Chapter Summary, Questions Answers - DNA Structure, Replication, and Repair

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Chapter: Biochemistry : DNA Structure, Replication, and Repair

DNA is a polymer of deoxyribonucleoside monophosphates covalently linked by 3I →5I -phosphodiester bonds.


CHAPTER SUMMARY

DNA is a polymer of deoxyribonucleoside monophosphates covalently linked by 3I →5I -phosphodiester bonds (Figure 29.32). The resulting long, unbranched chain has polarity, with both a 5I -end and a 3I -end. The sequence of nucleotides is read 5I →3I . 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 the 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 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 (pol) synthesize new DNA strands only in the 5I →3I direction. Therefore, one of the newly synthesized stretches of nucleotide chains must grow in the 5I →3I direction toward the replication fork (leading strand) and one in the 5I →3I direction away from the replication fork (lagging strand). DNA pols require a primer. The primer for de novo DNA synthesis is a short stretch of RNA synthesized by primase. The leading strand only needs one RNA primer, whereas the lagging strand needs many. In E. coli, DNA chain elongation is catalyzed by DNA pol III, using 5I -deoxyribonucleoside triphosphates as substrates. The enzyme “proofreads” the newly synthesized DNA, removing terminal mismatched nucleotides with its 3I →5I exonuclease activity. RNA primers are removed by DNA pol I, using its 5I →3I 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 polymerases. Pol α is a multisubunit enzyme, one subunit of which is a primase. Pol α 5I →3I polymerase activity adds a short piece of DNA to the RNA primer. Pol e completes DNA synthesis on the leading strand, whereas pol d elongates each lagging strand fragment. Pol β is involved with DNA repair, and pol γ replicates mitochondrial DNA. Pols e, d, and g u s e 3I →5I 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), telomerase employs its enzyme component reverse transcriptase to extend the telomeres, using its RNA as a template. There are five classes of positively charged histone (H) proteins. Two 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 in E. coli by DNA pol and joining by ligase. Ultraviolet light can cause thymine dimers that are recognized and removed by uvrABC proteins of nucleotide excision repair. Defects in the XP proteins needed for thymine dimer repair 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. Defective mismatch repair by homologous proteins in humans is associated with hereditary nonpolyposis colorectal cancer. Abnormal bases (such as uracil) are removed by glycosylases in base excision repair, and the sugar phosphate at the apyrimidinic or apurinic (AP) site is cut out. Double-strand breaks in DNA are repaired by nonhomologous end-joining (error prone) and homologous recombination.


Figure 29.32 Key concept map for DNA structure, replication, and repair. Key concept map for DNA structure, replication, and repair. NHEJ = nonhomologous end-joining; HR = homologous recombination; NER = nucleotide excision repair; MMR = mismatch repair; BER = base excision repair; UV = ultraviolet light.


 

Study Questions

Choose the ONE best answer.

 

29.1 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 3I -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

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 light–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 here does not cause XP.

 

29.2 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 3I →5I strand gets extended.

F. the direction of synthesis is 3I →5I .

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 5I →3I , and it is the 3I -end of the already longer 5I →3I strand that gets extended.

 

29.3 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 As, 25 Gs, 30 Cs, and 22 Ts. How many of each base is found in the complete double-stranded molecule?

A.A=40,G=50,C=60,T=44

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

E.A=42,G=55,C=55,T=42

Correct answer = E. 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 As on the first strand would be paired with 20 Ts on the second strand, the 25 Gs on the first strand would be paired with 25 Cs on the second strand, and so forth. When these are all added together, the correct numbers of each base are indicated in choice E. Notice that, in the correct answer, A = T and G = C.

 

29.4 List the order in which the following enzymes participate in prokaryotic replication.

A. Ligase

B. Polymerase I (3I →5I exonuclease activity)

C. Polymerase I (5I →3I exonuclease activity)

D. Polymerase I (5I →3I polymerase activity)

E. Polymerase III

F. Primase

Correct answer: F, E, C, D, B, A. Primase makes the RNA primer; Polymerase III extends the primer with DNA (and proofreads); polymerase I removes the primer with its 5I →3I exonuclease activity, fills in the gap with its 5I →3I polymerase activity, and removes errors with its 3I →5I exonuclease activity; and ligase makes the 5I →3I phosphodiester bond that links the DNA made by polymerase I and polymerase III.

 

29.5 Dideoxynucleotides lack a 3I -hydroxyl group. Why would incorporation of a dideoxynucleotide into DNA stop replication?

The lack of the 3I -OH group prevents formation of the 3I -hydroxyl → 5I -phosphate bond that links one nucleotide to the next in DNA.

 

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