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joining the cores is

complexed with the fifth type of histone, H1. Further compaction of chromatin occurs as the strings of

nucleosomes wind into helical tubular coils called solenoid structures.Although complexes of DNA and histones form the nucleosomal substructures of chromatin, other types of proteins are also associated with DNA in the nucleus. These proteins were given the

unimaginative name of nonhistone chromosomal proteins. The cells of different tissues contain different

amounts and types of these proteins, which include enzymes that act on DNA and factors that regulate

transcription.

If histones contain large amounts of arginine and lysine, will their net charge be positive or

negative?

At physiologic pH, arginine and lysine carry positive charges on their side chains; therefore, histones have a net positive charge. The arginine and lysine residues are clustered in regions of the histone molecules. These positively charged regions of the histones

interact with the negatively charged DNA phosphate groups. C. The Human Genome

The genome, or total genetic content, of a human haploid cell (a sperm or an egg) is distributed in 23

chromosomes. Haploid cells contain one copy of each chromosome. The haploid egg and haploid sperm

cells combine to form the diploid zygote, which continues to divide to form our other cells (mitosis), also

diploid. In diploid cells, there are thus 22 pairs of autosomal chromosomes, with each pair composed of

two homologous chromosomes containing a similar series of genes (Fig. 12.14). In addition to the

autosomal chromosomes, each diploid cell has two sex chromosomes, designated X and Y. A female has

two X chromosomes, and a male has one X and one Y chromosome. The total number of chromosomes

per diploid cell is 46.Genes are arranged linearly along each chromosome. A gene, in genetic terms, is the fundamental unit

of heredity. In structural terms, a gene encompasses the DNA sequence that encodes the structural

components of the gene product (whether it be a polypeptide chain or an RNA molecule) along with the

DNA sequences adjacent to the 5′-end of the gene that regulates its expression. A genetic locus is a

specific position or location on a chromosome. Each gene on a chromosome in a diploid cell is matched

by an alternate version of the gene at the same genetic locus on the homologous chromosome (Fig. 12.15).

These alternate versions of a gene are called alleles. We thus have two alleles of each gene, one from our

mother and one from our father. If the alleles are identical in base sequence, we are homozygous for this

gene. If the alleles differ, we are heterozygous for this gene and may produce two versions of the encoded

protein that differ somewhat in primary structure.

The genomes of prokaryotic and eukaryotic cells differ in size. The genome of the bacterium E. coli

contains approximately 3,000 genes. All of this bacterial DNA has a function; it either codes for proteins,rRNA, and tRNA, or it serves to regulate the synthesis of these gene products. In contrast, the genome of

the human haploid cell contains between 20,000 and 25,000 genes, about seven to eight times the number

in E. coli. The function of most of this extra DNA has not been determined (an issue considered in more

detail in Chapter 15). III. Structure of RNA

A. General Features of RNA

RNA is similar to DNA. Like DNA, it is composed of nucleotides joined by 3′- to 5′- phosphodiester

bonds, the purine bases adenine and guanine, and the pyrimidine base cytosine. However, its other

pyrimidine base is uracil rather than thymine. Uracil and thymine are identical bases except that thymine

has a methyl group at position 5 of the ring (Fig. 12.16). In RNA, the sugar is ribose, which contains a

hydroxyl group on the 2′-carbon (see Fig. 12.3; the prime refers to the position on the ribose ring). As

indicated previously, the presence of the 2′-hydroxyl is what renders RNA susceptible to alkaline

hydrolysis.

RNA chains are usually single-stranded and lack the continuous helical structure of double-stranded

DNA. However, RNA still has considerable secondary and tertiary structure because base pairs can form

in regions where the strand loops back on itself. As in DNA, pairing between the bases is complementary

and antiparallel. But in RNA, adenine pairs with uracil rather than thymine (see Fig. 12.16B). Base

pairing in RNA can be extensive, and the irregular looped structures that are generated are important for

the binding of molecules, such as enzymes, that interact with specific regions of the RNA.

Will S. has sickle cell anemia (see Chapters 6 and 7). He has two alleles for the β-globin

gene that both generate the mutated form of hemoglobin, HbS. His younger sister Amanda, a

carrier for sickle cell trait, has one normal allele (which produces HbA) and one that produces

HbS. A carrier would theoretically be expected to produce HbA:HbS in a 50:50 ratio. However,what is generally seen in electrophoresis is a 60:40 ratio of HbA:HbS. Dramatic deviations from

this ratio imply the occurrence of an additional hemoglobin mutation (e.g., thalassemia).

The three major types of RNA (mRNA, rRNA, and tRNA) participate directly in the process of

protein synthesis. Other, less abundant RNAs are involved in replication or in the processing of RNA; that

is, in the conversion of RNA precursors to their mature forms or destruction of existing RNA molecules

(see Chapter 17). Other forms of RNA are involved in gene regulation (e.g., microRNA; see Chapter 16).

Some RNA molecules are capable of catalyzing reactions. Thus, RNA, as well as protein, can have

enzymatic activity. Certain rRNA precursors can remove internal segments of themselves, splicing the

remaining fragments together. Because this RNA is changed by the reaction that it catalyzes, it is not truly

an enzyme and therefore has been termed a ribozyme. Other RNAs act as true catalysts, serving as

ribonucleases that cleave other RNA molecules or as a peptidyl transferase, the enzyme in protein

synthesis that catalyzes the formation of peptide bonds.

Clark T.’s original benign adenomatous polyp was located in the ascending colon. Because

Mr. T.’s father died of a cancer of the colon, his physician had warned him that his risk for

developing colon cancer was two to three times higher than for the general population.

Unfortunately, Mr. T. neglected to have a repeat colonoscopic examination after 3 years as

recommended, and he developed an adenocarcinoma that metastasized.

Mr. T. is being treated with several chemotherapy agents, including 5-FU, a pyrimidine base

similar to uracil and thymine. 5-FU inhibits the synthesis of the thymine nucleotides required for

DNA replication. Thymine is normally produced by a reaction catalyzed by thymidylate synthase,

an enzyme that converts deoxyuridine monophosphate (dUMP) to deoxythymidine monophosphate

(dTMP). 5-FU is converted in the body to 5-FdUMP, which binds tightly to thymidylate synthase

in a transition-state complex and inhibits the reaction (recall that thymine is 5-methyluracil). Thus,

thymine nucleotides cannot be generated for DNA synthesis, and the rate of cell proliferation

decreases.

B. Structure of mRNA

Each mRNA molecule contains a nucleotide sequence that is converted into the amino acid sequence of a

polypeptide chain in the process of translation. In eukaryotes, mRNA is transcribed from protein-coding

genes as a long primary transcript that is processed in the nucleus to form mRNA. The various processingintermediates, which are mRNA precursors, are called pre-mRNA or heterogenous nuclear RNA

(hnRNA). mRNA travels through nuclear pores to the cytoplasm, where it binds to ribosomes and tRNAs

and directs the sequential insertion of the appropriate amino acids into a polypeptide chain.

Eukaryotic mRNA consists of a leader sequence at the 5′-end, a coding region, and a trailer sequence

at the 3′-end (Fig. 12.17). The leader sequence begins with a guanosine cap structure at its 5′-end. The

coding region begins with a trinucleotide start codon that signals the beginning of translation, followed by

the trinucleotide codons for amino acids, and ends at a termination signal. The trailer sequence terminates

at its 3′-end with a poly(A) tail that may be up to 200 nucleotides long. Most of the leader sequence, all of

the coding region, and most of the trailer are formed by transcription of the complementary nucleotide

sequence in DNA. However, the terminal guanosine in the cap structure and the poly(A) tail do not have

complementary sequences; they are added after transcription has been completed (posttranscriptionally).

C. Structure of rRNA

Ribosomes are subcellular ribonucleoprotein complexes on which protein synthesis occurs. Different

types of ribosomes are found in prokaryotes and in the cytoplasm and mitochondria of eukaryotic cells

(Fig. 12.18). Prokaryotic ribosomes contain three types of rRNA molecules with sedimentation

coefficients of 16S, 23S, and 5S. A sedimentation coefficient is a measure of the rate of sedimentation of

a macromolecule in a high-speed centrifuge (ultracentrifuge). The units of sedimentation are expressed in

Svedberg units (S). The 30S ribosomal subunit contains the 16S rRNA complexed with proteins, and the

50S ribosomal subunit contains the 23S and 5S rRNAs complexed with proteins. The 30S and 50S

ribosomal subunits join to form the 70S ribosome, which participates in protein synthesis. Although larger

macromolecules generally have higher sedimentation coefficients than do smaller macromolecules,

sedimentation coefficients are not additive. Because frictional forces acting on the surface of a

macromolecule slow its migration through the solvent, the rate of sedimentation

depends not only on the

density of the macromolecule but also on its shape.Cytoplasmic ribosomes in eukaryotes contain four types of rRNA molecules of 18S, 28S, 5S, and

5.8S. The 40S ribosomal subunit contains the 18S rRNA complexed with proteins, and the 60S ribosomal

subunit contains the 28S, 5S, and 5.8S rRNAs complexed with proteins. In the cytoplasm, the 40S and

60S ribosomal subunits combine to form the 80S ribosomes that participate in protein synthesis.

Mitochondrial ribosomes, with a sedimentation coefficient of 55S, are smaller than cytoplasmic

ribosomes. Their properties are similar to those of the 70S ribosomes of bacteria. rRNAs contain many loops and exhibit extensive base pairing in the regions between the loops. The

sequences of the rRNAs of the smaller ribosomal subunits exhibit secondary structures that are common to

many different species. D. Structure of tRNA

During protein synthesis, tRNA molecules carry amino acids to ribosomes and ensure that they are

incorporated into the appropriate positions in the growing polypeptide chain. This is done through base

pairing in an antiparallel manner of three bases of the tRNA (the anticodon) with the three base codonswithin the coding region of the mRNA. Therefore, cells contain at least 20 different tRNA molecules that

differ somewhat in nucleotide sequence, one for each of the amino acids found in proteins. Many amino

acids have more than one tRNA.

tRNA molecules contain not only the usual nucleotides but also derivatives of these nucleotides that

are produced by posttranscriptional modifications. In eukaryotic cells, 10% to 20% of the nucleotides of

tRNA are modified. Most tRNA molecules contain ribothymidine (rT), in which a methyl group is added

to uridine to form ribothymidine. They also contain dihydrouridine (D), in which one of the double bonds

of the base is reduced, and pseudouridine (ψ), in which uracil is attached to ribose by a carbon–carbon

bond rather than a nitrogen–carbon bond (see Chapter 14). The base at the 5′-end of the anticodon of

tRNA is also frequently modified.

tRNA molecules are small compared with both mRNA and the large rRNA molecules. On average,

tRNA molecules contain about 80 nucleotides and have a sedimentation coefficient of 4S. Because of

their small size and high content of modified nucleotides, tRNAs were the first nucleic acids to be

sequenced. Since 1965, when Robert Holley deduced the structure of the first tRNA, the nucleotide

sequences of many different tRNAs have been determined. Although their primary sequences differ, all

tRNA molecules can form a structure resembling a cloverleaf (discussed in more detail in Chapter 14).

E. Other Types of RNA

In addition to the three major types of RNA described previously, other RNAs are present in cells. These

RNAs include the oligonucleotides that serve as primers for DNA replication and the RNAs in the small

nuclear ribonucleoproteins (snRNPs or snurps) that are involved in the splicing and modification

reactions that occur during the maturation of RNA precursors (see Chapter 14). Also included are

microRNAs, which participate in the regulation of gene expression (see Chapters 16 and 18).

Azithromycin, the antibiotic used to treat Paul T., inhibits protein synthesis on

prokaryotic

ribosomes but not on eukaryotic ribosomes. It binds to the 50S ribosomal subunit, which is

absent in eukaryotes. Therefore, it will selectively inhibit bacterial growth. However, because

mitochondrial ribosomes are similar to those of bacteria, mitochondrial protein synthesis can also

be inhibited. This fact is important in understanding some of the side effects of antibiotics that

work by inhibiting bacterial protein synthesis. CLINICAL COM M ENTS

Isabel S. Isabel S.’s infection with HIV was from needles contaminated with HIV. Without

treatment, her HIV will progress and will result in the development of AIDS. The progressive

immunologic deterioration that accompanies this disease ultimately results in life-threatening

opportunistic infections with fungi (e.g., Candida, Cryptococcus, Pneumocystis jirovecii [formerly

known as Pneumocystis carinii]), other viruses (e.g., cytomegalovirus, herpes simplex), and bacteria

(e.g., Mycobacterium, Salmonella). The immunologic incompetence also frequently results in the

development of certain neoplasms (e.g., Kaposi sarcoma, non-Hodgkin lymphoma) as well as meningitis,neuropathies, and neuropsychiatric disorders causing cognitive dysfunction. Although recent advances in

drug therapy can slow or stop the course of the disease, no cure is yet available. Clark T. Clark T.’s original benign adenomatous polyp was located on the right side of the colon,

which is less common than the left side but is increasing in incidence. Because Mr. T.’s father died

from a cancer of the colon, his physician had warned him that his risk for developing colon cancer was

two to three times higher than for the general population. Unfortunately, Mr. T. neglected to have his

annual colonoscopic examinations as prescribed, and he developed an adenocarcinoma that metastasized.

The most malignant characteristic of neoplasms is their ability to metastasize; that is, form a new

neoplasm at a noncontiguous site. The initial site of metastases for a tumor is usually at the first capillary

bed encountered by the malignant cells once they are released. Thus, cells from tumors of the

gastrointestinal tract often pass through the portal vein to the liver, which is Clark T.’s site of metastasis.

Because his adenocarcinoma has metastasized, there is little hope of eradicating it, and his therapy with

5-FU and other chemotherapy agents is palliative (directed toward reducing the severity of the disease

and alleviating the symptoms without actually curing the disease). Although if the metastatic disease in the

liver responds to the chemotherapy, those lesions may then be able to be resected, resulting in improved

survival.

Clark T. completed his first course of intravenous chemotherapy in the hospital. He tolerated the therapy with only mild anorexia and diarrhea and with only a mild leukopenia

(a decreased white blood cell count; leuko = white). Thirty days after the completion of the initial

course, these symptoms abated and he started his second course of chemotherapy, including 5-FU

as an outpatient.

Because 5-FU inhibits synthesis of thymine, DNA synthesis is affected in all cells in the human

body that are rapidly dividing, such as the cells in the bone marrow that produce

leukocytes and

the mucosal cells lining the intestines. Inhibition of DNA synthesis in rapidly dividing cells

contributes to the side effects of 5-FU and many other chemotherapeutic drugs. Paul T. Paul T.’s infection was treated with azithromycin, a macrolide antibiotic. Because this

agent can inhibit mitochondrial protein synthesis in eukaryotic cells, it has the potential to alter

host-cell function, leading to such side effects as epigastric distress, diarrhea, and, infrequently,

cholestatic jaundice. BIOCHEM ICAL COM M ENTS

Retroviruses. RNA also serves as the genome for certain types of viruses, including retroviruses.

HIV is an example of a retrovirus (Fig. 12.19). Because they are not capable of reproducing

independently, viruses must invade host cells to reproduce. Some viruses that are pathogenic to humans

contain DNA as their genetic material; others contain RNA as their genetic material. HIV invades cells of

the immune system and prevents the affected individual from mounting an adequate immune response to

combat infections.According to the “central dogma” proposed by Francis Crick, information flows from DNA to RNA to

proteins. For the most part, this concept holds true. However, retroviruses provide one violation of this

rule. When retroviruses invade cells, their RNA genome is transcribed to produce a DNA copy. The

enzyme that catalyzes this process is encoded in the viral RNA and is known as reverse transcriptase.

This DNA copy integrates into the genome of the infected cell, and enzymes of the host cell are used to

produce many copies of the viral RNA, as well as viral proteins, which can be packaged into new viral

particles. KEY CONCEPTS

The central dogma of molecular biology is that DNA is transcribed to RNA, which is translated to

protein.

Nucleotides, consisting of a nitrogenous base, a five-carbon sugar, and phosphate, are the

monomeric units of the nucleic acids DNA and RNA (see Chapter 5). DNA contains the sugar 2′-deoxyribose; RNA contains ribose.

DNA and RNA contain the purine bases adenine (A) and guanine (G).

DNA contains the pyrimidine bases cytosine (C) and thymine (T), whereas RNA contains C and

uracil (U).

DNA and RNA are linear sequences of nucleotides linked by phosphodiester bonds between the 3′-

sugar of one nucleotide and the 5′-sugar of the next nucleotide.

Genetic information is encoded by the sequence of the nucleotide bases in DNA.

DNA is double stranded; one strand runs in the 5′-to-3′ direction, whereas the other is antiparallel

and runs in the 3′-to-5′ direction.

The two strands of DNA wrap about each other to form a double helix and are held together by

hydrogen bonding between bases in each strand and by hydrophobic interactions between the

stacked bases in the core of the molecule.

The base adenine hydrogen-bonds to thymine, whereas cytosine hydrogen-bonds to guanine.

Transcription of a gene generates a single-stranded RNA; the three major types of RNA are

messenger RNA (mRNA), ribosomal RNA (rRNA), and transfer RNA (tRNA).

Eukaryotic mRNA is modified at both the 5′- and 3′-ends. In between, it contains a

coding region for

the synthesis of a protein.Codons within the coding region dictate the sequence of amino acids in a protein. Each codon is

three nucleotides long.

rRNA and tRNA are required for protein synthesis.

rRNA is complexed with proteins to form ribonucleoprotein particles called ribosomes, which

bind mRNA and tRNAs during translation.

tRNA contains an anticodon that binds to a complementary codon on mRNA, ensuring insertion

of the correct amino acid into the protein being synthesized.

The diseases discussed in this chapter are summarized in Table 12.2. REVIEW QUESTIONS—CHAPTER 12

Directions: For each question below, select the single best answer.

1.Viruses cause many human infections such as hepatitis, encephalitis, and the “common cold.” Which

one of the following is a common trait of all these viruses?

A. They are small, circular DNA molecules that enter bacteria and replicate outside the host

genome.

B. Upon infecting eukaryotic cells, they are called phages.

C. All the common cold viruses contain both DNA and RNA genomes.

D. In order to reproduce, they must use the infected cell’s DNA, RNA, and protein-synthesizing

machinery.

E. Upon becoming infectious, they are called plasmids.

2.Many drugs used to treat cancers inhibit DNA replication, whereas some will inhibit the pathways

required to synthesize proteins from certain genes. Which one of the following accurately describes

a step leading from DNA replication to the synthesis of a protein?

A. Translation of a gene generates a single-stranded RNA that is complementary to both strands of

the DNA.B. Transcription of a gene generates a single-stranded RNA that is complementary to one of the

DNA strands.

C. tRNA encodes the proteins during translation. D. mRNA encodes the proteins during transcription. E. rRNA encodes the proteins during transcription.

3.Gout is caused by the deposition of urate crystals in the joints and kidney. Purines are metabolized to

uric acid, whereas pyrimidines, when metabolized, do not generate uric acid. Which one of the

following should be restricted in the diet of a patient with gout? A. Cytosine

B. Guanine C. Thymine D. Uracil

E. Deoxyribose

4.A patient has a microcytic, hypochromic anemia. In order to ascertain the cause of the anemia, the

patient’s hemoglobin was isolated, and it was determined that there was much more β-chain present

than α-chain, indicating an α-thalassemia. To determine the genetic basis of the thalassemia, nucleic

acids were isolated from the blood of the patient. During the procedure of isolating the nucleic acids,

both heat and alkali treatment were required. The alkali and heat were included owing to which

ONE of the following?

A. Alkali causes the two strands of DNA and RNA to separate. B. Heat causes the two strands of DNA and RNA to separate.

C. Alkali cleaves the phosphodiester bonds of DNA and RNA, degrading them to nucleotides.

D. Alkali separates the strands of DNA and degrades RNA to nucleotides.

E. Alkali separates the strands of RNA and degrades DNA to nucleotides.

5.Targeting certain structural features of eukaryotic mRNA can result in an inhibition of protein

synthesis. Which one of the following describes a unique aspect of eukaryotic mRNA? A. A polyguanosine tail is found at the 3′-end.

B. The 5′-end begins with a leader sequence that contains an adenine cap. C. The cap and tail of the mRNA are added posttranscriptionally.

D. The leader sequence contains a guanosine cap at the 3′-end of the mRNA. E. The poly(A) tail is found at the 3′-end of the mRNA.

6.Some chemotherapeutic drugs alter the ability of DNA polymerase to faithfully replicate. For the

DNA sequence 5′–ATCGATCGATCGATCG–3,′ which one of the following represents the sequence

and polarity of the complementary strand? A. 5′–ATCTATCGATCGATCG–3′

B. 3′–ATCGATCGATCGATCG–5′ C. 5′–CGAUCGAUCAUCGAU–3′ D. 5′–CGATCGATCGATCGAT–3′ E. 3′–CGATCGATCGATCGAT–5′

7.Certain drugs inhibit bacterial RNA synthesis. If the DNA strand 5′– GCTATGCATCGTGATC

GAATTGCGT–3′ serves as a template for the synthesis of RNA, which one of the following choicesgives the sequence and polarity of the newly synthesized RNA? A. 5′–ACGCAATTCGATCACGATGCATAGC–3′

B. 5′–UGCGUUAAGCUAGUGCUACGUAUCG–3′ C. 5′–ACGCAAUUCGAUCACGAUGCAUAGC–3′ D. 5′–CGAUACGUAGCACUAGCUUAACGCA–3′ E. 5′–GCTATGCATCGTGATCGAATTGCGT–3′

8.Understanding the structure of DNA, and the process of replication, enabled various drugs to be

developed that interfered with DNA replication. In DNA, the bond between the deoxyribose sugar

and the phosphate is best described by which one of the following? A. A polar bond

B. An ionic bond C. A hydrogen bond D. A covalent bond

E. A van der Waals bond

9.Certain drugs can intercalate between DNA bases and alter the backbone of the DNA. The backbone

of a DNA strand is composed of which of the following? Choose the one best answer. A. Phosphates and sugars

B. Bases and phosphates C. Nucleotides and sugars

D. Phosphates and nucleosides E. Sugars and bases

10.Analysis of one strand of a double-stranded piece of DNA displayed 20 mol % A,

25mol % T, 30

mol % G, and 25 mol % C. Which one of the following accurately represents the composition of the

complementary strand?

A. A is 25 mol %, T is 20 mol %, G is 25 mol %, and C is 30 mol %.

B. [A] is 30 mol %, [T] is 25 mol %, [G] is 20 mol %, and [C] is 25 mol %. C. [U] is 25 mol %, [T] is 20 mol %, [G] is 25 mol %, and [C] is 30 mol %. D. [A] is 25 mol %, [T] is 25 mol %, [G] is 25 mol %, and [C] is 25 mol %.

E. The composition of the complementary strand cannot be determined from the data given.

ANSWERS TO REVIEW QUESTIONS

1.The answer is D. Viruses consist of either a DNA or RNA genome but not both. All the viruses

that cause the common cold are RNA viruses. When viruses infect bacteria (prokaryotes), they are

called bacteriophages or phages. Plasmids are not viruses and are not infectious. Plasmids are

small, circular DNA molecules that can replicate autonomously, whereas viruses

cannot and must

use the host cell’s DNA, RNA, and protein-synthesizing machinery.

2. The answer is B. rRNA and tRNA are part of the apparatus for protein synthesis, but the

sequences of rRNA and tRNA do not encode proteins. mRNA carries the genetic information that

is converted into the amino acid sequence of a protein, but that is used in the process of

translation. Transcription of a gene from DNA generates RNA that is complementary to only one

strand of DNA.3. The answer is B. Adenine and guanine are purines, which form urate during their metabolism.

Cytosine, thymine, and uracil are pyrimidines that follow different metabolic pathways and do not

form uric acid. Deoxyribose is a component of DNA, but it is a sugar and not a purine base, and it

is not converted to uric acid.

4.The answer is D. Both alkali and heat cause the two strands of DNA to separate. Alkali does not

break the phosphodiester bonds of DNA but does cleave the phosphodiester bonds of RNA,

degrading RNA to nucleotides. RNA is single stranded, not double stranded. In the analysis of

DNA, many techniques call for its separation from RNA (the alkali treatment) and its denaturation

(separation of the double strands).

5.The answer is E. The leader sequence begins with an N7-methylguanosine cap at the 5′-end of the

mRNA. The coding region of the mRNA then follows. The 3′-end of the mRNA contains a polyadenine tail, which aids in the stability of the mRNA. Only the tail of the mRNA is added

after transcription of the mRNA.

6.The answer is D. The complementary strand must run in the opposite direction, so the 5′-end must

base-pair with the G at the 3′-end of the given strand. Therefore, the 5′-end of the complementary

strand must be C. G would then base-pair to C, A to T, and T to A. Answer B is incorrect because

the bases do not base-pair with each other with the sequences indicated, C is incorrect because U

is not found in DNA, and E is incorrect because it has the wrong polarity (the 5′-T in answer E

would not base-pair with the 3′-G in the given sequence).

7.The answer is C. The RNA strand must be complementary to the DNA strand, and A in

DNA

base-pairs with U in RNA, whereas T in DNA base-pairs with A in RNA, G in DNA base-pairs

with C in RNA, and C in DNA base-pairs with G in RNA. Answers A and E are incorrect because

they contain T, which is found in DNA, not RNA. Answer B is incorrect because the base-pairing

rules are broken when the strands are aligned in antiparallel fashion. Answers D is incorrect

because the polarity of the strand is incorrect (if one were to switch the 5′- and 3′-ends, the

answer would be correct).

8.The answer is D. The phosphate is in an ester bond between two deoxyribose groups, generating

the phosphodiester bond in the DNA backbone (these are covalent bonds). None of the other types

of bonds is correct.

9.The answer is A. The DNA backbone is composed of the phosphates and deoxyribose

in

phosphodiester linkages. The bases are internal to the backbone, base-paired to bases in the

complementary strand, and form stacking interactions within the double helix.

10. The answer is A. The base pairs in double-stranded DNA require that [A] = [T], and [C] = [G].

Therefore, if the concentration of A in one strand is 20 mol %, the concentration of T in the

complementary strand must also be 20 mol %. For the example given, then, [A] would be 25 mol

%, [T] would be 20 mol %, [G] would be 25 mol %, and [C] would be 30 mol %.13 Synthesis of DNA

For additional ancillary materials related to this chapter, please visit thePoint. DNA synthesis occurs by the process of replication. During replication, each of the two parental strands

of DNA serves as a template for the synthesis of a complementary strand. Thus, each DNA molecule

generated by the replication process contains one intact parental strand and one newly synthesized strand

(Fig. 13.1). In eukaryotes, DNA replication occurs during the S phase of the cell cycle, which is

followed by the G2 phase. The cell divides during the next phase (M), and each daughter cell receives an

exact copy of the DNA of the parent cell.

The Replication Fork. In both prokaryotes and eukaryotes, the site at which replication is occurring at

any given moment is called the replication fork. As replication proceeds, the two parental strandsseparate in front of the fork. Behind the fork, each newly synthesized strand of DNA base-pairs with its

complementary parental template strand. A complex of proteins is involved in replication. Helicases and

topoisomerases unwind the parental strands, and single-strand binding proteins prevent them from

reannealing.

The major enzyme involved in replication is a DNA polymerase that copies each parental template

strand in the 3′-to-5′ direction, producing new strands in a 5′-to-3′ direction. Deoxyribonucleoside

triphosphates serve as the precursors. One strand of newly synthesized DNA grows continuously,

whereas the other strand is synthesized discontinuously in short segments known as Okazaki fragments.

These fragments are subsequently joined by DNA ligase.

Initiation. DNA polymerase cannot initiate the synthesis of new strands. Therefore, a short primer is

produced which contains ribonucleotides (RNA). DNA polymerase can add deoxyribonucleotides to the

3′-end of this primer. This RNA primer is subsequently removed and replaced by deoxyribonucleotides.

Telomeres. The ends of linear chromosomes are called telomeres. The enzyme telomerase, an RNAdependent DNA polymerase that carries its own RNA template, is required for their replication.

Errors and Repair. Errors that occur during replication can lead to deleterious mutations. However,

many errors are corrected by enzyme activities associated with the complex at the replication fork. The

error rate is thus kept at a very low level.

Damage to DNA molecules also causes mutations. Repair mechanisms correct DNA damage, usually

by removing and replacing the damaged region. The intact, undamaged strand serves as a template for the

DNA polymerase involved in the repair process.

Recombination. Although cells have mechanisms to correct replication errors and to repair DNA

damage, some genetic change is desirable. It produces new proteins or variations of proteins that may

increase the survival rate of the species. Genetic change is produced by unrepaired mutations and by a