript by joining the complementary ribonucleotides. The primary mRNA transcript consists of sequences called *exons* that provide coding information and *introns* that are excised from mRNA. Introns may contain important information, and their functions remain under investigation.

Primary mRNA segments are composed of introns and exons. Introns are untranslated intervening sequences located within the coding portions of genes. Their functions remain unclear, although they may play a role in regulation of gene expression.⁵ Exons are the sequences that encode the gene product. Before mRNA can serve as a translation template, the introns must be excised from the primary transcript and the exons adjoined. The mature mRNA is completed by the addition of a 5′ cap and a tail of many repeated adenine nucleotides.⁶ The mRNA leaves the nucleus and enters cytoplasmic ribosomes to be translated.

Ribosomes translate the mRNA code into a peptide sequence. Complexes of proteins and structural ribosomal RNAs (rRNAs) form both large and small ribosome subunits. Mature cytoplasmic mRNA is bound by the small ribosomal subunit at the translation start site. At this point, another series of elements is introduced, transfer RNAs (tRNAs), each bound to its specific amino acid. Because there are 20 natural amino acids, there are 20 tRNAs. Each tRNA has a specific nucleic acid sequence located at the point of interaction with the mRNA, complementary to the nucleotide sequence of the mRNA. Each tRNA interacting sequence (anticodon) complements a specific three-nucleotide sequence (codon) of the mRNA.

The mRNA codon AUG is the most common translation start site and codes for the amino acid methionine. The first step in translation is hydrogen bonding of the appropriately charged tRNA (with a bound methionine) to the start codon of the mRNA. The appropriate tRNA is then bonded to the adjacent codon and a peptide bond is catalyzed between the two amino acids. The peptide bond forms between the carboxyl terminus of the methionine in the existing peptide chain and the amino terminus of the amino acid to be added. Hydrogen bonding of tRNAs to the codons and the formation of the peptide bonds are mediated by the ribosome. With addition of more amino acids, translation proceeds until a termination site (nonsense codon) is reached. Three codons exist that do not code for any amino acid: UAA, UAG, and UGA. These terminate translation. The ribosome then dissociates, and the peptide folds to its functional shape.

DNA Replication and the Cell Cycle

After cells carry out their functions, they either divide via mitosis or die via apoptosis, also called programmed cell death. The cell cycle progresses through a sequence (Figure 32-11). Interphase is made up of the G₁, S, and G₂ phases. During the G₁ phase, the cell grows rapidly and performs its cellular functions. S phase is the synthesis stage, in which DNA is replicated. The G₂ phase is the period when the cell produces materials essential for cell division. The M phase refers to mitosis, during which two identical daughter cells are produced, each of which receives one entire set of the DNA that was replicated during S phase. Some cells exit the cell cycle during the G₁ phase and enter a phase called G₀. Cells in G₀ normally do not reenter the cell cycle and remain alive performing their function until apoptosis occurs.

Figure 32-11 The cell cycle consists of interphase and mitosis. Interphase is divided into G₁, S, and G₂ phases. Cell growth occurs during G₁. During the S phase, DNA synthesis (replication) occurs. The cell prepares for mitosis during the G₂ phase. During mitosis the cell divides, producing two identical daughter cells. Cells may also enter a quiescent phase called *G₀.* During G₀, the cell performs its function but does not replicate.

DNA replication during the S phase requires a complex orchestration of events; this discussion focuses on those events that are exploited for molecular diagnostic testing. Contained within the double-stranded DNA helix are multiple origins of replication. At each origin, the enzyme helicase disrupts and untwists the hydrogen bonds, separating the DNA strands and producing two replication forks. Here a deoxyribonucleotide (deoxynucleotide triphosphate, or dNTP) polymerizes to form new complementary strands (Figure 32-12). DNA replication occurs bidirectionally from the two replication origin sites. Each DNA strand in the replication fork serves as a template for the formation of a daughter or complementary strand through the activity of DNA polymerase.⁷ The DNA polymerase substrate is the free hydroxyl group located on the 3′ carbon of a deoxyribonucleotide. DNA polymerase recognizes the group and catalyzes the joining of the complementary deoxyribonucleotide. DNA is read 3′ to 5′ by DNA polymerase, and the complementary strand is synthesized 5′ to 3′.

Figure 32-12 DNA replication occurs at replication origin sites throughout the DNA. Helicases separate the DNA strands, producing replication forks to the left and right of the origins.

A primer provides the free 3′ hydroxyl group required for DNA polymerase activity. Primers are short nucleotide polymers complementary to the template. The hybridization of the primer to the template requires the enzyme primase. At the replication origin, primase joins a primer to the 3′ end of the 5′-to-3′ (top) template strand (Figure 32-13). Then DNA polymerase recognizes the free hydroxyl group on the 3′ carbon of the last nucleotide in the primer and catalyzes the formation of phosphodiester bonds between the correct complementary nucleotide triphosphate and the primer, releasing the β- and γ-phosphate groups. DNA polymerase continues adding deoxyribonucleotides along the replication fork, going to the left of the replication origin, producing the complementary strand called the leading strand.

Figure 32-13 A, Primases join primers to the single-stranded template strands. The primers must be oriented in such a way that the hydroxyl group on the 3′ end of the primers is available for deoxyribonucleotide addition by DNA polymerase. B, DNA polymerase extends the primer located on the 5′-to-3′ template strand, producing the complementary leading strand (*blue*). On the 3′-to-5′ template strand, DNA polymerase extends the primers, producing Okazaki fragments. The primer ribonucleotides (*red*) are replaced with deoxynucleotides by DNA polymerase to produce the complementary lagging strand (*green*).

The second template strand, called the lagging strand, is also read in the 3′-to-5′ direction. To form a complementary strand, a primer hybridizes to the exposed 3′ end of the replication fork. To proceed in the 5′-to-3′ direction, nucleotides are added in fragments toward the origin of replication. As the left replication fork extends to open more of the template strands for replication, additional primers are hybridized, and DNA polymerase uses the primers to initiate the formation of the complementary strand, continuing until it meets a previously hybridized primer.

DNA polymerase not only joins nucleotides, it also degrades the RNA primers and fills in the correct complementary deoxyribonucleotides. Because the replication of the lagging strand produces many small fragments, it is called discontinuous replication, and the fragments are called Okazaki fragments. Finally, the enzyme ligase joins the discontinuous fragments. The replication fork to the right (downstream) is replicated in the same fashion, although the lagging strand is now formed complementary to the top (5′-to-3′) strand, and the leading strand is formed from the 3′-to-5′ strand; the opposite of the situation described occurs for the left replication fork (Figure 32-14).

Figure 32-14 Bidirectional DNA replication. The 5′-to-3′ parent strand serves as the template for producing the continuous leading strands on a replication fork to the left of an origin. The 3′-to-5′ parent strand is the template for the lagging strands, which are produced in a discontinuous manner. The continuous and discontinuous strands are reversed on the replication fork to the right.

The cell cycle is highly regulated. At certain critical points within the cycle, decisions are made to continue or begin cell death via apoptosis. This decision may depend on the state of the DNA replicated (Figure 32-15). Normally, the cell detects errors made during replication and either corrects them or begins apoptosis. This prevents the persistence of daughter cells with genetic errors. If the sensing molecules fail, cell division may continue. Debilitating mutations that mediate cell cycle control may result in tumor formation.

Figure 32-15 Toward the end of the G₁ phase, the first critical point in the cell cycle is reached. At this critical point, the cell will either continue into the S phase or go through apoptosis. The second critical point is located at the end of the G₂ phase. At this point, the cell will either continue into mitosis or initiate apoptosis.

One protein responsible for signaling damaged DNA is p53, a tumor suppressor protein. Damaged cells with increased p53 arrest cell division at G₁, which allows time for DNA repair (Figure 32-16). Cells with mutant p53 are unable to arrest cells in G₁; they continue the process of cell division with damaged DNA.^8–10 If the cell can repair the DNA damage, the cell cycle continues. If the cell damage is too severe, the cell undergoes apoptosis. Hematologic malignancies, such as 21% of chronic myelogenous leukemias (CML),^11–13 23% of chronic lymphocytic leukemias (CLL),^14–16 and 17% of acute lymphoblastic leukemias (ALL),^17–19 are associated with a p53 mutation or deletion (see Box 32-3). In summary, DNA synthesis and accurate cell cycle control demand that the integrity of the nucleotide sequence be maintained during DNA replication.