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Biochemistry_Lippinco. and Pi by increasing bone resorption and intestinal absorption and renal reabsorption of Ca2+ and Pi (Fig. 29.2). PTH (from the parathyroid glands) increases serum Ca2+ by increasing bone resorption, increasing renal reabsorption of Ca2+ , and activating the renal 1-hydroxylase that produces calcitriol from calcidiol (see p. 390) (Fig. 29.3). In contrast to calcitriol, PTH decreases Pi reabsorption in the kidneys, lowering serum Pi. [Note: High serum Pi increases PTH and decreases calcitriol.] A third hormone, calcitonin (from the C cells of the thyroid gland), responds to elevated serum Ca2+ levels by promoting bone mineralization and increasing renal excretion of Ca2+ (and Pi).


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Biochemistry_Lippinco. B. Magnesium About 60% of the body’s Mg2+ is in bone, but it accounts for just 1% of the bone mass. The mineral is required by a variety of enzymatic reactions, including phosphorylation by kinases (Mg2+ binds the ATP cosubstrate) and phosphodiester bond formation by DNA and RNA polymerases. Mg2+ is widely distributed in foods, but the average intake in the United States is below the recommended level. Hypomagnesemia can result from decreased absorption or increased excretion of Mg2+ . Symptoms include hyperexcitability of skeletal muscles and nerves and cardiac arrhythmias. With hypermagnesemia, hypotension is seen. [Note: Magnesium sulfate is used in the treatment of preeclampsia, a hypertensive disorder of pregnancy.] C. Sodium, chloride, and potassium


Biochemistry_Lippinco

Biochemistry_Lippinco. These macrominerals are considered together because they play important roles in several physiologic processes. For example, they maintain water balance, osmotic equilibrium, acid–base balance (pH), and the electrical gradients across cell membranes (membrane potential) that are essential for the functioning of neurons and myocytes. [Note: These processes are discussed in Lippincott’s Illustrated Reviews: Physiology.] 1. Sodium and chloride: Na+ and Cl− are primarily extracellular electrolytes. They are readily absorbed from foods containing salt (NaCl), much of which comes from processed foods. [Note: Na+ is required for the intestinal absorption (and renal reabsorption) of glucose and galactose (see p. 87) and free amino acids (see p. 249) by Na+-linked transporters. Cl− is used to form hydrochloric acid required for digestion (see p. 248).] In the United States, the average daily consumption of NaCl is 1.5–3 times the adequate intake (AI) of 3.8 mg/day (UL = 5.8 g/day). Dietary


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Biochemistry_Lippinco. hydrochloric acid required for digestion (see p. 248).] In the United States, the average daily consumption of NaCl is 1.5–3 times the adequate intake (AI) of 3.8 mg/day (UL = 5.8 g/day). Dietary deficiency is rare.


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Biochemistry_Lippinco. a. Hypertension: Na+ intake is related to blood pressure (BP). Ingestion of Na+ stimulates thirst centers in the brain and secretion of antidiuretic hormone from the pituitary, leading to water retention. This results in an increase in plasma volume and, consequently, an increase in BP. Chronic hypertension can damage the heart, kidneys, and blood vessels. Modest reductions in Na+ intake have been shown to result in modest reductions in BP. [Note: Some populations (for example, African Americans) are “salt sensitive” and have larger responses to Na+.] b.


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Biochemistry_Lippinco. Hyper-and hyponatremia: Hypernatremia, typically caused by excess water loss, and hyponatremia, typically caused by decreased ability to excrete water, can result in severe brain damage. [Note: Chronic hyponatremia increases Ca2+ excretion and can result in osteoporosis (low bone mass).] 2. Potassium: In contrast to Na+, K+ is primarily an intracellular electrolyte. [Note: The concentration differential of Na+ and K+ across the cell membrane is maintained by the Na+/K+ ATPase (Fig. 29.4).] In contrast to Na+ and Cl−, K+ (like Mg2+) is underingested in Western diets because its primary sources, fruits and vegetables, are underingested. [Note: Increasing dietary K+ decreases BP by increasing Na+ excretion.] There is a narrow range for normal serum K+ levels, and even modest changes (up or down, resulting in hyper-or hypokalemia) can result in cardiac arrhythmias and skeletal muscle weakness. [Note: Hypokalemia can result from the inappropriate use of laxatives to lose weight.] No UL for


Biochemistry_Lippinco

Biochemistry_Lippinco. resulting in hyper-or hypokalemia) can result in cardiac arrhythmias and skeletal muscle weakness. [Note: Hypokalemia can result from the inappropriate use of laxatives to lose weight.] No UL for K+ has been established.


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Biochemistry_Lippinco. III. MICROMINERALS (TRACE MINERALS) The trace minerals include copper (Cu), iron (Fe), manganese (Mn), and zinc (Zn). They are required by adults in amounts between 1 and 100 mg/day. A. Copper Cu is a key component of several enzymes that play critical functions in the body (Fig. 29.5). These include ferroxidases such as the ceruloplasmin and hephaestin involved in the oxidation of ferrous iron (Fe2+) to the ferric form (Fe3+) that is required for its intracellular storage or transport through blood (see B.1. below). Meat, shellfish, nuts, and whole grains are good dietary sources of Cu. Dietary deficiency is uncommon. If a deficiency does develop, anemia may be seen because of the effect on Fe metabolism. Toxicity from dietary sources is rare (UL = 10 mg/day). Menkes syndrome and Wilson disease are genetic causes of Cu deficiency and Cu overload, respectively. 1.


Biochemistry_Lippinco

Biochemistry_Lippinco. 1. Menkes syndrome: In Menkes syndrome (“kinky hair” disease), a rare X-linked (1:140,000 males) disorder, efflux of dietary Cu out of intestinal enterocytes into the circulation by a Cu-transporting ATPase (ATP7A) is impaired. This results in systemic Cu deficiency. Consequently, urinary and serum free (unbound) Cu are low, as is the concentration of ceruloplasmin, which carries over 90% of the Cu in the circulation (Fig. 29.6). Progressive neurologic degeneration and connective tissue disorders are seen, as are changes to hair. Parenteral administration of Cu has been used as a treatment with varying success. [Note: The mildest form of Menkes syndrome is called occipital horn syndrome.] 2.


Biochemistry_Lippinco

Biochemistry_Lippinco. Wilson disease: In Wilson disease, an autosomal-recessive (AR) disorder affecting 1:35,000 live births, efflux of excess Cu from the liver by ATP7B is impaired. Cu accumulates in the liver; leaks into the blood; and is deposited in the brain, eyes, kidneys, and skin. In contrast to Menkes syndrome, urinary and serum free Cu are high (see Fig. 29.6). Hepatic dysfunction and neurologic and psychiatric symptoms are seen. Kayser-Fleischer rings (corneal deposits of Cu) may be present (Fig. 29.7). Life-long use of Cu-chelating agents, such as penicillamine, is the treatment. The bioavailability (percent of the amount ingested that is able to be absorbed) of a mineral can be influenced by other minerals. For example, excess Zn decreases the absorption of Cu, and Cu is needed for the absorption of Fe. B. Iron


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Biochemistry_Lippinco. B. Iron The adult body typically contains 3–4 g of Fe. It is a component of many proteins, both catalytic (for example, hydroxylases such as prolyl hydroxylase, see p. 47) and noncatalytic. Iron can be linked to sulfur (S) as seen in the Fe–S proteins of the electron transport chain (see p. 75), or it can be part of the heme prosthetic group (see p. 25) in proteins such as hemoglobin (~70% of all Fe), myoglobin, and the cytochromes. [Note: Free ionic Fe is toxic because it can cause production of the hydroxyl radical, a reactive oxygen species (ROS).] Dietary Fe is available as Fe2+ in heme (animal sources) and Fe3+ in nonheme sources (plants). Heme iron is less abundant, but it is better absorbed. Meat, poultry, some shellfish, ready-toeat cereals, lentils, and molasses are good dietary sources of Fe. About 10% of ingested Fe is absorbed. This amount, ~1−2 mg/day, is sufficient to replace Fe lost from the body primarily by the sloughing of cells. 1.


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Biochemistry_Lippinco. Absorption, storage, and transport: Intestinal uptake of heme is by a heme carrier protein (Fig. 29.8). Within the enterocytes, heme oxygenase releases Fe2+ from heme (see p. 282). Nonheme Fe is taken up via the apical membrane protein divalent metal ion transporter-1 (DMT-1). [Note: Vitamin C enhances absorption of nonheme Fe because it is the coenzyme for duodenal cytochrome b (Dcytb), a ferrireductase that reduces Fe3+ to Fe2+.] Absorbed Fe2+ from heme and nonheme sources has two possible fates: It can be 1) oxidized to Fe3+ and stored by the intracellular protein ferritin (up to 4,500 Fe3+/ferritin) or 2) transported out of the enterocyte by the basolateral membrane protein ferroportin, oxidized by the Cu-containing membrane protein hephaestin, and taken up by the plasma transport protein transferrin (2 Fe3+/transferrin), as shown in Figure 29.8. [Note: Cells other than enterocytes use the Cu-containing plasma protein ceruloplasmin in place of hephaestin.] In normal individuals,


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Biochemistry_Lippinco. transferrin (2 Fe3+/transferrin), as shown in Figure 29.8. [Note: Cells other than enterocytes use the Cu-containing plasma protein ceruloplasmin in place of hephaestin.] In normal individuals, transferrin (Tf) is about one third saturated with Fe3+ . Ferroportin, the only known exporter of Fe from cells to the blood in humans, is regulated by the hepatic peptide hepcidin that induces internalization and lysosomal degradation of ferroportin. Therefore, hepcidin is the central molecule in Fe homeostasis. [Note: Transcription of hepcidin is suppressed when Fe is deficient.] 2.


Biochemistry_Lippinco

Biochemistry_Lippinco. Recycling: Macrophages phagocytose old and/or damaged red blood cells (RBC), freeing heme Fe that is sent out of the cells via ferroportin, oxidized by ceruloplasmin, and transported by Tf as described above. This recycled Fe meets ~90% of our daily need, which is predominantly for erythropoiesis. 3. Uptake: Tf-bound Fe3+ from enterocytes and macrophages binds to receptors (TfR) on erythroblasts and other Fe-requiring cells and is taken up by receptor-mediated endocytosis. The Fe3+ is released from Tf for use (or stored on ferritin), and the TfR (and Tf) is recycled in a process similar to the receptor-mediated endocytosis seen with low-density lipoprotein particles (see p. 231). [Note: Regulation of the translation of the messenger RNA for ferritin and the TfR by iron regulatory proteins and iron-responsive elements is discussed on p. 474.] 4.


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Biochemistry_Lippinco. Deficiency: Fe deficiency can result in a microcytic, hypochromic anemia (Fig. 29.9), the most common anemia in the United States, as a result of decreased hemoglobin synthesis and, consequently, decreased RBC size. Treatment is the administration of Fe.


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Biochemistry_Lippinco. 5. Excess: Fe overload can occur with accidental ingestion. [Note: Acute Fe poisoning is the most common cause of poisoning deaths of children age <6 years (UL = 40 mg/day for children, 45 mg/day for adults).] Treatment is use of an Fe chelator. Overload can also occur with genetic defects. An example is hereditary hemochromatosis (HH), an AR disorder of Fe overload found primarily in those of Northern European ancestry. It is most commonly caused by mutations to the HFE (high iron) gene. Hyperpigmentation with hyperglycemia (“bronze diabetes”) and damage to the liver (a major storage site for Fe), pancreas, and heart may be seen. In HH, serum Fe and Tf saturation are elevated. Treatment is phlebotomy or use of Fe chelators. [Note: Fe overload is seen with mutations to proteins of Fe metabolism that result in inappropriately low levels of hepcidin. It can result in hemosiderosis (the deposition of hemosiderin, an intracellular, insoluble storage form of Fe).] C. Manganese


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Biochemistry_Lippinco. C. Manganese Mn is important for the function of several enzymes (Fig. 29.10). Whole grains, legumes (for example, beans and peas), nuts, and tea (especially green tea) are good sources of the mineral. Consequently, Mn deficiency in humans is rare. Toxicity from foods and/or supplements is also rare (UL = 11 mg/day for adults). D. Zinc


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Biochemistry_Lippinco. Zn plays important structural and catalytic functions in the body. Zinc fingers are supersecondary structures (motifs, see p. 18) in proteins (for example, transcription factors) that bind to DNA and regulate gene expression (Fig. 29.11). Hundreds of enzymes require Zn for activity. Examples include alcohol dehydrogenase, which oxidizes ethanol to acetaldehyde (see p. 317); carbonic anhydrase, which is important in the bicarbonate buffer system (see p. 30); porphobilinogen synthase of heme synthesis, which is inhibited by lead (lead replaces the zinc; see p. 279); and the nonmitochondrial isoform of superoxide dismutase (SOD), which also requires Cu (see Fig. 29.5). Dietary sources of Zn include meat, fish, eggs, and dairy products. Phytates (phosphate storage molecules in some plant products) irreversibly bind Zn in the intestine, decreasing its absorption, and can result in a deficiency. [Note: Phytates may also bind Ca2+ and nonheme Fe.] Several drugs (for example, penicillamine)


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Biochemistry_Lippinco. irreversibly bind Zn in the intestine, decreasing its absorption, and can result in a deficiency. [Note: Phytates may also bind Ca2+ and nonheme Fe.] Several drugs (for example, penicillamine) chelate metals, and their use may cause Zn deficiency. [Note: Severe deficiency is seen with a defect in the intestinal transporter for Zn that results in the malabsorption disorder acrodermatitis enteropathica. Symptoms include rashes, slowed growth and development, diarrhea, and immune deficiencies. Vision problems may also occur because Zn is needed in the metabolism of vitamin A.]


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Biochemistry_Lippinco. Eukaryotic cells infected with bacteria can restrict availability of the essential micronutrients Fe, Mn, and Zn to the pathogens. This decreases the intracellular survival of the pathogen and is known as “nutritional immunity.” E. Other microminerals Chromium (Cr) and fluorine (F) also play roles in the body. Cr potentiates the action of insulin by an unknown mechanism. It is found in fruits, vegetables, dairy products, and meat. F (as fluoride [F−]) is added to water in many parts of the world to reduce the incidence of dental caries (Fig. 29.12). F− replaces the hydroxyl group of hydroxylapatite, forming fluoroapatite that is more resistant to the enamel-dissolving acid produced by mouth bacteria. IV. ULTRATRACE MINERALS The ultratrace minerals include iodine (I), selenium (Se), and molybdenum (Mo). They are required by adults in amounts <1 mg/day. A. Iodine


Biochemistry_Lippinco

Biochemistry_Lippinco. A. Iodine I is utilized in the synthesis of the thyroid hormones triiodothyronine (T3) and thyroxine (T4) that are required for development, growth, and metabolism. Circulating iodide (I−) is taken up (“trapped”) and concentrated in the epithelial follicular cells of the thyroid gland. It then is sent into the colloid of the follicular lumen where it is oxidized to iodine (I2) by thyroperoxidase (TPO), as shown in Figure 29.13. TPO then uses I2 to iodinate selected tyrosine residues in thyroglobulin (Tg), forming monoiodinated tyrosine (MIT) and diiodinated tyrosine (DIT), as shown in Figure 29.14. [Note: Tg is synthesized and secreted into colloid by follicular cells.] The coupling of two DIT on Tg gives T4, whereas coupling one MIT and one DIT gives T3. The iodinated Tg is endocytosed and stored in follicular cells until needed, at which time it is proteolytically digested to release T3 and T4, which are secreted into the circulation (see Fig. 29.13).


Biochemistry_Lippinco

Biochemistry_Lippinco. Under normal conditions, ~90% of secreted thyroid hormone is T4 that is carried by transthyretin. In target tissues (for example, the liver and developing brain), T4 is converted to T3 (the more active form) by Se-containing deiodinases. T3 binds to a nuclear receptor that binds DNA at thyroid response elements and functions as a transcription factor. [Note: Thyroid hormone production is controlled by thyrotropin (thyroid-stimulating hormone ([TSH]) from the anterior pituitary. TSH secretion is itself controlled by thyrotropin-releasing hormone (TRH) from the hypothalamus.] 1.


Biochemistry_Lippinco

Biochemistry_Lippinco. Hypothyroidism: Underingestion of I can result in goiter (enlargement of the thyroid in response to excessive stimulation by TSH), as shown in Figure 29.15. More severe deficiency results in hypothyroidism that is characterized by fatigue, weight gain, decreased thermogenesis, and decreased metabolic rate (see p. 359). If hormone deficiency occurs during fetal and infant development (congenital hypothyroidism), irreversible intellectual disability (formerly called “cretinism”), hearing loss, spasticity, and short stature can result. In the United States, dairy products, seafood, and meat are the primary sources of I. The use of iodized salt has greatly reduced dietary I deficiency. [Note: Autoimmune destruction of TPO is a cause of Hashimoto thyroiditis (a primary hypothyroidism).] 2.


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Biochemistry_Lippinco. Hyperthyroidism: This condition is the result of overproduction of thyroid hormone. Although it can be caused by overingestion of I-containing supplements (UL = 1.1 g/day for adults), the most common cause of hyperthyroidism is Graves disease, in which an antibody that mimics the effect of TSH is produced, resulting in dysregulated production of thyroid hormone. This can cause nervousness, weight loss, increased perspiration and heart rate, protruding eyes (exophthalmos, Fig. 29.16), and goiter. B. Selenium


Biochemistry_Lippinco

Biochemistry_Lippinco. B. Selenium Se is present in ~25 human proteins (selenoproteins) as a constituent of the amino acid selenocysteine, which is derived from serine (see p. 268). Selenoproteins include glutathione peroxidase that oxidizes glutathione in the reduction of hydrogen peroxide, a ROS, to water (see p. 148); thioredoxin reductase that reduces thioredoxin, a coenzyme of ribonucleotide reductase (see p. 297); and deiodinases that remove I from thyroid hormones. Meat, dairy products, and grains are important dietary sources. Keshan disease, first identified in China, is a cardiomyopathy caused by eating foods produced from Se-deficient soil. Toxicity (selenosis) caused by overingestion of supplements causes brittle nails and hair. Cutaneous and neurologic effects may also be seen (UL = 400 µg in adults). C. Molybdenum


Biochemistry_Lippinco

Biochemistry_Lippinco. C. Molybdenum Mo functions as a cofactor for a small number of mammalian oxidases (Fig. 29.17). Legumes are important dietary sources. No dietary deficiency syndromes are known. Mo has low toxicity in humans (UL = 2 mg/day in adults). Cobalt (Co), an ultratrace mineral, is a component of vitamin B12 (cobalamin, see p. 379), which is required as methylcobalamin in the remethylation of homocysteine to methionine (see p. 264) or adenosylcobalamin in the isomerization of methylmalonyl coenzyme A (CoA) to succinyl CoA (see p. 194). No Recommended Dietary Allowance or Daily Reference Intake (see p. 358) has been established for Co. V. CHAPTER SUMMARY The minerals are summarized in Figure 29.18 on p. 408. For Questions 29.1–29.7, match the mineral to the most appropriate description. A. Calcium B. Chloride C. Copper D. Iodine E. Iron F. Magnesium G. Manganese H. Molybdenum I. Phosphorus J. Potassium K. Selenium L. Sodium


Biochemistry_Lippinco

Biochemistry_Lippinco. A. Calcium B. Chloride C. Copper D. Iodine E. Iron F. Magnesium G. Manganese H. Molybdenum I. Phosphorus J. Potassium K. Selenium L. Sodium M. Zinc 9.1. Elevated levels of which mineral may result in hypertension in certain populations? 9.2. Which mineral is the major extracellular anion? 9.3. A decrease of which mineral is seen in refeeding syndrome and with overuse of aluminum-containing antacids? 9.4. Which mineral is a constituent of some amino acids found in proteins involved in antioxidant defense, thyroid hormone metabolism, and redox reactions? 9.5. Which mineral is required for the formation of a supersecondary protein structure that allows binding to DNA? (Its deficiency can result in a dermatitis.) 9.6. Deficiency of which mineral can cause bone pain, tetany (intermittent muscle spasms), paresthesia (a “pins and needles” sensation), and an increased tendency to bleed? 9.7. Deficiency of which mineral can result in goiter and a decreased metabolic rate?


Biochemistry_Lippinco

Biochemistry_Lippinco. Correct answers = L, B, I, K, M, A, D. Hypernatremia (elevation of serum sodium) can lead to water retention that can cause hypertension in salt-sensitive populations (for example, African Americans). Chloride is the major extracellular anion. [Note: Sodium is the major extracellular cation, potassium is the major intracellular cation, and phosphate is the major intracellular anion. The concentration differential across the membrane is maintained by active transport.] Carbohydrate metabolism involves the generation of phosphorylated intermediates. Refeeding severely malnourished individuals traps phosphate and results in hypophosphatemia. Muscle weakness is a common symptom. Selenocysteine, an amino acid formed from serine and selenium, is found in proteins (selenoproteins) such as glutathione peroxidase, deiodinases, and thioredoxin reductase. Zinc fingers are a type of structural motif found in proteins (for example, transcription factors) that bind to DNA. Severe deficiency of zinc


Biochemistry_Lippinco

Biochemistry_Lippinco. peroxidase, deiodinases, and thioredoxin reductase. Zinc fingers are a type of structural motif found in proteins (for example, transcription factors) that bind to DNA. Severe deficiency of zinc as a result of mutations to its intestinal transporter can result in acrodermatitis enteropathica, which is characterized by dermatitis, diarrhea, and alopecia. Calcium is required for bone mineralization, muscle contraction, nerve conduction, and blood clotting. Its deficiency will affect all of these processes. Thyroid hormones are iodinated tyrosines released by proteolytic digestion of thyroglobulin. Underingestion of iodine causes enlargement of the thyroid in an attempt to increase hormone synthesis. [Note: Goiter can also result if too much hormone is made, as in Graves disease, or if too little is made, as in Hashimoto disease. Both are autoimmune diseases.] Thyroid hormone increases the resting metabolic rate.


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Biochemistry_Lippinco. 9.8. DiGeorge syndrome is a congenital condition that results in structural anomalies and failure of the thymus and parathyroid glands to develop. Clinical manifestations include recurrent infections as a consequence of a deficiency in T cells. Which one of the following is an expected clinical consequence of the deficiency in parathyroid hormone? A. Increased bone resorption B. Increased calcium reabsorption in the kidney C. Increased serum calcitriol D. Increased serum phosphate Correct answer = D. Parathyroid hormone (PTH) increases bone resorption (demineralization) resulting in the release of calcium and phosphate. It also increases the renal reabsorption of calcium, because PTH activates the renal hydroxylase that converts calcidiol to calcitriol. PTH also increases the renal excretion of phosphate. With the hypoparathyroidism of DiGeorge syndrome, all of these activities of PTH are impaired. Consequently, hypocalcemia and hyperphosphatemia are seen.


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Biochemistry_Lippinco. For questions 29.9 and 29.10, match the signs and symptoms to the pathology. A. Graves disease B. Hereditary hemochromatosis C. Hypercalcemia D. Hyperphosphatemia E. Keshan disease F. Menkes syndrome G. Selenosis H. Wilson disease 9.9. A 28-year-old male is seen for complaints of recent, severe, upper-rightquadrant pain. He also reports some difficulty with fine motor tasks. No jaundice is observed on physical examination. Laboratory tests were remarkable for elevated liver function tests (serum aspartate and alanine aminotransferases) and elevated urinary calcium and phosphate. Ophthalmology consult revealed Kayser-Fleischer rings in the cornea. The patient was started on penicillamine and zinc.


Biochemistry_Lippinco

Biochemistry_Lippinco. Correct answer = H. The patient has Wilson disease, an autosomal-recessive disorder that decreases copper efflux from the liver because of mutations to the hepatic copper transport protein ATP7B. Some copper leaks into the blood and is deposited in the brain, eyes, kidney, and skin. This results in liver and kidney damage, neurologic effects, and corneal changes caused by the excess copper. Administration of the metal chelator penicillamine is the treatment. [Note: Because zinc is also chelated, supplementation with zinc is common.] Graves disease results in hyperthyroidism. Hereditary hemochromatosis is a disorder of iron overload. Keshan disease is the result of selenium deficiency, whereas selenosis is caused by selenium excess. Menkes syndrome is the result of a systemic deficiency in copper as a result of mutations to ATP7A, an intestinal copper transport protein.


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Biochemistry_Lippinco. 9.10. A 52-year-old female is seen because of unplanned changes in the pigmentation of her skin that give her a tanned appearance. Physical examination shows hyperpigmentation, hepatomegaly, and mild scleral icterus. Laboratory tests are remarkable for elevated serum transaminases (liver function tests) and fasting blood glucose. Results of other tests are pending. Correct answer = B. The patient has hereditary hemochromatosis, a disease of iron overload that results from inappropriately low levels of hepcidin caused primarily by mutations to the HFE (high iron) gene. Hepcidin regulates ferroportin, the only known iron export protein in humans, by increasing its degradation. The increase in iron with hepcidin deficiency causes hyperpigmentation and hyperglycemia (“bronze diabetes”). Phlebotomy or use of iron chelators is the treatment. [Note: Pending lab tests would show an increase in serum iron and transferrin saturation.] UNIT VII Storage and Expression of Genetic Information


Biochemistry_Lippinco

Biochemistry_Lippinco. UNIT VII Storage and Expression of Genetic Information DNA Structure, Replication, and Repair For additional ancillary materials related to this chapter, please visit thePoint. I. OVERVIEW


Biochemistry_Lippinco

Biochemistry_Lippinco. Nucleic acids are required for the storage and expression of genetic information. There are two chemically distinct types of nucleic acids: deoxyribonucleic acid (DNA) and ribonucleic acid ([RNA] see Chapter 31). DNA, the repository of genetic information (or, genome), is present not only in chromosomes in the nucleus of eukaryotic organisms, but also in mitochondria and the chloroplasts of plants. Prokaryotic cells, which lack nuclei, have a single chromosome but may also contain nonchromosomal DNA in the form of plasmids. The genetic information found in DNA is copied and transmitted to daughter cells through DNA replication. The DNA contained in a fertilized egg encodes the information that directs the development of an organism. This development may involve the production of billions of cells. Each cell is specialized, expressing only those functions that are required for it to perform its role in maintaining the organism. Therefore, DNA must be able not only to replicate


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Biochemistry_Lippinco. of cells. Each cell is specialized, expressing only those functions that are required for it to perform its role in maintaining the organism. Therefore, DNA must be able not only to replicate precisely each time a cell divides, but also to have the information that it contains be selectively expressed. Transcription (RNA synthesis) is the first stage in the expression of genetic information (see Chapter 31). Next, the code contained in the nucleotide sequence of messenger RNA molecules is translated (protein synthesis; see Chapter 32), thus completing gene expression. The regulation of gene expression is discussed in Chapter 33.


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Biochemistry_Lippinco. The flow of information from DNA to RNA to protein is termed the “central dogma” of molecular biology (Fig. 30.1) and is descriptive of all organisms, with the exception of some viruses that have RNA as the repository of their genetic information. II. DNA STRUCTURE DNA is a polymer of deoxyribonucleoside monophosphates (dNMP) covalently linked by 3′→5′-phosphodiester bonds. With the exception of a few viruses that contain single-stranded DNA (ssDNA), DNA exists as a double-stranded molecule (dsDNA), in which the two strands wind around each other, forming a double helix. [Note: The sequence of the linked dNMP is primary structure, whereas the double helix is secondary structure.] In eukaryotic cells, DNA is found associated with various types of proteins (known collectively as nucleoprotein) present in the nucleus, whereas the protein–DNA complex is present in a non–membrane-bound region known as the nucleoid in prokaryotes. A. 3′→5′-Phosphodiester bonds


Biochemistry_Lippinco

Biochemistry_Lippinco. A. 3′→5′-Phosphodiester bonds Phosphodiester bonds join the 3′-hydroxyl group of the deoxypentose of one nucleotide to the 5′-hydroxyl group of the deoxypentose of an adjacent nucleotide through a phosphoryl group (Fig. 30.2). The resulting long, unbranched chain has polarity, with both a 5′-end (the end with the free phosphate) and a 3′-end (the end with the free hydroxyl) that are not attached to other nucleotides. By convention, the bases located along the resulting deoxyribose-phosphate backbone are always written in sequence from the 5′-end of the chain to the 3′-end. For example, the sequence of bases in the DNA shown in Figure 30.2D (5′-TACG-3′) is read “thymine, adenine, cytosine, guanine.” Phosphodiester linkages between nucleotides can be hydrolyzed enzymatically by a family of nucleases, deoxyribonucleases for DNA and ribonucleases for RNA, or cleaved hydrolytically by chemicals. [Note: Only RNA is cleaved by alkali.] B. Double helix


Biochemistry_Lippinco

Biochemistry_Lippinco. In the double helix, the two chains are coiled around a common axis called the helical axis. The chains are paired in an antiparallel manner (that is, the 5′-end of one strand is paired with the 3′-end of the other strand), as shown in Figure 30.3. In the DNA helix, the hydrophilic deoxyribose-phosphate backbone of each chain is on the outside of the molecule, whereas the hydrophobic bases are stacked inside. The overall structure resembles a twisted ladder. The spatial relationship between the two strands in the helix creates a major (wide) groove and a minor (narrow) groove. These grooves provide access for the binding of regulatory proteins to their specific recognition sequences along the DNA chain. [Note: Certain anticancer drugs, such as dactinomycin (actinomycin D), exert their cytotoxic effect by intercalating into the narrow groove of the DNA double helix, thereby interfering with DNA (and RNA) synthesis.] 1. Base-pairing: The bases of one strand of DNA are paired with the


Biochemistry_Lippinco

Biochemistry_Lippinco. effect by intercalating into the narrow groove of the DNA double helix, thereby interfering with DNA (and RNA) synthesis.] 1. Base-pairing: The bases of one strand of DNA are paired with the bases of the second strand, so that an adenine (A) is always paired with a thymine (T), and a cytosine (C) is always paired with a guanine (G). [Note: The base pairs are perpendicular to the helical axis (see Fig. 30.3).] Therefore, one polynucleotide chain of the DNA double helix is always the complement of the other. Given the sequence of bases on one chain, the sequence of bases on the complementary chain can be determined (Fig. 30.4). [Note: The specific base-pairing in DNA leads to the Chargaff rule, which states that in any sample of dsDNA, the amount of A equals the amount of T, the amount of G equals the amount of C, and the total amount of purines (A + G) equals the total amount of pyrimidines (T + C).] The base pairs are held together by hydrogen bonds: two between A and T and three


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Biochemistry_Lippinco. of G equals the amount of C, and the total amount of purines (A + G) equals the total amount of pyrimidines (T + C).] The base pairs are held together by hydrogen bonds: two between A and T and three between G and C (Fig. 30.5). These hydrogen bonds, plus the hydrophobic interactions between the stacked bases, stabilize the structure of the double helix.


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Biochemistry_Lippinco. 2. DNA strand separation: The two strands of the double helix separate when hydrogen bonds between the paired bases are disrupted. Disruption can occur in the laboratory if the pH of the DNA solution is altered so that the nucleotide bases ionize, or if the solution is heated. [Note: Covalent phosphodiester bonds are not broken by such treatment.] When DNA is heated, the temperature at which one half of the helical structure is lost is defined as the melting temperature (Tm). The loss of helical structure in DNA, called denaturation, can be monitored by measuring its absorbance at 260 nm. [Note: ssDNA has a higher relative absorbance at this wavelength than does dsDNA.] Because there are three hydrogen bonds between G and C but only two between A and T, DNA that contains high concentrations of A and T denatures at a lower temperature than does G-and C-rich DNA (Fig. 30.6). Under appropriate conditions, complementary DNA strands can reform the double helix by the process called


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Biochemistry_Lippinco. of A and T denatures at a lower temperature than does G-and C-rich DNA (Fig. 30.6). Under appropriate conditions, complementary DNA strands can reform the double helix by the process called renaturation (or, reannealing). [Note: Separation of the two strands over short regions occurs during both DNA and RNA synthesis.] 3. Structural forms: There are three major structural forms of DNA: the B form (described by Watson and Crick in 1953), the A form, and the Z form. The B form is a right-handed helix with 10 base pairs (bp) per 360° turn (or twist) of the helix, and with the planes of the bases perpendicular to the helical axis. Chromosomal DNA is thought to consist primarily of B-DNA (Fig. 30.7 shows a space-filling model of B-DNA). The A form is produced by moderately dehydrating the B form. It is also a right-handed helix, but there are 11 bp per turn, and the planes of the base pairs are tilted 20° away from the perpendicular to the helical axis. The conformation found in DNA–RNA


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Biochemistry_Lippinco. form. It is also a right-handed helix, but there are 11 bp per turn, and the planes of the base pairs are tilted 20° away from the perpendicular to the helical axis. The conformation found in DNA–RNA hybrids (see p. 418) or RNA–RNA double-stranded regions is probably very close to the A form. Z-DNA is a left-handed helix that contains 12 bp per turn (see Fig. 30.7). [Note: The deoxyribose-phosphate backbone zigzags, hence, the name Z-DNA.] Stretches of Z-DNA can occur naturally in regions of DNA that have a sequence of alternating purines and pyrimidines (for example, poly GC). Transitions between the B and Z helical forms of DNA may play a role in regulating gene expression.


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C. Linear and circular DNA molecules

Biochemistry_Lippinco. C. Linear and circular DNA molecules


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Biochemistry_Lippinco. Each chromosome in the nucleus of a eukaryote consists of one long, linear molecule of dsDNA, which is bound by a complex mixture of proteins (histone and nonhistone, see p. 425) to form chromatin. Eukaryotes have closed, circular, dsDNA molecules in their mitochondria, as do plant chloroplasts. A prokaryotic organism typically contains a single, circular, dsDNA molecule. [Note: Circular DNA is “supercoiled,” that is, the double helix crosses over on itself one or more times. Supercoiling can result in overwinding (positive supercoiling) or underwinding (negative supercoiling) of DNA. Supercoiling, a type of tertiary structure, compacts DNA.] Each prokaryotic chromosome is associated with nonhistone proteins that help compact the DNA to form a nucleoid. In addition, most species of bacteria also contain small, circular, extrachromosomal DNA molecules called plasmids. Plasmid DNA carries genetic information and undergoes replication that may or may not be synchronized to chromosomal


Biochemistry_Lippinco

Biochemistry_Lippinco. also contain small, circular, extrachromosomal DNA molecules called plasmids. Plasmid DNA carries genetic information and undergoes replication that may or may not be synchronized to chromosomal division. [Note: The use of plasmids as vectors in recombinant DNA technology is described in Chapter 34.]


Biochemistry_Lippinco

Biochemistry_Lippinco. Plasmids may carry genes that convey antibiotic resistance to the host bacterium and may facilitate the transfer of genetic information from one bacterium to another. III. STEPS IN PROKARYOTIC DNA REPLICATION


Biochemistry_Lippinco

Biochemistry_Lippinco. When the two strands of dsDNA are separated, each can serve as a template for the replication (synthesis) of a new complementary strand. This produces two daughter molecules, each of which contains two DNA strands (one old, one new) in an antiparallel orientation (see Fig. 30.3). This process is called semiconservative replication because, although the parental duplex is separated into two halves (and, therefore, is not conserved as an entity), each of the parental strands remains intact in one of the two new duplexes (Fig. 30.8). The enzymes involved in DNA replication are template-directed, magnesium (Mg2+)requiring polymerases that can synthesize the complementary sequence of each strand with extraordinary fidelity. The reactions described in this section were first known from studies of the bacterium Escherichia coli (E. coli), and the description given below refers to the process in prokaryotes. DNA synthesis in higher organisms is more complex but involves the same types of


Biochemistry_Lippinco

Biochemistry_Lippinco. of the bacterium Escherichia coli (E. coli), and the description given below refers to the process in prokaryotes. DNA synthesis in higher organisms is more complex but involves the same types of mechanisms. In either case, initiation of DNA replication commits the cell to continue the process until the entire genome has been replicated.


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Biochemistry_Lippinco. A. Complementary strand separation In order for the two complementary strands of the parental dsDNA to be replicated, they must first separate (or “melt”) over a small region, because the polymerases use only ssDNA as a template. In prokaryotic organisms, DNA replication begins at a single, unique nucleotide sequence, a site called the origin of replication, or ori (oriC in E. coli), as shown in Figure 30.9A. [Note: This sequence is referred to as a consensus sequence, because the order of nucleotides is essentially the same at each site.] The ori includes short, AT-rich segments that facilitate melting. In eukaryotes, replication begins at multiple sites along the DNA helix (Fig. 30.9B). Having multiple origins of replication provides a mechanism for rapidly replicating the great length of eukaryotic DNA molecules. B. Replication fork formation


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Biochemistry_Lippinco. B. Replication fork formation As the two strands unwind and separate, synthesis occurs at two replication forks that move away from the origin in opposite directions (bidirectionally), generating a replication bubble (see Fig. 30.9). [Note: The term “replication fork” derives from the Y-shaped structure in which the tines of the fork represent the separated strands (Fig. 30.10).] 1. Required proteins: Initiation of DNA replication requires the recognition of the origin (start site) by a group of proteins that form the prepriming complex. These proteins are responsible for melting at the ori, maintaining the separation of the parental strands, and unwinding the double helix ahead of the advancing replication fork. In E. coli, these proteins include the following. a.


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Biochemistry_Lippinco. a. DnaA protein: DnaA protein initiates replication by binding to specific nucleotide sequences (DnaA boxes) within oriC. Binding causes an AT-rich region (the DNA unwinding element) in the origin to melt. Melting (strand separation) results in a short, localized region of ssDNA. b. DNA helicases: These enzymes bind to ssDNA near the replication fork and then move into the neighboring double-stranded region, forcing the strands apart (in effect, unwinding the double helix). Helicases require energy provided by ATP hydrolysis (see Fig. 30.10). Unwinding at the replication fork causes supercoiling in other regions of the DNA molecule. [Note: DnaB is the principal helicase of replication in E. coli. Binding of this hexameric protein to DNA requires DnaC.] c.


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Biochemistry_Lippinco. Single-stranded DNA–binding protein: This protein binds to the ssDNA generated by helicases (see Fig. 30.10). Binding is cooperative (that is, the binding of one molecule of single-stranded binding [SSB] protein makes it easier for additional molecules of SSB protein to bind tightly to the DNA strand). The SSB proteins are not enzymes, but rather serve to shift the equilibrium between dsDNA and ssDNA in the direction of the single-stranded forms. These proteins not only keep the two strands of DNA separated in the area of the replication origin, thus providing the single-stranded template required by polymerases, but also protect the DNA from nucleases that degrade ssDNA.


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Biochemistry_Lippinco. 2. Solving the problem of supercoils: As the two strands of the double helix are separated, a problem is encountered, namely, the appearance of positive supercoils in the region of DNA ahead of the replication fork as a result of overwinding (Fig. 30.11) and negative supercoils in the region behind the fork. The accumulating positive supercoils interfere with further unwinding of the double helix. [Note: Supercoiling can be demonstrated by tightly grasping one end of a helical telephone cord while twisting the other end. If the cord is twisted in the direction of tightening the coils, the cord will wrap around itself in space to form positive supercoils. If the cord is twisted in the direction of loosening the coils, the cord will wrap around itself in the opposite direction to form negative supercoils.] To solve this problem, there is a group of enzymes called DNA topoisomerases, which are responsible for removing supercoils in the helix by transiently cleaving one or both of the DNA


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Biochemistry_Lippinco. supercoils.] To solve this problem, there is a group of enzymes called DNA topoisomerases, which are responsible for removing supercoils in the helix by transiently cleaving one or both of the DNA strands.


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Biochemistry_Lippinco. a. Type I DNA topoisomerases: These enzymes reversibly cleave one strand of the double helix. They have both strand-cutting and strand-resealing activities. They do not require ATP, but rather appear to store the energy from the phosphodiester bond they cleave, reusing the energy to reseal the strand (Fig. 30.12). Each time a transient nick is created in one DNA strand, the intact DNA strand is passed through the break before it is resealed, thus relieving (relaxing) accumulated supercoils. Type I topoisomerases relax negative supercoils (that is, those that contain fewer turns of the helix than does relaxed DNA) in E. coli and both negative and positive supercoils (that is, those that contain fewer or more turns of the helix than does relaxed DNA) in many prokaryotic cells (but not E. coli) and in eukaryotic cells. b.


Biochemistry_Lippinco

Biochemistry_Lippinco. b. Type II DNA topoisomerases: These enzymes bind tightly to the DNA double helix and make transient breaks in both strands. The enzyme then causes a second stretch of the DNA double helix to pass through the break and, finally, reseals the break (Fig. 30.13). As a result, both negative and positive supercoils can be relieved by this ATP-requiring process. DNA gyrase, a type II topoisomerase found in bacteria and plants, has the unusual property of being able to introduce negative supercoils into circular DNA using energy from the hydrolysis of ATP. This facilitates the replication of DNA because the negative supercoils neutralize the positive supercoils introduced during opening of the double helix. It also aids in the transient strand separation required during transcription (see p. 436).


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Biochemistry_Lippinco. Anticancer agents, such as the camptothecins, target human type I topoisomerases, whereas etoposide targets human type II topoisomerases. Bacterial DNA gyrase is a unique target of a group of antimicrobial agents called fluoroquinolones (for example, ciprofloxacin). C. Direction of DNA replication The DNA polymerases (DNA pols) responsible for copying the DNA templates are only able to read the parental nucleotide sequences in the 3′→5′ direction, and they synthesize the new DNA strands only in the 5′→3′ (antiparallel) direction. Therefore, beginning with one parental double helix, the two newly synthesized stretches of nucleotide chains must grow in opposite directions, one in the 5′→3′ direction toward the replication fork and one in the 5′→3′ direction away from the replication fork (Fig. 30.14). This feat is accomplished by a slightly different mechanism on each strand. 1.


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Biochemistry_Lippinco. 1. Leading strand: The strand that is being copied in the direction of the advancing replication fork is synthesized continuously and is called the leading strand. 2. Lagging strand: The strand that is being copied in the direction away from the replication fork is synthesized discontinuously, with small fragments of DNA being copied near the replication fork. These short stretches of discontinuous DNA, termed Okazaki fragments, are eventually joined (ligated) by ligase to become a single, continuous strand. The new strand of DNA produced by this mechanism is termed the lagging strand. D. RNA primer


Biochemistry_Lippinco

Biochemistry_Lippinco. D. RNA primer DNA pols cannot initiate synthesis of a complementary strand of DNA on a totally single-stranded template. Rather, they require an RNA primer, which is a short piece of RNA base-paired to the DNA template, thereby forming a double-stranded DNA–RNA hybrid. The free hydroxyl group on the 3′end of the RNA primer serves as the first acceptor of a deoxynucleotide by action of a DNA pol (Fig. 30.15). [Note: Recall that glycogen synthase also requires a primer (see p. 126).] 1.


Biochemistry_Lippinco

Biochemistry_Lippinco. Primase: A specific RNA polymerase, called primase (DnaG), synthesizes the short stretches of RNA (~10 nucleotides long) that are complementary and antiparallel to the DNA template. In the resulting hybrid duplex, the U (uracil) in RNA pairs with A in DNA. As shown in Figure 30.16, these short RNA sequences are constantly being synthesized at the replication fork on the lagging strand, but only one RNA sequence at the origin of replication is required on the leading strand. The substrates for this process are 5′-ribonucleoside triphosphates, and pyrophosphate is released as each ribonucleoside monophosphate is added through formation of a 3′→5′-phosphodiester bond. [Note: The RNA primer is later removed, as described in F. below.] 2.


Biochemistry_Lippinco

Biochemistry_Lippinco. Primosome: The addition of primase converts the prepriming complex of proteins required for DNA strand separation (see p. 415) to a primosome. The primosome makes the RNA primer required for leading-strand synthesis and initiates Okazaki fragment formation in discontinuous lagging-strand synthesis. As with DNA synthesis, the direction of synthesis of the primer is 5′→3′. E. Chain elongation Prokaryotic (and eukaryotic) DNA pols elongate a new DNA strand by adding deoxyribonucleotides, one at a time, to the 3′-end of the growing chain (see Fig. 30.16). The sequence of nucleotides that are added is dictated by the base sequence of the template strand with which the incoming nucleotides are paired.


Biochemistry_Lippinco

Biochemistry_Lippinco. 1. DNA polymerase III: DNA chain elongation is catalyzed by the multisubunit enzyme, DNA pol III. Using the 3′-hydroxyl group of the RNA primer as the acceptor of the first deoxyribonucleotide, DNA pol III begins to add nucleotides along the single-stranded template that specifies the sequence of bases in the newly synthesized chain. DNA pol III is a highly processive enzyme (that is, it remains bound to the template strand as it moves along and does not diffuse away and then rebind before adding each new nucleotide). The processivity of DNA pol III is the result of the β subunits of the holoenzyme forming a ring that encircles and moves along the template strand of the DNA, thus serving as a sliding DNA clamp. [Note: Clamp formation is facilitated by a protein complex, the clamp loader, and ATP hydrolysis.] The new (daughter) strand grows in the 5′→3′ direction, antiparallel to the parental strand (see Fig. 30.16). The nucleotide substrates are 5′deoxyribonucleoside triphosphates.


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Biochemistry_Lippinco. and ATP hydrolysis.] The new (daughter) strand grows in the 5′→3′ direction, antiparallel to the parental strand (see Fig. 30.16). The nucleotide substrates are 5′deoxyribonucleoside triphosphates. Pyrophosphate (PPi) is released when each new deoxynucleoside monophosphate is added to the free 3′hydroxyl group of the growing chain through a 3′→5′-phosphodiester bond (see Fig. 30.15). Hydrolysis of PPi to 2 Pi by pyrophosphatase means that a total of two high-energy bonds are used to drive the addition of each deoxynucleotide.


Biochemistry_Lippinco

Biochemistry_Lippinco. The production of PPi with subsequent hydrolysis to 2 Pi is a common theme in biochemistry. Removal of the PPi product drives a reaction in the forward direction, making it essentially irreversible. All four substrates (deoxyadenosine triphosphate [dATP], deoxythymidine triphosphate [dTTP], deoxycytidine triphosphate [dCTP], and deoxyguanosine triphosphate [dGTP]) must be present for DNA elongation to occur. If one of the four is in short supply, DNA synthesis stops when that nucleotide is depleted.


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Biochemistry_Lippinco. 2. Proofreading newly synthesized DNA: It is highly important for the survival of an organism that the nucleotide sequence of DNA be replicated with as few errors as possible. Misreading of the template sequence could result in deleterious, perhaps lethal, mutations. To insure replication fidelity, DNA pol III has a proofreading activity (3′→5′ exonuclease, Fig. 30.17) in addition to its 5′→3′ polymerase activity. As each nucleotide is added to the chain, DNA pol III checks to make certain the base of the newly added nucleotide is, in fact, the complement of the base on the template strand. If it is not, the 3′→5′ exonuclease activity removes the error in the direction opposite to polymerization. [Note: Because the enzyme requires an improperly base-paired 3′hydroxy terminus, it does not degrade correctly paired nucleotide sequences.] For example, if the template base is C and the enzyme inserts an A instead of a G into the new chain, the 3′→5′ exonuclease activity hydrolytically


Biochemistry_Lippinco

Biochemistry_Lippinco. not degrade correctly paired nucleotide sequences.] For example, if the template base is C and the enzyme inserts an A instead of a G into the new chain, the 3′→5′ exonuclease activity hydrolytically removes the misplaced nucleotide. The 5′→3′ polymerase activity then replaces it with the correct nucleotide containing G (see Fig. 30.17). [Note: The 5′→3′ polymerase and 3′→5′ exonuclease domains are located on different subunits of DNA pol III.]


Biochemistry_Lippinco

Biochemistry_Lippinco. F. RNA primer excision and replacement by DNA DNA pol III continues to synthesize DNA on the lagging strand until it is blocked by proximity to an RNA primer. When this occurs, the RNA is excised and the gap filled by DNA pol I. 1.


Biochemistry_Lippinco

Biochemistry_Lippinco. 5′→3′ Exonuclease activity: In addition to having the 5′→3′ polymerase activity that synthesizes DNA and the 3′→5′ exonuclease activity that proofreads the newly synthesized DNA like DNA pol III, monomeric DNA pol I also has a 5′→3′ exonuclease activity that is able to hydrolytically remove the RNA primer. [Note: Exonucleases remove nucleotides from the end of the DNA chain, rather than cleaving the chain internally as do endonucleases (Fig. 30.18).] First, DNA pol I locates the space (nick) between the 3′-end of the DNA newly synthesized by DNA pol III and the 5′-end of the adjacent RNA primer. Next, DNA pol I hydrolytically removes the RNA nucleotides ahead of itself, moving in the 5′→3′ direction (5′→3′ exonuclease activity). As it removes ribonucleotides, DNA pol I replaces them with deoxyribonucleotides, synthesizing DNA in the 5′→3′ direction (5′→3′ polymerase activity). As it synthesizes the DNA, it also proofreads using its 3′→5′ exonuclease activity to remove errors. This


Biochemistry_Lippinco

Biochemistry_Lippinco. deoxyribonucleotides, synthesizing DNA in the 5′→3′ direction (5′→3′ polymerase activity). As it synthesizes the DNA, it also proofreads using its 3′→5′ exonuclease activity to remove errors. This removal/synthesis/proofreading continues until the RNA primer is totally degraded, and the gap is filled with DNA (Fig. 30.19). [Note: DNA pol I uses its 5′→3′ polymerase activity to fill in gaps generated during most types of DNA repair (see p. 428).] 2.


Biochemistry_Lippinco

Biochemistry_Lippinco. Comparison of 5′→3′ and 3′→5′ exonuclease activities: The 5′→3′ exonuclease activity of DNA pol I allows the polymerase, moving 5′→3′, to hydrolytically remove one or more nucleotides at a time from the 5′-end of the ~10 nucleotide–long RNA primer. In contrast, the 3′→5′ exonuclease activity of DNA pol I and pol III allows these polymerases, moving 3′→5′, to hydrolytically remove one misplaced nucleotide at a time from the 3′-end of a growing DNA strand, increasing the fidelity of replication such that newly replicated DNA has one error per 107 nucleotides. G. DNA ligase The final phosphodiester linkage between the 5′-phosphate group on the DNA synthesized by DNA pol III and the 3′-hydroxyl group on the DNA made by DNA pol I is catalyzed by DNA ligase (Fig. 30.20). The joining of these two stretches of DNA requires energy, which in most organisms is provided by the cleavage of ATP to adenosine monophosphate + PPi. H. Termination


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Biochemistry_Lippinco. H. Termination Replication termination in E. coli is mediated by sequence-specific binding of the protein Tus (terminus utilization substance) to replication termination (ter) sites on the DNA, stopping the movement of the replication fork. IV. EUKARYOTIC DNA REPLICATION The process of eukaryotic DNA replication closely follows that of prokaryotic DNA synthesis. Some differences, such as the multiple origins of replication in eukaryotic cells versus single origins of replication in prokaryotes, have already been noted. Eukaryotic origin recognition proteins, ssDNA-binding proteins, and ATP-dependent DNA helicases have been identified, and their functions are analogous to those of the prokaryotic proteins previously discussed. In contrast, RNA primers are removed by RNase H and flap endonuclease 1 (FEN1) rather than by a DNA pol (Fig. 30.21). A. Eukaryotic cell cycle


Biochemistry_Lippinco

Biochemistry_Lippinco. A. Eukaryotic cell cycle The events surrounding eukaryotic DNA replication and cell division (mitosis) are coordinated to produce the cell cycle (Fig. 30.22). The period preceding replication is called the G1 phase (Gap 1). DNA replication occurs during the S (synthesis) phase. Following DNA synthesis, there is another phase (G2, or Gap 2) before mitosis (M). Cells that have stopped dividing, such as mature T lymphocytes, are said to have gone out of the cell cycle into the G0 phase. Such quiescent cells can be stimulated to reenter the G1 phase to resume division. [Note: The cell cycle is controlled at a series of checkpoints that prevent entry into the next phase of the cycle until the preceding phase has been completed. Two key classes of proteins that control the progress of a cell through the cell cycle are the cyclins and cyclin-dependent kinases (Cdk).] B. Eukaryotic DNA polymerases


Biochemistry_Lippinco

Biochemistry_Lippinco. B. Eukaryotic DNA polymerases At least five high-fidelity eukaryotic DNA pols have been identified and categorized on the basis of molecular weight, cellular location, sensitivity to inhibitors, and the templates or substrates on which they act. They are designated by Greek letters rather than by Roman numerals (Fig. 30.23). 1. Pol α: Pol α is a multisubunit enzyme. One subunit has primase activity, which initiates strand synthesis on the leading strand and at the beginning of each Okazaki fragment on the lagging strand. The primase subunit synthesizes a short RNA primer that is extended by the 5′→3′ polymerase activity of pol α, generating a short piece of DNA. [Note: Pol α is also referred to as pol α/primase.] 2.


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Biochemistry_Lippinco. Pol ε and pol δ: Pol ε is recruited to complete DNA synthesis on the leading strand, whereas pol δ elongates the Okazaki fragments of the lagging strand, each using 3′→5′ exonuclease activity to proofread the newly synthesized DNA. [Note: DNA pol ε associates with proliferating cell nuclear antigen (PCNA), a protein that serves as a sliding DNA clamp in much the same way the β subunits of DNA pol III do in E. coli, thus insuring high processivity.] 3. Pol β and pol γ: Pol β is involved in gap filling in DNA repair. Pol γ replicates mitochondrial DNA. C. Telomeres


Biochemistry_Lippinco

Biochemistry_Lippinco. Pol β and pol γ: Pol β is involved in gap filling in DNA repair. Pol γ replicates mitochondrial DNA. C. Telomeres Telomeres are complexes of DNA plus proteins (collectively known as shelterin) located at the ends of linear chromosomes. They maintain the structural integrity of the chromosome, preventing attack by nucleases, and allow repair systems to distinguish a true end from a break in dsDNA. In humans, telomeric DNA consists of several thousand tandem repeats of a noncoding hexameric sequence, AGGGTT, base-paired to a complementary region containing C and A. The G-rich strand is longer than its C-rich complement, leaving ssDNA a few hundred nucleotides in length at the 3′end. The single-stranded region is thought to fold back on itself, forming a loop structure that is stabilized by protein. 1.


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Biochemistry_Lippinco. 1. Telomere shortening: Eukaryotic cells face a special problem in replicating the ends of their linear DNA molecules. Following removal of the RNA primer from the extreme 5′-end of the lagging strand, there is no way to fill in the remaining gap with DNA. Consequently, in most normal human somatic cells, telomeres shorten with each successive cell division. Once telomeres are shortened beyond some critical length, the cell is no longer able to divide and is said to be senescent. In germ cells and stem cells, as well as in cancer cells, telomeres do not shorten and the cells do not senesce. This is a result of the ribonucleoprotein telomerase, which maintains telomeric length in these cells. 2.


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Biochemistry_Lippinco. 2. Telomerase: This complex contains a protein (Tert) that acts as a reverse transcriptase and a short piece of RNA (Terc) that acts as a template. The C-rich RNA template base-pairs with the G-rich, single-stranded 3′end of telomeric DNA (Fig. 30.24). The reverse transcriptase uses the RNA template to synthesize DNA in the usual 5′→3′ direction, extending the already longer 3′-end. Telomerase then translocates to the newly synthesized end, and the process is repeated. Once the G-rich strand has been lengthened, primase activity of DNA pol α can use it as a template to synthesize an RNA primer. The primer is extended by DNA pol α and then removed by nucleases. Telomeres may be viewed as mitotic clocks in that their length in most cells is inversely related to the number of times the cells have divided. The study of telomeres provides insight into the biology of normal aging, diseases of premature aging (the progerias), and cancer. D. Reverse transcriptases


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Biochemistry_Lippinco. D. Reverse transcriptases As seen with telomerase, reverse transcriptases are RNA-directed DNA pols. A reverse transcriptase is involved in the replication of retroviruses, such as human immunodeficiency virus (HIV). These viruses carry their genome in the form of ssRNA molecules. Following infection of a host cell, the viral enzyme reverse transcriptase uses the viral RNA as a template for the 5′→3′ synthesis of viral DNA, which then becomes integrated into host chromosomes. Reverse transcriptase activity is also seen with transposons, DNA elements that can move about the genome (see p. 477). In eukaryotes, most transposons are transcribed to RNA, the RNA is used as a template for DNA synthesis by a reverse transcriptase encoded by the transposon, and the DNA is randomly inserted into the genome. [Note: Transposons that involve an RNA intermediate are called retrotransposons or retroposons.] E. DNA replication inhibition by nucleoside analogs


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Biochemistry_Lippinco. E. DNA replication inhibition by nucleoside analogs DNA chain growth can be blocked by the incorporation of certain nucleoside analogs that have been modified on the sugar portion (Fig. 30.25). For example, removal of the hydroxyl group from the 3′-carbon of the deoxyribose ring as in 2′,3′-dideoxyinosine ([ddI] also known as didanosine), or conversion of the deoxyribose to another sugar, such as arabinose, prevents further chain elongation. By blocking DNA replication, these compounds slow the division of rapidly growing cells and viruses. Cytosine arabinoside (cytarabine, or araC) has been used in anticancer chemotherapy, whereas adenine arabinoside (vidarabine, or araA) is an antiviral agent. Substitution on the sugar moiety, as seen in azidothymidine (AZT), also called zidovudine (ZDV), also terminates DNA chain elongation. [Note: These drugs are generally supplied as nucleosides, which are then converted to nucleotides by cellular kinases.] V. EUKARYOTIC DNA ORGANIZATION


Biochemistry_Lippinco

Biochemistry_Lippinco. A typical (diploid) human somatic cell contains 46 chromosomes, whose total DNA is ~2 m long! It is difficult to imagine how such a large amount of genetic material can be effectively packaged into a volume the size of a cell nucleus so that it can be efficiently replicated and its genetic information expressed. To do so requires the interaction of DNA with a large number of proteins, each of which performs a specific function in the ordered packaging of these long molecules of DNA. Eukaryotic DNA is associated with tightly bound basic proteins, called histones. These serve to order the DNA into fundamental structural units, called nucleosomes, which resemble beads on a string. Nucleosomes are further arranged into increasingly more complex structures that organize and condense the long DNA molecules into chromosomes that can be segregated during cell division. [Note: The complex of DNA and protein found inside the nuclei of eukaryotic cells is called chromatin.]


Biochemistry_Lippinco

Biochemistry_Lippinco. A. Histones and nucleosome formation There are five classes of histones, designated H1, H2A, H2B, H3, and H4. These small, evolutionally conserved proteins are positively charged at physiologic pH as a result of their high content of lysine and arginine. Because of their positive charge, they form ionic bonds with negatively charged DNA. Histones, along with ions such as Mg2+ , help neutralize the negatively charged DNA phosphate groups.


Biochemistry_Lippinco

Biochemistry_Lippinco. 1. Nucleosomes: Two molecules each of H2A, H2B, H3, and H4 form the octameric core of the individual nucleosome “beads.” Around this structural core, a segment of dsDNA is wound nearly twice (Fig. 30.26). Winding eliminates a helical turn, causing negative supercoiling. [Note: The N-terminal ends of these histones can be acetylated, methylated, or phosphorylated. These reversible covalent modifications influence how tightly the histones bind to the DNA, thereby affecting the expression of specific genes. Histone modification is an example of epigenetics, or heritable changes in gene expression caused without alteration of the nucleotide sequence.] Neighboring nucleosomes are joined by linker DNA ~50 bp long. H1 is not found in the nucleosome core, but instead binds to the linker DNA chain between the nucleosome beads. H1 is the most tissue specific and species specific of the histones. It facilitates the packing of nucleosomes into more compact structures.


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Biochemistry_Lippinco. 2. Higher levels of organization: Nucleosomes can be packed more tightly (stacked) to form a nucleofilament. This structure assumes the shape of a coil, often referred to as a 30-nm fiber. The fiber is organized into loops that are anchored by a nuclear scaffold containing several proteins. Additional levels of organization lead to the final chromosomal structure (Fig. 30.27). B. Nucleosome fate during DNA replication Parental nucleosomes are disassembled to allow access to DNA during replication. Once DNA is synthesized, nucleosomes form rapidly. Their histone proteins come both from de novo synthesis and from the transfer of parental histones. VI. DNA REPAIR


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Biochemistry_Lippinco. Despite the elaborate proofreading system employed during DNA synthesis, errors (including incorrect base-pairing or insertion of one to a few extra nucleotides) can occur. In addition, DNA is constantly being subjected to environmental insults that cause the alteration or removal of nucleotide bases. The damaging agents can be either chemicals (for example, nitrous acid, which can deaminate bases) or radiation (for example, nonionizing ultraviolet [UV] radiation, which can fuse two pyrimidines adjacent to each other in the DNA, and high-energy ionizing radiation, which can cause double-strand breaks). Bases are also altered or lost spontaneously from mammalian DNA at a rate of many thousands per cell per day. If the damage is not repaired, a permanent change (mutation) is introduced that can result in any of a number of deleterious effects, including loss of control over the proliferation of the mutated cell, leading to cancer. Luckily, cells are remarkably efficient at repairing


Biochemistry_Lippinco

Biochemistry_Lippinco. can result in any of a number of deleterious effects, including loss of control over the proliferation of the mutated cell, leading to cancer. Luckily, cells are remarkably efficient at repairing damage done to their DNA. Most of the repair systems involve recognition of the damage (lesion) on the DNA, removal or excision of the damage, replacement or filling the gap left by excision using the sister strand as a template for DNA synthesis, and ligation. These excision repair systems remove one to tens of nucleotides. [Note: Repair synthesis of DNA can occur outside of the S phase.]


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Biochemistry_Lippinco. A. Mismatch repair Sometimes replication errors escape the proofreading activity during DNA synthesis, causing a mismatch of one to several bases. In E. coli, mismatch repair (MMR) is mediated by a group of proteins known as the Mut proteins (Fig. 30.28). Homologous proteins are present in humans. [Note: MMR occurs within minutes of replication and reduces the error rate of replication from 1 in 107 to 1 in 109 nucleotides.] 1.


Biochemistry_Lippinco

Biochemistry_Lippinco. Mismatched strand identification: When a mismatch occurs, the Mut proteins that identify the mispaired nucleotide(s) must be able to discriminate between the correct strand and the strand with the mismatch. In prokaryotes, discrimination is based on the degree of methylation. GATC sequences, which are found once every thousand nucleotides, are methylated on the adenine (A) residue by DNA adenine methylase (DAM). This methylation is not done immediately after synthesis, so the DNA is hemimethylated (that is, the parental strand is methylated, but the daughter strand is not). The methylated parental strand is assumed to be correct, and it is the daughter strand that gets repaired. [Note: The exact mechanism by which the daughter strand is identified in eukaryotes is not yet known, but likely involves recognition of nicks in the newly synthesized strand.] 2.


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Biochemistry_Lippinco. Repair procedure: When the strand containing the mismatch is identified, an endonuclease nicks the strand, and the mismatched nucleotide(s) is/are removed by an exonuclease. Additional nucleotides at the 5′-and 3′-ends of the mismatch are also removed. The gap left by removal of the nucleotides is filled, using the sister strand as a template, by a DNA pol, typically DNA pol III. The 3′-hydroxyl of the newly synthesized DNA is joined to the 5′-phosphate of the remaining stretch of the original DNA strand by DNA ligase. Mutation to the proteins involved in MMR in humans is associated with hereditary nonpolyposis colorectal cancer (HNPCC), also known as Lynch syndrome. Although HNPCC confers an increased risk for developing colon cancer (as well as other cancers), only about 5% of all colon cancer is the result of mutations in MMR. B. Nucleotide excision repair


Biochemistry_Lippinco

Biochemistry_Lippinco. B. Nucleotide excision repair Exposure of a cell to UV radiation can result in the covalent joining of two adjacent pyrimidines (usually thymines), producing a dimer. These intrastrand cross-links prevent DNA pol from replicating the DNA strand beyond the site of dimer formation. Thymine dimers are excised in bacteria by UvrABC proteins in a process known as nucleotide excision repair (NER), as illustrated in Figure 30.29. A related pathway is present in humans (see 2. below). [Note: Transcription-coupled repair, a type of NER, fixes DNA lesions encountered during RNA synthesis.] 1. Recognition and excision of UV-induced dimers: A UV-specific endonuclease (called uvrABC excinuclease) recognizes the bulky dimer and cleaves the damaged strand on both the 5′-side and 3′-side of the lesion. A short oligonucleotide containing the dimer is excised, leaving a gap in the DNA strand. This gap is filled in using a DNA pol I and DNA ligase. NER occurs throughout the cell cycle. 2.


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Biochemistry_Lippinco. 2. UV radiation and cancer: Pyrimidine dimers can be formed in the skin cells of humans exposed to UV radiation in unfiltered sunlight. In the rare genetic disease xeroderma pigmentosum (XP), the cells cannot repair the damaged DNA, resulting in extensive accumulation of mutations and, consequently, early and numerous skin cancers (Fig. 30.30). XP can be caused by defects in any of the several genes that code for the XP proteins required for NER of UV damage in humans. C. Base excision repair


Biochemistry_Lippinco

Biochemistry_Lippinco. C. Base excision repair DNA bases can be altered, either spontaneously, as is the case with cytosine, which slowly undergoes deamination (the loss of its amino group) to form uracil, or by the action of deaminating or alkylating compounds. For example, nitrous acid, which is formed by the cell from precursors such as the nitrates, deaminates cytosine, adenine (to hypoxanthine), and guanine (to xanthine). Dimethyl sulfate can alkylate (methylate) adenine. Bases can also be lost spontaneously. For example, ~10,000 purine bases are lost this way per cell per day. Lesions involving base alterations or loss can be corrected by base excision repair ([BER], Fig. 30.31). 1.


Biochemistry_Lippinco

Biochemistry_Lippinco. 1. Abnormal base removal: In BER, abnormal bases, such as uracil, which can occur in DNA by either deamination of cytosine or improper use of dUTP instead of dTTP during DNA synthesis, are recognized by specific DNA glycosylases that hydrolytically cleave them from the deoxyribosephosphate backbone of the strand. This leaves an apyrimidinic site, or apurinic if a purine was removed, both referred to as AP sites. 2. AP site recognition and repair: Specific AP endonucleases recognize that a base is missing and initiate the process of excision and gap filling by making an endonucleolytic cut just to the 5′-side of the AP site. A deoxyribose phosphate lyase removes the single, base-free, sugar phosphate residue. DNA pol I and DNA ligase complete the repair process. D. Double-strand break repair


Biochemistry_Lippinco

Biochemistry_Lippinco. Ionizing radiation, chemotherapeutic agents such as doxorubicin, and oxidative free radicals (see p. 148) can cause double-strand breaks in DNA that can be lethal to the cell. [Note: Such breaks also occur naturally during genetic recombination.] dsDNA breaks cannot be corrected by the previously described strategy of excising the damage on one strand and using the undamaged strand as a template for replacing the missing nucleotide(s). Instead, they are repaired by one of two systems. The first is nonhomologous end joining (NHEJ), in which a group of proteins mediates the recognition, processing, and ligation of the ends of two DNA fragments. However, some DNA is lost in the process. Consequently, NHEJ is error prone and mutagenic. Defects in NHEJ are associated with a predisposition to cancer and immunodeficiency syndromes. The second repair system, homologous recombination (HR), uses the enzymes that normally perform genetic recombination between homologous chromosomes during


Biochemistry_Lippinco

Biochemistry_Lippinco. to cancer and immunodeficiency syndromes. The second repair system, homologous recombination (HR), uses the enzymes that normally perform genetic recombination between homologous chromosomes during meiosis. This system is much less error prone (“error-free”) than NHEJ because any DNA that was lost is replaced using homologous DNA as a template. HR occurs in late S and G2 of the cell cycle, whereas NHEJ can occur anytime.


Biochemistry_Lippinco

Biochemistry_Lippinco. [Note: Mutations to the proteins BRCA1 or BRCA2 (breast cancer 1 or 2), which are involved in HR, increase the risk for developing breast and ovarian cancer.] VII. CHAPTER SUMMARY


Biochemistry_Lippinco

Biochemistry_Lippinco. DNA is a polymer of deoxynucleoside monophosphates covalently linked by 3′→5′-phosphodiester bonds (Fig. 30.32). The resulting long, unbranched chain has polarity, with both a 5′-end (free phosphate) and a 3′end (free hydroxyl). The sequence of nucleotides is read 5′→3′. DNA exists as a double-stranded molecule, in which the two chains are paired in an antiparallel manner and wind around each other, forming a double helix. Adenine pairs with thymine, and cytosine pairs with guanine. Each strand of the double helix serves as a template for constructing a complementary daughter strand (semiconservative replication). DNA replication occurs in the S phase of the cell cycle and begins at an origin of replication. As the two strands unwind and separate, synthesis occurs at two replication forks that move away from the origin in opposite directions (bidirectionally). Helicase unwinds the double helix. As the two strands of the double helix are separated, positive supercoils are produced in


Biochemistry_Lippinco

Biochemistry_Lippinco. that move away from the origin in opposite directions (bidirectionally). Helicase unwinds the double helix. As the two strands of the double helix are separated, positive supercoils are produced in the region of DNA ahead of the replication fork and negative supercoils behind the fork. DNA topoisomerases types I and II remove supercoils. DNA polymerases (pols) synthesize new DNA strands only in the 5′→3′ direction. Therefore, one of the newly synthesized stretches of nucleotide chains must grow in the 5′→3′ direction toward the replication fork (leading strand) and one in the 5′→3′ direction away from the replication fork (lagging strand). DNA pols require a primer, a short stretch of RNA synthesized by primase. Leading-strand synthesis needs only one RNA primer (continuous synthesis), whereas the lagging strand needs many (discontinuous synthesis involving Okazaki fragments). In Escherichia coli (E. coli), DNA chain elongation is catalyzed by DNA pol III, using 5′-deoxyribonucleoside


Biochemistry_Lippinco

Biochemistry_Lippinco. the lagging strand needs many (discontinuous synthesis involving Okazaki fragments). In Escherichia coli (E. coli), DNA chain elongation is catalyzed by DNA pol III, using 5′-deoxyribonucleoside triphosphates as substrates. The enzyme proofreads the newly synthesized DNA, removing terminal mismatched nucleotides with its 3′→5′ exonuclease activity. RNA primers are removed by DNA pol I, using its 5′→3′ exonuclease activity. This enzyme fills the gaps with DNA, proofreading as it synthesizes. The final phosphodiester linkage is catalyzed by DNA ligase. There are at least five high-fidelity eukaryotic DNA pols. Pol α is a multisubunit enzyme, one subunit of which is a primase. Pol α 5′→3′ polymerase activity adds a short piece of DNA to the RNA primer. Pol ε completes DNA synthesis on the leading strand, whereas pol δ elongates each lagging strand fragment. Pol β is involved with DNA repair, and pol γ replicates mitochondrial DNA. Pols ε, δ, and γ use 3′→5′ exonuclease activity to


Biochemistry_Lippinco

Biochemistry_Lippinco. the leading strand, whereas pol δ elongates each lagging strand fragment. Pol β is involved with DNA repair, and pol γ replicates mitochondrial DNA. Pols ε, δ, and γ use 3′→5′ exonuclease activity to proofread. Nucleoside analogs containing modified sugars can be used to block DNA chain growth. They are useful in anticancer and antiviral chemotherapy. Telomeres are stretches of highly repetitive DNA complexed with protein that protect the ends of linear chromosomes. As most cells divide and age, these sequences are shortened, contributing to senescence. In cells that do not senesce (for example, germline and cancer cells), the ribonucleoprotein telomerase employs its protein component reverse transcriptase to extend the telomeres, using its RNA component as a template. There are five classes of positively charged histone (H) proteins. Two of each of histones H2A, H2B, H3, and H4 form an octameric structural core around which DNA is wrapped, creating a nucleosome. The DNA connecting


Biochemistry_Lippinco

Biochemistry_Lippinco. of positively charged histone (H) proteins. Two of each of histones H2A, H2B, H3, and H4 form an octameric structural core around which DNA is wrapped, creating a nucleosome. The DNA connecting the nucleosomes, called linker DNA, is bound to H1. Nucleosomes can be packed more tightly to form a nucleofilament. Additional levels of organization create a chromosome. Most DNA damage can be corrected by excision repair involving recognition and removal of the damage by repair proteins, followed by replacement by DNA pols and joining by ligase. Ultraviolet radiation can cause thymine dimers that are recognized and removed in E. coli by uvrABC proteins of nucleotide excision repair. Defects in the XP proteins needed for nucleotide excision repair of thymine dimers in humans result in xeroderma pigmentosum. Mismatched bases are repaired by a similar process of recognition and removal by Mut proteins in E. coli. The extent of methylation is used for strand identification in prokaryotes.


Biochemistry_Lippinco

Biochemistry_Lippinco. pigmentosum. Mismatched bases are repaired by a similar process of recognition and removal by Mut proteins in E. coli. The extent of methylation is used for strand identification in prokaryotes. Defective mismatch repair by homologous proteins in humans is associated with hereditary nonpolyposis colorectal cancer. Abnormal bases (such as uracil) are removed by DNA N-glycosylases in base excision repair, and the sugar phosphate at the apyrimidinic or apurinic site is cut out. Double-strand breaks in DNA are repaired by nonhomologous end joining (error prone) and template-requiring homologous recombination (“error-free”).


Biochemistry_Lippinco

Biochemistry_Lippinco. Choose the ONE best answer. 0.1. A 10-year-old girl is brought by her parents to the dermatologist. She has many freckles on her face, neck, arms, and hands, and the parents report that she is unusually sensitive to sunlight. Two basal cell carcinomas are identified on her face. Based on the clinical picture, which of the following processes is most likely to be defective in this patient? A. Repair of double-strand breaks by error-prone homologous recombination B. Removal of mismatched bases from the 3′-end of Okazaki fragments by a methyl-directed process C. Removal of pyrimidine dimers from DNA by nucleotide excision repair D. Removal of uracil from DNA by base excision repair


Biochemistry_Lippinco

Biochemistry_Lippinco. C. Removal of pyrimidine dimers from DNA by nucleotide excision repair D. Removal of uracil from DNA by base excision repair Correct answer = C. The sensitivity to sunlight, extensive freckling on parts of the body exposed to the sun, and presence of skin cancer at a young age indicate that the patient most likely suffers from xeroderma pigmentosum (XP). These patients are deficient in any one of several XP proteins required for nucleotide excision repair of pyrimidine dimers in ultraviolet radiation– damaged DNA. Double-strand breaks are repaired by nonhomologous end joining (error prone) or homologous recombination (“error free”). Methylation is not used for strand discrimination in eukaryotic mismatch repair. Uracil is removed from DNA molecules by a specific glycosylase in base excision repair, but a defect in this process does not cause XP.


Biochemistry_Lippinco

Biochemistry_Lippinco. 0.2. Telomeres are complexes of DNA and protein that protect the ends of linear chromosomes. In most normal human somatic cells, telomeres shorten with each division. In stem cells and in cancer cells, however, telomeric length is maintained. In the synthesis of telomeres: A. telomerase, a ribonucleoprotein, provides both the RNA and the protein needed for synthesis. B. the RNA of telomerase serves as a primer. C. the RNA of telomerase is a ribozyme. D. the protein of telomerase is a DNA-directed DNA polymerase. E. the shorter 3′→5′ strand gets extended. F. the direction of synthesis is 3′→5′.


Biochemistry_Lippinco

Biochemistry_Lippinco. C. the RNA of telomerase is a ribozyme. D. the protein of telomerase is a DNA-directed DNA polymerase. E. the shorter 3′→5′ strand gets extended. F. the direction of synthesis is 3′→5′. Correct answer = A. Telomerase is a ribonucleoprotein particle required for telomere maintenance. Telomerase contains an RNA that serves as the template, not the primer, for the synthesis of telomeric DNA by the reverse transcriptase of telomerase. Telomeric RNA has no catalytic activity. As a reverse transcriptase, telomerase synthesizes DNA using its RNA template and so is an RNA-directed DNA polymerase. The direction of synthesis, as with all DNA synthesis, is 5′→3′, and it is the 3′-end of the already longer 5′→3′ strand that gets extended.


Biochemistry_Lippinco

Biochemistry_Lippinco. 0.3. While studying the structure of a small gene that was sequenced during the Human Genome Project, an investigator notices that one strand of the DNA molecule contains 20 A, 25 G, 30 C, and 22 T. How many of each base is found in the complete double-stranded molecule? A. A=40,G =50,C =60,T =44 E. A=42,G =55,C =55,T =42 B. A=44,G =60,C =50,T =40 C. A=45,G =45,C =52,T =52 D. A=50,G =47,C =50,T =47 Correct answer = B. The two DNA strands are complementary to each other, with A base-paired with T and G base-paired with C. So, for example, the 20 A on the first strand would be paired with 20 T on the second strand, the 25 G on the first strand would be paired with 25 C on the second strand, and so forth. When these are all added together, the correct numbers of each base are indicated in choice B. Notice that, in the correct answer, A = T and G = C. 0.4. List the order in which the following enzymes participate in prokaryotic replication. A. Ligase


Biochemistry_Lippinco

Biochemistry_Lippinco. 0.4. List the order in which the following enzymes participate in prokaryotic replication. A. Ligase B. Polymerase I (3′→5′ exonuclease activity) C. Polymerase I (5′→3′ exonuclease activity) D. Polymerase I (5′→3′ polymerase activity) E. Polymerase III F. Primase Correct answer: F, E, C, D, B, A. Primase makes the RNA primer; polymerase (pol) III extends the primer with DNA (and proofreads); pol I removes the primer with its 5′→3′ exonuclease activity, fills in the gap with its 5′→3′ polymerase activity, and removes errors with its 3′→5′ exonuclease activity; and ligase makes the 5′→3′-phosphodiester bond that links the DNA made by pols I and III. 0.5. Dideoxynucleotides lack a 3′-hydroxyl group. Why would incorporation of a dideoxynucleotide into DNA stop replication? The lack of the 3′-OH group prevents formation of the 3′-hydroxyl → 5′phosphate bond that links one nucleotide to the next in DNA. RNA Structure, Synthesis, and Processing 31


Biochemistry_Lippinco

Biochemistry_Lippinco. The lack of the 3′-OH group prevents formation of the 3′-hydroxyl → 5′phosphate bond that links one nucleotide to the next in DNA. RNA Structure, Synthesis, and Processing 31 For additional ancillary materials related to this chapter, please visit thePoint. I. OVERVIEW


Biochemistry_Lippinco

Biochemistry_Lippinco. The genetic master plan of an organism is contained in the sequence of deoxyribonucleotides in its DNA. However, it is through ribonucleic acid (RNA), the “working copies” of DNA, that the master plan is expressed (Fig. 31.1). The copying process, during which a DNA strand serves as a template for the synthesis of RNA, is called transcription. Transcription produces messenger RNA (mRNA), which are translated into sequences of amino acids (proteins), and ribosomal RNA (rRNA), transfer RNA (tRNA), and additional RNA molecules that perform specialized structural, catalytic, and regulatory functions and are not translated. That is, they are noncoding RNA (ncRNA). Therefore, the final product of gene expression can be RNA or protein, depending upon the gene. [Note: Only ~2% of the genome encodes proteins.] A central feature of transcription is that it is highly selective. For example, many transcripts are made of some regions of the DNA. In other regions, few or no transcripts are made.


Biochemistry_Lippinco

Biochemistry_Lippinco. proteins.] A central feature of transcription is that it is highly selective. For example, many transcripts are made of some regions of the DNA. In other regions, few or no transcripts are made. This selectivity is due, at least in part, to signals embedded in the nucleotide sequence of the DNA. These signals instruct the RNA polymerase where to start, how often to start, and where to stop transcription. Several regulatory proteins are also involved in this selection process. The biochemical differentiation of an organism’s tissues is ultimately a result of the selectivity of the transcription process. [Note: This selectivity of transcription is in contrast to the “all-or-none” nature of genomic replication.] Another important feature of transcription is that many RNA transcripts that initially are faithful copies of one of the two DNA strands may undergo various modifications, such as terminal additions, base modifications, trimming, and internal segment removal, which convert the


Biochemistry_Lippinco

Biochemistry_Lippinco. are faithful copies of one of the two DNA strands may undergo various modifications, such as terminal additions, base modifications, trimming, and internal segment removal, which convert the inactive primary transcript into a functional molecule. The transcriptome is the complete set of RNA transcripts expressed by a genome.


Biochemistry_Lippinco

Biochemistry_Lippinco. II. RNA STRUCTURE There are three major types of RNA that participate in the process of protein synthesis: rRNA, tRNA, and mRNA. Like DNA, these RNA are unbranched polymeric molecules composed of nucleoside monophosphates joined together by 3′→5′-phosphodiester bonds (see p. 412). However, they differ from DNA in several ways. For example, they are considerably smaller than DNA, contain ribose instead of deoxyribose and uracil instead of thymine, and exist as single strands that are capable of folding into complex structures. The three major types of RNA also differ from each other in size, function, and special structural modifications. [Note: In eukaryotes, additional small ncRNA molecules found in the nucleolus (snoRNA), nucleus (snRNA), and cytoplasm (microRNA [miRNA]) perform specialized functions as described on pp. 441, 442, and 475.]


Biochemistry_Lippinco

Biochemistry_Lippinco. A. Ribosomal RNA rRNA are found in association with several proteins as components of the ribosomes, the complex structures that serve as the sites for protein synthesis (see p. 451). Prokaryotic cells contain three distinct size species of rRNA (23S, 16S, and 5S, where S is the Svedberg unit for sedimentation rate that is determined by the size and shape of the particle), as shown in Figure 31.2. Eukaryotic cells contain four rRNA species (28S, 18S, 5.8S, and 5S). Together, rRNA make up ~80% of the total RNA in the cell. [Note: Some RNA function as catalysts, for example, an rRNA in protein synthesis (see p. 455). RNA with catalytic activity is termed a ribozyme.]


Biochemistry_Lippinco

Biochemistry_Lippinco. B. Transfer RNA tRNA are the smallest (4S) of the three major types of RNA molecules. There is at least one specific type of tRNA molecule for each of the 20 amino acids commonly found in proteins. Together, tRNA make up ~15% of the total RNA in the cell. The tRNA molecules contain a high percentage of unusual (modified) bases, for example, dihydrouracil (see Fig. 22.2, p. 292), and have extensive intrachain base-pairing (Fig. 31.3) that leads to characteristic secondary and tertiary structure. Each tRNA serves as an adaptor molecule that carries its specific amino acid, covalently attached to its 3′-end, to the site of protein synthesis. There, it recognizes the genetic code sequence on an mRNA, which specifies the addition of that amino acid to the growing peptide chain (see p. 447).


Biochemistry_Lippinco

Biochemistry_Lippinco. C. Messenger RNA mRNA comprises only ~5% of the RNA in a cell, yet is by far the most heterogeneous type of RNA in size and base sequence. mRNA is coding RNA in that it carries genetic information from DNA for use in protein synthesis. In eukaryotes, this involves transport of mRNA out of the nucleus and into the cytosol. An mRNA carrying information from more than one gene is polycistronic (cistron = gene). Polycistronic mRNA is characteristic of prokaryotes. An mRNA carrying information from only one gene is monocistronic and is characteristic of eukaryotes. In addition to the protein-coding regions that can be translated, mRNA contains untranslated regions at its 5′-and 3′-ends (Fig. 31.4). Special structural characteristics of eukaryotic (but not prokaryotic) mRNA include a long sequence of adenine nucleotides (a poly-A tail) on the 3′-end of the RNA, plus a cap on the 5′-end consisting of a molecule of 7-methylguanosine attached through an unusual (5′→5′) triphosphate linkage.


Biochemistry_Lippinco

Biochemistry_Lippinco. of adenine nucleotides (a poly-A tail) on the 3′-end of the RNA, plus a cap on the 5′-end consisting of a molecule of 7-methylguanosine attached through an unusual (5′→5′) triphosphate linkage. The mechanisms for modifying mRNA to create these special structural characteristics are discussed on pp. 441–442.


Biochemistry_Lippinco

Biochemistry_Lippinco. III. PROKARYOTIC GENE TRANSCRIPTION The structure of magnesium-requiring RNA polymerase (RNA pol), the signals that control transcription, and the varieties of modification that RNA transcripts can undergo differ among organisms, particularly from prokaryotes to eukaryotes. Therefore, the discussions of prokaryotic and eukaryotic transcription are presented separately. A. Prokaryotic RNA polymerase


Biochemistry_Lippinco

Biochemistry_Lippinco. In bacteria, one species of RNA pol synthesizes all of the RNA except for the short RNA primers needed for DNA replication [Note: RNA primers are synthesized by the specialized, monomeric enzyme primase (see p. 418).] RNA pol is a multisubunit enzyme that recognizes a nucleotide sequence (the promoter region) at the beginning of a length of DNA that is to be transcribed. It next makes a complementary RNA copy of the DNA template strand and then recognizes the end of the DNA sequence to be transcribed (the termination region). RNA is synthesized from its 5′-end to its 3′-end, antiparallel to its DNA template strand (see p. 415). The template is copied as it is in DNA synthesis, in which a guanine (G) on the DNA specifies a cytosine (C) in the RNA, a C specifies a G, a thymine (T) specifies an adenine (A), but an A specifies a uracil (U) instead of a T (Fig. 31.5). The RNA, then, is complementary to the DNA template (antisense, minus) strand and identical to the coding (sense, plus)


Biochemistry_Lippinco

Biochemistry_Lippinco. an adenine (A), but an A specifies a uracil (U) instead of a T (Fig. 31.5). The RNA, then, is complementary to the DNA template (antisense, minus) strand and identical to the coding (sense, plus) strand, with U replacing T. Within the DNA molecule, regions of both strands can serve as templates for transcription. For a given gene, however, only one of the two DNA strands can be the template. Which strand is used is determined by the location of the promoter for that gene. Transcription by RNA pol involves a core enzyme and several auxiliary proteins.


Biochemistry_Lippinco

Biochemistry_Lippinco. 1. Core enzyme: Five of the enzyme’s peptide subunits, 2 α, 1 β, 1 β′, and 1 Ω, are required for enzyme assembly (α, Ω), template binding (β′), and the 5′→3′ polymerase activity (β) and together are referred to as the core enzyme (Fig. 31.6). However, this enzyme lacks specificity (that is, it cannot recognize the promoter region on the DNA template). 2. Holoenzyme: The σ subunit (sigma factor) enables RNA pol to recognize promoter regions on the DNA. The σ subunit plus the core enzyme make up the holoenzyme. [Note: Different σ factors recognize different groups of genes, with σ70 predominating.] B. Steps in RNA synthesis The process of transcription of a typical gene of Escherichia coli (E. coli) can be divided into three phases: initiation, elongation, and termination. A transcription unit extends from the promoter to the termination region, and the initial product of transcription by RNA pol is termed the primary transcript.


Biochemistry_Lippinco

Biochemistry_Lippinco. 1. Initiation: Transcription begins with the binding of the RNA pol holoenzyme to a region of the DNA known as the promoter, which is not transcribed. The prokaryotic promoter contains characteristic consensus sequences (Fig. 31.7). [Note: Consensus sequences are idealized sequences in which the base shown at each position is the base most frequently (but not necessarily always) encountered at that position.] Those that are recognized by prokaryotic RNA pol σ factors include the following. a.


Biochemistry_Lippinco

Biochemistry_Lippinco. a. –35 Sequence: A consensus sequence (5′-TTGACA-3′), centered about 35 bases to the left of the transcription start site (see Fig. 31.7), is the initial point of contact for the holoenzyme, and a closed complex is formed. [Note: By convention, the regulatory sequences that control transcription are designated by the 5′→3′ nucleotide sequence on the coding strand. A base in the promoter region is assigned a negative number if it occurs prior to (to the left of, toward the 5′-end of, or “upstream” of) the transcription start site. Therefore, the TTGACA sequence is centered at approximately base −35. The first base at the transcription start site is assigned a position of +1. There is no base designated “0”.] b.


Biochemistry_Lippinco

Biochemistry_Lippinco. Pribnow box: The holoenzyme moves and covers a second consensus sequence (5′-TATAAT-3′), centered at about −10 (see Fig. 31.7), which is the site of melting (unwinding) of a short stretch (~14 base pairs) of DNA. This initial melting converts the closed initiation complex to an open complex known as a transcription bubble. [Note: A mutation in either the −10 or the −35 sequence can affect the transcription of the gene controlled by the mutant promoter.] 2.


Biochemistry_Lippinco

Biochemistry_Lippinco. Elongation: Once the promoter has been recognized and bound by the holoenzyme, local unwinding of the DNA helix continues (Fig. 31.8), mediated by the polymerase. [Note: Unwinding generates supercoils in the DNA that can be relieved by DNA topoisomerases (see p. 417).] RNA pol begins to synthesize a transcript of the DNA sequence, and several short pieces of RNA are made and discarded. The elongation phase begins when the transcript (typically starting with a purine) exceeds 10 nucleotides in length. Sigma is then released, and the core enzyme is able to leave (clear) the promoter and move along the template strand in a processive manner, serving as its own sliding clamp. During transcription, a short DNA–RNA hybrid helix is formed (see Fig. 31.8). Like DNA pol, RNA pol uses nucleoside triphosphates as substrates and releases pyrophosphate each time a nucleoside monophosphate is added to the growing chain. As with replication, transcription is always in the 5′→3′ direction. In