End of Chapter ?'s Test 3

10.4 A DNA template plus primer with the structure 3′ P —TGCGAATTAGCGACAT— P 5′ 5′ P —ATCGGTACGACGCTTAAC—OH 3′ (where P = a phosphate group) is placed in an in vitro DNA synthesis system (Mg2+, an excess of the four deoxyribonucleoside triphosphates, etc.) containing a mutant form of E. coli DNA polymerase I that lacks exonuclease activity. The polymerase and exonuclease activities of this aberrant enzyme are identical to those of normal E. coli DNA polymerase I. It simply has no exonuclease activity. (a) What will be the structure of the final product? (b) What will be the first step in the reaction sequence?

(a) note that DNA synthesis will not occur on the left end since the 3'-terminus of the potential primer strand is blocked with a phosphate group—all DNA polymerases require a free 3'-OH terminus. (b) The first step will be the removal of the mismatched C (exiting as dCMO) from the 3'-OH primer terminus by the 3' 5' exonuclease ("proofreading") activity.

9.26 (a) What functions do (1) centromeres, (2) telomeres provide? (b) Do telomeres have any unique structural features? (c) What is the function of telomerase? (d) When chromosomes are broken by exposure to high-energy radiation such as X rays, the resulting broken ends exhibit a pronounced tendency to stick to each other and fuse. Why might this occur?

(a) (1) Centromeres function as spindle-fiber attachment sites on chromosomes; they are required for the separation of homologous chromosomes to opposite poles of the spindle during anaphase I of meiosis and for the separation of sister chromatids during anaphase of mitosis and anaphase II of meiosis. (2) Telomeres provide at least three important functions: (i) prevention of exonucleolytic degredation of the ends of the linear DNA molecules in eukaryotic chromosomes, (ii) prevention of the fusion of ends of DNA molecules of different chromosomes, and (iii) provision of a mechanism for replication of the distal tips of linear DNA molecules in eukaryotic chromosomes. (b) Yes. Most telomeres studied to date contain DNA sequence repeat units (for example, TTAGGG in human chromosomes), and at least in some species, telomeres terminate with single-stranded 3' overhangs that form "hairpin" structures. The bases in these hairpins exhibit unique patterns of methylation that presumably contribute to the structure and stability of telomeres. (c) Telemerase adds the terminal DNA sequences, telomeres, to the linear chromosomes in eukaryotes. (d) The broken ends resulting from irradiation will not contain telomeres; as a result, the free ends of the DNA molecules are apparently subject to the activities of enzymes such as exonucleases, ligases, and the like, which modify the ends. They can regain stability by fusing to broken ends of other DNA molecules that contain terminal telomere sequences.

10.14 The bacteriophage lambda chromosome has several A:T-rich segments that denature when exposed to pH 11.05 for 10 minutes. After such partial denaturation, the linear packaged form of the lambda DNA molecule has the structure shown in Figure 10.7a. Following its injection into an E. coli cell, the lambda DNA molecule is converted to a covalently closed circular molecule by hydrogen bonding between its complementary single-stranded termini and the action of DNA ligase. It then replicates as a -shaped structure. The entire lambda chromosome is 17.5 μm long. It has a unique origin of replication located 14.3 μm from the left end of the linear form shown in Figure 10.7a. Draw the structure that would be observed by electron microscopy after both (1) replication of an approximately 6 μm-long segment of the lambda chromosomal DNA molecule (in vivo) and (2) exposure of this partially replicated DNA molecule to pH 11.05 for 10 minutes (in vitro), (a) if replication had proceeded bidirectionally from the origin, and (b) if replication had proceeded unidirectionally from the origin. No answer

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10.7 The Boston teaberry is an imaginary plant with a diploid chromosome number of 4, and Boston teaberry cells are easily grown in suspended cell cultures. 3H-Thymidine was added to the culture medium in which a G1-stage cell of this plant was growing. After one cell generation of growth in 3H-thymidine-containing medium, colchicine was added to the culture medium. The medium now contained both 3H-thymidine and colchicine. After two "generations" of growth in 3H-thymidine-containing medium (the second "generation" occurring in the presence of colchicine as well), the two progeny cells (each now containing eight chromosomes) were transferred to culture medium containing nonradioactive thymidine (1H-thymidine) plus colchicine. Note that a "generation" in the presence of colchicine consists of a normal cell cycle's chromosomal duplication but no cell division. The two progeny cells were allowed to continue to grow, proceeding through the "cell cycle," until each cell contained a set of metaphase chromosomes that looked like the following. If autoradiography were carried out on these metaphase chromosomes (four large plus four small), what pattern of radioactivity (as indicated by silver grains on the autoradiograph) would be expected? (Assume no recombination between DNA molecules.)

???

10.8 Suppose that the experiment described in Problem 10.7 was carried out again, except this time replacing the 3H-thymidine with nonradioactive thymidine at the same time that the colchicine was added (after one cell generation of growth in 3H-thymidine-containing medium). The cells were then maintained in colchicine plus nonradioactive thymidine until the metaphase shown in Problem 10.7 occurred. What would the autoradiographs of these chromosomes look like?

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11.9 For several decades, the dogma in biology has been that molecular reactions in living cells are catalyzed by enzymes composed of polypeptides. We now know that the introns of some precursor RNA molecules such as the rRNA precursors in Tetra-hymena are removed autocatalytically ("self-spliced") with no involvement of any catalytic protein. What does the demonstration of autocatalytic splicing indicate about the dogma that biological reactions are always catalyzed by proteinaceous enzymes?

ANS: "Self-splicing" of RNA precursors demonstrates that RNA molecules can also contain catalytic sites; this property is not restricted to proteins.

10.6 E. coli cells contain five different DNA polymerases—I, II, III, IV, and V. Which of these enzymes catalyzes the semiconservative replication of the bacterial chromosome during cell division? What are the functions of the other four DNA polymerases in E. coli?

ANS: (1) DNA Pol III (2) see p256

10.29 In what ways does chromosomal DNA replication in eukaryotes differ from DNA replication in prokaryotes?

ANS: (1) DNA replication usually occurs continuously in rapidly growing prokaryotic cells but is restricted to the S phase of the cell cycle in eukaryotes. (2) Most eukaryotic chromosomes contain multiple origins of replication, whereas most prokaryotic chromosomes contain a single origin of replication. (3) Prokaryotes utilize two catalytic complexes that contain the same DNA polymerase to replicate the leading and lagging strands, whereas eukaryotes utilize two or three distinct DNA polymerases for leading and lagging strand synthesis. (4) Replication of eukaryotic chromosomes requires the partial disassembly and reassembly of nucleosomes as replisomes move along parental DNA molecules. In prokaryotes, replication probably involves a similar partial disassembly/reassembly of nucleosome-like structures. (5) Most prokaryotic chromosomes are circular and thus have no ends. Most eukaryotic chromosomes are linear and have unique termini called telomeres that are added to replicating DNA molecules by a unique, RNA-containing enzyme called telomerase.

11.6. List three ways in which the mRNAs of eukaryotes differ from the mRNAs of prokaryotes.

ANS: (1) Eukaryotes have a 5′ cap on their mRNAs; prokaryotes do not. (2) Messenger RNAs of eukaryotes generally have a 3′ poly-A tail, prokaryotic mRNAs never do. (3) Messenger RNA formation in eukaryotes involves removal of introns (when present) and the splicing together of exons. Prokaryotic genes (with very rare exceptions) do not have introns.

10.16 One species of tree has a very large genome consisting of 2.0 × 1010 base pairs of DNA. (a) If this DNA was organized into a single linear molecule, how long (meters) would this molecule be? (b) If the DNA is evenly distributed among 10 chromosomes and each chromosome has one origin of DNA replication, how long would it take to complete the S phase of the cell cycle, assuming that DNA polymerase can synthesize 2 × 104 bp of DNA per minute? (c) An actively growing cell can complete the S phase of the cell cycle in approximately 300 minutes. Assuming that the origins of replication are evenly distributed, how many origins of replication are present on each chromosome? (d) What is the average number of base pairs between adjacent origins of replication?

ANS: (a) (34 nm/100bp) (2 × 1010 bp) = 6.8 × 109 = 6.8 meters. (b) 2 × 1010/10 chromosomes = 2 × 109 bp 4 × 104 bp/min (bidirectional); 2 × 109/4 × 104 = 5 × 104 min = 50,000 min. (c) (50,000 min) (1 ori) = (300 min) (X ori); X = 50,000/300 = 167 ori. (d) (2 × 109 bp/chrom.)/(167 ori/chrom.) = 1.2 × 107 bp/ori.

10.2 Escherichia coli cells are grown for many generations in a medium in which the only available nitrogen is the heavy isotope 15N. They are then transferred to a medium containing 14N as the only source of nitrogen. (a) What distribution of 15N and 14N would be expected in the DNA molecules of cells that had grown for one generation in the 14N-containing medium assuming that DNA replication was (i) conservative, (ii) semiconservative, or (iii) dispersive? (b) What distribution would be expected after two generations of growth in the 14N-containing medium assuming (i) conservative, (ii) semiconservative, or (iii) dispersive replication?

ANS: (a) (i) One-half of the DNA molecules with 15N in both strands and ½ with 14N in both strands; (ii) all DNA molecules with one strand containing 15N and the complementary strand containing 14N; (iii) all DNA molecules with both strands containing roughly equal amounts of 15N and 14N. (b) (i) ¼ of the DNA molecules with 15N in both strands and ¾ with 14N in both strands; (ii) ½ of the DNA molecules with one strand containing 15N and the complementary strand containing 14N and the other ½ with 14N in both strands; (iii) all DNA molecules with both strands containing about ¼ 15N and ¾ 14N. See Milestone Figure 1.

10.30 (a)The chromosome of the bacterium Salmonella typhimurium contains about 4 × 106 nucleotide pairs. Approximately how many Okazaki fragments are produced during one complete replication of the S. typhimurium chromosome? (b) The largest chromosome of D. melanogaster contains approximately 6 × 107 nucleotide pairs. About how many Okazaki fragments are produced during the replication of this chromosome?

ANS: (a) 2,000 to 4,000 Okazaki fragments. (b) 300,000 to 600,000 Okazaki fragments.

9.20 A diploid rye plant, Secale cereale, has 2n = 14 chromosomes and approximately 1.6 × 1010 bp of DNA. How much DNA is in a nucleus of a rye cell at (a) mitotic metaphase, (b) meiotic metaphase I, (c) mitotic telophase, and (d) meiotic telophase II?

ANS: (a) 3.2 1010 bp. (b) 3.2 1010 bp. (c) 1.6 1010 bp. (d) 0.8 1010 bp.

9.10 (a) If a virus particle contained double-stranded DNA with 200,000 base pairs, how many nucleotides would be present? (b) How many complete spirals would occur on each strand? (c) How many atoms of phosphorus would be present? (d) What would be the length of the DNA configuration in the virus?

ANS: (a) 400,000; (b) 20,000; (c) 400,000; (d) 68,000 nm.

9.19 The relationship between the melting Tm and GC content can be expressed, in its much simplified form, by the formula Tm = 69 + 0.41 (% GC). (a) Calculate the melting temperature of E. coli DNA that has about 50% GC. (b) Estimate the % GC of DNA from a human kidney cell where Tm = 85C.

ANS: (a) 89.50 C (b) about 39%

9.9 (a) Why did Watson and Crick choose a double helix for their model of DNA structure? (b) Why were hydrogen bonds placed in the model to connect the bases?

ANS: (a) A multistranded, spiral structure was suggested by the X-ray diffraction patterns. A double-stranded helix with specific base-pairing nicely fits the 1:1 stoichiometry observed for A:T and G:C in DNA. (b) Use of the known hydrogen-bonding potential of the bases provided a means of holding the two complementary strands in a stable configuration in such a double helix.

10.1 DNA polymerase I of E. coli is a single polypeptide of molecular weight 103,000. (a) What enzymatic activities other than polymerase activity does this polypeptide possess? (b) What are the in vivo functions of these activities? (c) Are these activities of major importance to an E. coli cell? Why?

ANS: (a) Both 3′ → 5′ and 5′ → 3′ exonuclease activities. (b) The 3′ → 5′ exonuclease "proofreads" the nascent DNA strand during its synthesis. If a mismatched base pair occurs at the 3′-OH end of the primer, the 3′ → 5′ exonuclease removes the incorrect terminal nucleotide before polymerization proceeds again. The 5′ → 3′ exonuclease is responsible for the removal of RNA primers during DNA replication and functions in pathways involved in the repair of damaged DNA (see Chapter 13). (c) Yes, both exonuclease activities appear to be very important. Without the 3′ → 5′ proofreading activity during replication, an intolerable mutation frequency would occur. The 5′ → 3′ exonuclease activity is essential to the survival of the cell. Conditional mutations that alter the 5′ → 3′ exonuclease activity of DNA polymerase I are lethal to the cell under conditions where the exonuclease is nonfunctional.

0.15 What enzyme activity catalyzes each of the following steps in the semiconservative replication of DNA in prokaryotes? (a) The formation of negative supercoils in progeny DNA molecules. (b) The synthesis of RNA primers. (c) The removal of RNA primers. (d) The covalent extension of DNA chains at the 3′-OH termini of primer strands. (e) Proofreading of the nucleotides at the 3′-OH termini of DNA primer strands?

ANS: (a) DNA gyrase; (b) primase; (c) the 5′ → 3′ exonuclease activity of DNA polymerase I; (d) the 5′ → 3′ polymerase activity of DNA polymerase III; (e) the 3′ → 5′ exonuclease activity of DNA polymerase III.

8.15 (a) Of what use are F' factors in genetic analysis? (b) How are F' factors formed? (c) By what mechanism does sexduction occur?

ANS: (a) F' factors are useful for genetic analyses where two copies of a gene must be present in the same cell, for example, in determining dominance relationships. (b) F' factors are formed by abnormal excision of F factors from Hfr chromosomes (see Fig. 8.31). (c) By the conjugative transfer of an F' factor from a donor cell to a recipient (F-) cell.

8.14 (a) What are the genotypic differences between F cells, F+ cells, and Hfr cells? (b) What are the phenotypic differences? (c) By what mechanism are F cells converted to F+ cells? F+ cells to Hfr cells? Hfr cells to F+ cells?

ANS: (a) F- cells, no F factor present; F+ cells, autonomous F factor; Hfr cells, integrated F factor (see Fig. 8.22). (b) F+ and Hfr cells have F pili; F- cells do not. (c) F- cells are converted to F+ cells by the conjugative transfer of F factors from F+ cells. Hfr cells are formed when F factors in F+ cells become integrated into the chromosomes of these cells. Hfr cells become F+ cells when the integrated F factors exit the chromosome and become autonomous (self-replicating) genetic elements.

10.12 The E. coli chromosome contains approximately 4 × 106 nucleotide pairs and replicates as a single bidirectional replicon in approximately 40 minutes under a wide variety of growth conditions. The largest chromosome of D. melanogaster contains about 6 × 107 nucleotide pairs. (a) If this chromosome contains one giant molecule of DNA that replicates bidirectionally from a single origin located precisely in the middle of the DNA molecule, how long would it take to replicate the entire chromosome if replication in Drosophila occurred at the same rate as replication in E. coli? (b) Actually, replication rates are slower in eukaryotes than in prokaryotes. If each replication bubble grows at a rate of 5000 nucleotide pairs per minute in Drosophila and 100,000 nucleotide pairs per minute in E. coli, how long will it take to replicate the largest Drosophila chromosome if it contains a single bidirectional replicon as described in (a) above? (c) During the early cleavage divisions in Drosophila embryos, the nuclei divide every 9 to 10 minutes. Based on your calculations in (a) and (b) above, what do these rapid nuclear divisions indicate about the number of replicons per chromosome in Drosophila?

ANS: (a) Given bidirectional replication of a single replicon, each replication fork must traverse 2 106 nucleotide pairs in E. coli and 3 107 nucleotide pairs in the largest Drosphila chromosome. If the rates were the same in both species, it would take 15 times [Insert equation from previous edition answer 10.12] as long to replicate the Drosphila chromosome or 10 hours (40 minutes 15 = 600 minutes). (b) 10 hours x 2 = 20 hours (c) If it would take 20 hours to replicate the Drosophila chromosome from one origin, then it would require approximately (1200min/9 or 10min) 120-130 replicons per chromosome.

9.1 (a) How did the transformation experiments of Griffith differ from those of Avery and his associates? (b) What was the significant contribution of each? (c) Why was Griffith's work not evidence for DNA as the genetic material, whereas the experiments of Avery and coworkers provided direct proof that DNA carried the genetic information?

ANS: (a) Griffith's in vivo experiments demonstrated the occurrence of transformation in pneumococcus. They provided no indication as to the molecular basis of the transformation phenomenon. Avery and colleagues carried out in vitro experiments, employing biochemical analyses to demonstrate that transformation was mediated by DNA. (b) Griffith showed that a transforming substance existed; Avery et al. defined it as DNA. (c) Griffith's experiments did not include any attempt to characterize the substance responsible for transformation. Avery et al. isolated DNA in "pure" form and demonstrated that it could mediate transformation.

9.30 (a) Which class of chromosomal proteins, histones or nonhistones, is the more highly conserved in different eukaryotic species? Why might this difference be expected? (b) If one compares the histone and nonhistone chromosomal proteins of chromatin isolated from different tissues or cell types of a given eukaryotic organism, which class of proteins will exhibit the greater heterogeneity? Why are both classes of proteins not expected to be equally homogeneous in chromosomes from different tissues or cell types?

ANS: (a) Histones have been highly conserved throughout the evolution of eukaryotes. A major function of histones is to package DNA into nucleosomes and chromatin fibers. Since DNA is composed of the same four nucleotides and has the same basic structure in all eukaryotes, one might expect that the proteins that play a structural role in packaging this DNA would be similarly conserved. (b) The nonhistone chromosomal proteins exhibit the greater heterogeneity in chromatin from different tissues and cell types of an organism. The histone composition is largely the same in all cell types within a given species—consistent with the role of histones in packaging DNA into nucleosomes. The nonhistone chromosomal proteins include proteins that regulate gene expression. Because different sets of genes are transcribed in different cell types, one would expect heterogeneity in some of the nonhistone chromosomal proteins of different tissues.

9.32 (a) If the haploid human genome contains 3 × 109 nucleotide pairs and the average molecular weight of a nucleotide pair is 660, how many copies of the human genome are present, on average, in 1 mg of human DNA? (b) What is the weight of one copy of the human genome? (c) If the haploid genome of the small plant Arabidopsis thaliana contains 7.7 × 107 nucleotide pairs, how many copies of the A. thaliana genome are present, on average, in 1 mg of A. thaliana DNA? (d) What is the weight of one copy of the A. thaliana genome? (e) Of what importance are calculations of the above type to geneticists?

ANS: (a) One g of human DNA will contain, on average, 3.04 105 copies of the genome. Using an average molecular weight per nucleotide pair of 660, the molecular weight of the entire human genome is 1.98 1012 (3 109 660). Thus 1.98 1012 grams (1 "mole" = number of grams equivalent to the "molecular" weight) of human DNA will contain, on average, 6.02 1023 molecules [Avogadro's number = number of molecules (here, copies of the genome) present in one "mole" of a substance.] One gram will contain, on average, copies of the genome; thus 1 g will contain, on average, 3.04 105 copies of the human genome. (b) One copy of the human genome weighs approximately (c) By analogous calculations, 1 g of Arabidopsis thaliana DNA contains, on average, 1.18 107 copies of the genome. (d) Similarly, one copy of the A. thaliana genome weighs approximately 8.4 10-8 g. (e) In carrying out molecular analyses of the structures of genomes, geneticists frequently need to know how many copies of a genome are present, on average, in a given quantity of DNA.

11.1 Distinguish between DNA and RNA (a) chemically, (b) functionally, and (c) by location in the cell.

ANS: (a) RNA contains the sugar ribose, which has an hydroxyl (OH) group on the 2-carbon; DNA contains the sugar 2-deoxyribose, with only hydrogens on the 2-carbon. RNA usually contains the base uracil at positions where thymine is present in DNA. However, some DNAs contain uracil, and some RNAs contain thymine. DNA exists most frequently as a double helix (double-stranded molecule); RNA exists more frequently as a single-stranded molecule. But, some DNAs are single-stranded and some RNAs are double-stranded. (b) The main function of DNA is to store genetic information and to transmit that information from cell to cell and from generation to generation. RNA stores and transmits genetic information in some viruses that contain no DNA. In cells with both DNA and RNA: (1) mRNA acts as an intermediary in protein synthesis, carrying the information from DNA in the chromosomes to the ribosomes (sites at which proteins are synthesized). (2) tRNAs carry amino acids to the ribosomes and function in codon recognition during the synthesis of polypeptides. (3) rRNA molecules are essential components of the ribosomes. (4) snRNAs are important components of spliceosomes, and (5) miRNAs play key roles in regulating gene expression (see Chapter 20). (c) DNA is located primarily in the chromosomes, which are found in the nucleus of eukaryotic cells; however, some DNA is also found in cytoplasmic organelles, such as mitochondria and chloroplasts. RNA is located throughout cells.

9.7 (a) What distinguishes a viroid from an RNA virus? (b) What distinguishes a prion from a viroid? (c) In what ways are viroids and prions similar?

ANS: (a) The RNA genome of a virus is packaged inside a protein coat, whereas a viroid is composed of naked RNA. (b) A prion is composed of protein, whereas a viroid is RNA. (c) Viroids and prions are both infectious; they can maintain the phenotype they produce from generation to generation.

9.8 (a) What background material did Watson and Crick have available for developing a model of DNA? (b) What was their contribution to building the model?

ANS: (a) The ladderlike pattern was known from X-ray diffraction studies. Chemical analyses had shown that a 1:1 relationship existed between the organic bases adenine and thymine and between cytosine and guanine. Physical data concerning the length of each spiral and the stacking of bases were also available. (b) Watson and Crick developed the model of a double helix, with the rigid strands of sugar and phosphorus forming spirals around an axis, and hydrogen bonds connecting the complementary bases in nucleotide pairs.

9.5 (a) What was the objective of the experiment carried out by Hershey and Chase? (b) How was the objective accomplished? (c) What is the significance of this experiment?

ANS: (a) The objective was to determine whether the genetic material was DNA or protein. (b) By labeling phosphorus, a constituent of DNA, and sulfur, a constituent of protein, in a virus, it was possible to demonstrate that only the labeled phosphorus was introduced into the host cell during the viral reproductive cycle. The DNA was enough to produce new phages. (c) Therefore DNA, not protein, is the genetic material.

9.28 How many DNA molecules are present in (a) the axial regions, (b) the lateral loops of lampbrush chromosomes?

ANS: (a) Two. The axial region of a "lampbrush" chromosome contains the two chromatids of one homologous chromosome (postreplication). (b) One. Each lateral loop of a "lampbrush" chromosome is a segment of a single-chromatid.

9.16 The nucleic acids from various viruses were extracted and examined to determine their base composition. Given the following results, what can you hypothesize about the physical nature of the nucleic acids from these viruses? (a) 35% A, 35% T, 15% G, and 15% C. (b) 35% A, 15% T, 25% G, and 25% C. (c) 35% A, 30% U, 30% G, and 5% C.

ANS: (a) double-stranded DNA; (b) single-stranded DNA; (c) single-stranded RNA.

9.15 Indicate whether each of the following statements about the structure of DNA is true or false. (Each letter is used to refer to the concentration of that base in DNA.) (a) A + T = G + C (b) A = G; C = T (c) (d) (e) A + G = C + T (f) (g) A = T within each single strand. (h) Hydrogen bonding provides stability to the double helix in aqueous cytoplasms. (i) Hydrophobic bonding provides stability to the double helix in aqueous cytoplasms. (j) When separated, the two strands of a double helix are identical. (k) Once the base sequence of one strand of a DNA double helix is known, the base sequence of the second strand can be deduced. (l) The structure of a DNA double helix is invariant. (m) Each nucleotide pair contains two phosphate groups, two deoxyribose molecules, and two bases.

ANS: (a) false (b) false (c) true (d) true (e) true (f) true (g) false (h) true (i) true (j) false (k) true (l) false (m) true.

9.2 A cell-free extract is prepared from Type IIIS pneumococcal cells. What effect will treatment of this extract with (a) protease, (b) RNase, and (c) DNase have on its subsequent capacity to transform recipient Type IIR cells to Type IIIS? Why?

ANS: (a) no effect; (b) no effect; (c) DNase will destroy the capacity of the extract to transform type IIR cells to Type IIIS by degrading the DNA in the extract. Protease and RNase will degrade the proteins and RNA, respectively, in the extract. They will have no effect, since the proteins and RNA are not involved in transformation.

9.14 If one strand of DNA in the Watson-Crick double helix has a base sequence of 5-GTCATGAC-3, what is the base sequence of the complementary strand?

ANS: 3'-C A G T A C T G-5'

11.2. What bases in the mRNA transcript would represent the following DNA template sequence: 5′-TGCAGACA-3′?

ANS: 3′-ACGUCUGU-5′

11.3 What bases in the transcribed strand of DNA would give rise to the following mRNA base sequence: 5′-CUGAU-3′?

ANS: 3′—GACTA—5′

9.4 What is the macromolecular composition of a bacterial virus or bacteriophage such as phage T2?

ANS: About ½ protein, ½ DNA. A single long molecule of DNA is enclosed within a complex "coat" composed of many proteins.

8.3 In what ways do the life cycles of bacteriophages T4 and differ? In what respects are they the same?

ANS: Bacteriophage T4 is a virulent phage. When it infects a host cell, it reproduces and kills the host cell in the process. Bacteriophage lambda can reproduce and kill the host bacterium—the lytic response—just like phage T4, or it can insert its chromosome into the chromosome of the host and remain there in a dormant state—the lysogenic response.

8.2 How do bacteriophages differ from other viruses?

ANS: Bacteriophages reproduce in bacteria; other viruses reproduce in protists, plants, and animals.

10.20 Many of the origins of replication that have been characterized contain A:T-rich core sequences. Are these A:T-rich cores of any functional significance? If so, what?

ANS: Because A:T base pairs are held together by only two hydrogen bonds instead of the three hydrogen bonds present in G:C base pairs, the two strands of A:T-rich regions of double helices are separated more easily, providing the single-stranded template regions required for DNA replication.

11.7 What different types of RNA molecules are present in prokaryotic cells? in eukaryotic cells? What roles do these different classes of RNA molecules play in the cell?

ANS: Both prokaryotic and eukaryotic organisms contain messenger RNAs, transfer RNAs, and ribosomal RNAs. In addition, eukaryotes contain small nuclear RNAs and micro RNAs. Messenger RNA molecules carry genetic information from the chromosomes (where the information is stored) to the ribosomes in the cytoplasm (where the information is expressed during protein synthesis). The linear sequence of triplet codons in an mRNA molecule specifies the linear sequence of amino acids in the polypeptides produced during translation of that mRNA. Transfer RNA molecules are small (about 80 nucleotides long) molecules that carry amino acids to the ribosomes and provide the codon-recognition specificity during translation. Ribosomal RNA molecules provide part of the structure and function of ribosomes; they represent an important part of the machinery required for the synthesis of polypeptides. Small nuclear RNAs are structural components of spliceosomes, which excise noncoding intron sequences from nuclear gene transcripts. Micro RNAs are involved in the regulation of gene expression.

8.18 How can bacterial genes be mapped by interrupted mating experiments?

ANS: By interrupting conjugation at various times after the donor and recipient cells are mixed (using a blender or other form of agitation), one can determine the length of time required to transfer a given genetic marker from an Hfr cell to an F-.

8.19 What does the term cotransduction mean? How can cotransduction frequencies be used to map genetic markers?

ANS: Cotransduction refers to the simultaneous transduction of two different genetic markers to a single recipient cell. Since bacteriophage particles can package only 1/100 to 1/50 of the total bacterial chromosome, only markers that are relatively closely linked can be cotransduced. The frequency of cotransduction of any two markers will be an inverse function of the distance between them on the chromosome. As such, this frequency can be used as an estimate of the linkage distance. Specific cotransduction-linkage functions must be prepared for each phage-host system studied.

10.11 Seven distinct DNA polymerases—, , , , , ζ, and η—have been characterized in mammals. What are the intracellular locations and functions of these polymerases?

ANS: Current evidence suggests that polymerases α, δ, and/or ε are required for the replication of nuclear DNA. Polymerase δ and/or ε are thought to catalyze the continuous synthesis of the leading strand, and polymerase α is believed to function as a primase in the discontinuous synthesis of the lagging strand. Polymerase γ catalyzes replication of organellar chromosomes. Polymerases β, ζ, η, θ, ι, κ, λ, μ and σ function in various DNA repair pathways (see Chapter 13).

9.21 The available evidence indicates that each eukaryotic chromosome (excluding polytene chromosomes) contains a single giant molecule of DNA. What different levels of organization of this DNA molecule are apparent in chromosomes of eukaryotes at various times during the cell cycle?

ANS: DNA during interphase is not yet organized into individual chromosomes, but consists of a series of ellipsoidal "beads on a string" that form an 11-nm fiber. Here, 146 bp of DNA is wrapped 1.65 turns around the nucleosome core of eight histones. However, during metaphase of meiosis and mitosis, DNA becomes organized into chromosomes. The 11-nm fiber is folded and supercoiled to produce a 30-nm chromatin fiber, the basic structural unit of the metaphase chromosome. A third and final level of packaging involves nonhistone chromosomal proteins that form a scaffold to condense the 30-nm fibers into tightly packaged metaphase chromosomes, the highest level of DNA condensation observed.

9.11 What are the differences between DNA and RNA?

ANS: DNA has one atom less of oxygen than RNA in the sugar part of the molecule; the sugar in DNA is 2-deoxyribose, whereas the sugar in RNA is ribose. In DNA, thymine replaces the uracil that is present in RNA. (In certain bacteriophages, DNA-containing uracil is present.) DNA is most frequently double-stranded, but bacteriophages such as ΦX174 contain single-stranded DNA. RNA is most frequently single-stranded. Some viruses, such as the Reoviruses, however, contain double-stranded RNA chromosomes.

10.23 In E. coli, three different proteins are required to unwind the parental double helix and keep the unwound strands in an extended template form. What are these proteins, and what are their respective functions?

ANS: DNA helicase unwinds the DNA double helix, and single-strand DNA-binding protein coats the unwound strands, keeping them in an extended state. DNA gyrase catalyzes the formation of negative supercoiling in E. coli DNA, and this negative supercoiling behind the replication forks is thought to drive the unwinding process because superhelical tension is reduced by unwinding the complementary strands.

10.24 How similar are the structures of DNA polymerase I and DNA polymerase III in E. coli? What is the structure of the DNA polymerase III holoenzyme? What is the function of the dnaN gene product in E. coli?

ANS: DNA polymerase I is a single polypeptide of molecular weight 109,000, whereas DNA polymerase III is a complex multimeric protein. The DNA polymerase holoenzyme has a molecular mass of about 900,000 daltons and is composed of at least 20 different polypeptides. The dnaN gene product, the subunit of DNA polymerase III, forms a dimeric clamp that encircles the DNA molecule and prevents the enzyme from dissociating from the template DNA during replication.

10.22 DNA polymerase I is needed to remove RNA primers during chromosome replication in E. coli. However, DNA polymerase III is the true replicase in E. coli. Why doesn't DNA polymerase III remove the RNA primers?

ANS: DNA polymerase III does not have a 5 3 exonuclease activity that acts on double-stranded nucleic acids. Thus, it cannot excise RNA primer strands from replicating DNA molecules. DNA polymerase I is present in cells at much higher concentrations and functions as a monomer. Thus DNA polymerase I is able to catalyze the removal of RNA primers from the vast number of Okazaki fragments formed during the discontinuous replication of the lagging strand.

10.25 The dnaA gene product of E. coli is required for the initiation of DNA synthesis at oriC. What is its function? How do we know that the DnaA protein is essential to the initiation process?

ANS: DnaA protein initiates the formation of the replication bubble by binding to the 9-bp repeats of OriC. DnaA protein is known to be required for the initiation process because bacteria with temperature-sensitive mutations in the dnaA gene cannot initiate DNA replication at restrictive temperatures.

8.16 What are the basic differences between generalized transduction and specialized transduction?

ANS: Generalized transduction: (1) transducting particles often contain only host DNA; (2) transducing particles may carry any segment of the host chromosome. Thus, all host genes are transduced. Specialized transduction: (1) transducing particles carry a recombinant chromosome, which contains both phage DNA and host DNA; (2) only host genes that are adjacent to the prophage integration site are transduced.

8.6 How are genetic crosses performed with viruses? In what ways do these crosses differ from crosses in eukaryotes such as Drosophila?

ANS: Genetic crosses between viruses are performed by simultaneously infecting host cells with viruses of different genotypes and examining the progeny viruses for recombinant genotypes, rather than by mating two eukaryotic organisms, such as Drosophila. Recombination occurs between viral chromosomes in infected cells; this recombination occurs within a population of replicating viral chromosomes rather than between the paired homologous chromosomes during meiosis in eukaryotes.

8.17 What roles do IS elements play in the integration of F factors?

ANS: IS elements (or insertion sequences) are short (800-1400 nucleotide pairs) DNA sequences that are transposable—that is, capable of moving from one position in a chromosome to another position or from one chromosome to another chromosome. IS elements mediate recombination between nonhomologous DNA molecules—for example, between F factors and bacterial chromosomes.

10.5 How might continuous and discontinuous modes of DNA replication be distinguished experimentally?

ANS: If nascent DNA is labeled by exposure to 3H-thymidine for very short periods of time, continuous replication predicts that the label would be incorporated into chromosome-sized DNA molecules, whereas discontinuous replication predicts that the label would first appear in small pieces of nascent DNA (prior to covalent joining, catalyzed by polynucleotide ligase).

10.17 Why must each of the giant DNA molecules in eukaryotic chromosomes contain multiple origins of replication?

ANS: In eukaryotes, the rate of DNA synthesis at each replication fork is about 2500 to 3000 nucleotide pairs per minute. Large eukaryotic chromosomes often contain 107 to 108 nucleotide pairs. A single replication fork could not replicate the giant DNA in one of these large chromosomes fast enough to permit the observed cell generation times.

9.22 A diploid nucleus of Drosophila melanogaster contains about 3.4 × 108 nucleotide pairs. Assume (1) that all the nuclear DNA is packaged in nucleosomes and (2) that an average internucleosome linker size is 60 nucleotide pairs. How many nucleosomes would be present in a diploid nucleus of D. melanogaster? How many molecules of histone H2a, H2b, H3, and H4 would be required?

ANS: In the diploid nucleus of D. melanogaster, 1.65 106 nucleosomes would be present; these would contain 3.3 106 molecules of each histone, H2a, H2b, H3, and H4.

9.25 Are eukaryotic chromosomes metabolically most active during prophase, metaphase, anaphase, telophase, or interphase?

ANS: Interphase. Chromosomes are for the most part metabolically inactive (exhibiting little transcription) during the various stages of condensation in mitosis and meiosis.

9.24 Experimental evidence indicates that most highly repetitive DNA sequences in the chromosomes of eukaryotes do not produce any RNA or protein products. What does this indicate about the function of highly repetitive DNA?

ANS: It indicates that most highly repetitive DNA sequences do not contain structural genes specifying RNA and polypeptide gene products.

8.7 Bacteriophage T4 has a linear chromosome but a circular genetic map. How can the two structures be reconciled?

ANS: Linkage maps are based on population averages. With circularly permuted chromosomes, genes that are close together on some chromosomes are at opposite ends of other chromosomes. At the population level, these circularly permuted chromosomes yield a circular genetic map.

10.19 Other polA mutants of E. coli lack the 3 5 exonuclease activity of DNA polymerase I. Will the rate of DNA synthesis be altered in these mutants? What effect(s) will these polA mutations have on the phenotype of the organism?

ANS: No, the rate of DNA synthesis will not be altered. E. coli strains carrying polA mutations that eliminate the 3′ → 5′ exonuclease activity of DNA polymerase I will exhibit unusually high mutation rates.

9.13 RNA was extracted from TMV (tobacco mosaic virus) particles and found to contain 20 percent cytosine (20 percent of the bases were cytosine). With this information, is it possible to predict what percentage of the bases in TMV are adenine? If so, what percentage? If not, why not?

ANS: No. TMV RNA is single-stranded. Thus the base-pair stoichiometry of DNA does not apply.

9.29 Are the scaffolds of eukaryotic chromosomes composed of histone or nonhistone chromosomal proteins? How has this been determined experimentally?

ANS: Nonhistone chromosomal proteins. The "scaffold" structures of metaphase chromosomes can be observed by light microscopy after removal of the histones by differential extraction procedures.

10.27 The chromosomal DNA of eukaryotes is packaged into nucleosomes during the S phase of the cell cycle. What obstacles do the size and complexity of both the replisome and the nucleosome present during the semiconservative replication of eukaryotic DNA? How might these obstacles be overcome?

ANS: Nucleosomes and replisomes are both large macromolecular structures, and the packaging of eukaryotic DNA into nucleosomes raises the question of how a replisome can move past a nucleosome and replicate the DNA in the nucleosome in the process. The most obvious solution to this problem would be to completely or partially disassemble the nucleosome to allow the replisome to pass. The nucleosome would then reassemble after the replisome had passed. One popular model has the nucleosome partially disassembling, allowing the replisome to move past it (see Figure 10.31b).

10.3 A culture of bacteria is grown for many generations in a medium in which the only available nitrogen is the heavy isotope (15N). The culture is then switched to a medium containing only 14N for one generation of growth; it is then returned to a 15N-containing medium for one final generation of growth. If the DNA from these bacteria is isolated and centrifuged to equilibrium in a CsCl density gradient, how would you predict the DNA to band in the gradient?

ANS: One-half of the DNA molecules fully heavy (15N in both strands); the other half of the molecules "hybrid" (15N in one strand, 14N in the complementary strand).

8.13 Assume that you have just demonstrated genetic recombination (e.g., when a strain of genotype a b+ is present with a strain of genotype a+ b, some recombinant genotypes, a+ b+ and a b, are formed) in a previously unstudied species of bacteria. How would you determine whether the observed recombination resulted from transformation, conjugation, or transduction?

ANS: Perform two experiments: (1) determine whether the process is sensitive to DNase, and (2) determine whether cell contact is required for the process to take place. The cell contact requirement can be tested by a U-tube experiment (see Fig. 8.17). If the process is sensitive to DNase, it is similar to transformation. If cell contact is required, it is similar to conjugation. If it is neither sensitive to DNase nor requires cell contact, it is similar to transduction.

11.5 At what locations in a eukaryotic cell does protein synthesis occur?

ANS: Protein synthesis occurs on ribosomes. In eukaryotes, most of the ribosomes are located in the cytoplasm and are attached to the extensive membranous network of endoplasmic reticulum. Some protein synthesis also occurs in cytoplasmic organelles such as chloroplasts and mitochondria.

9.3 How could it be demonstrated that the mixing of heat-killed Type III pneumococcus with live Type II resulted in a transfer of genetic material from Type III to Type II rather than a restoration of viability to Type III by Type II?

ANS: Purified DNA from Type III cells was shown to be sufficient to transform Type II cells. This occurred in the absence of any dead Type III cells.

10.21 (a) Why isn't DNA primase activity required to initiate rolling-circle replication? (b) DNA primase is required for the discontinuous synthesis of the lagging strand, which occurs on the single-stranded tail of the rolling circle. Why?

ANS: Rolling-circle replication begins when an endonuclease cleaves one strand of a circular DNA double helix. This cleavage produces a free 3′-OH on one end of the cut strand, allowing it to function as a primer. The discontinuous synthesis of the lagging strand requires the de novo initiation of each Okazaki fragment, which requires DNA primase activity.

11.10. What role(s) do spliceosomes play in pathways of gene expression? What is their macromolecular structure?

ANS: Spliceosomes excise intron sequences from nuclear gene transcripts to produce the mature mRNA molecules that are translated on ribosomes in the cytoplasm. Spliceosomes are complex macromolecular structures composed of snRNA and protein molecules (see Figure 11.28).

10.13 What experimental techniques can be used to separate DNA molecules of mass 3 × 107 daltons isolated from lambda bacteriophage and DNA molecules of mass 1.3 × 108 daltons isolated from T2 bacteriophage?

ANS: Sucrose velocity density gradient centrifugation is the standard technique for separating DNA molecules in this size range. Pulsed-field gel electrophoresis (Chapter 9) could also be used.

10.9 Suppose that the DNA of cells (growing in a cell culture) in a eukaryotic species was labeled for a short period of time by the addition of 3H-thymidine to the medium. Next assume that the label was removed and the cells were resuspended in nonradioactive medium. After a short period of growth in nonradioactive medium, the DNA was extracted from these cells, diluted, gently layered on filters, and autoradiographed. If autoradiographs of the type were observed, what would this indicate about the nature of DNA replication in these cells? Why?

ANS: That DNA replication was unidirectional rather than bidirectional. As the intracellular pools of radioactive 3H-thymidine are gradually diluted after transfer to nonradioactive medium, less and less 3H-thymidine will be incorporated into DNA at each replicating fork. This will produce autoradiograms with tails of decreasing grain density at each growing point. Since such tails appear at only one end of each track, replication must be unidirectional. Bidirectional replication would produce such tails at both ends of an autoradiographic track (see Figure 10.29).

10.18 In E. coli, viable polA mutants have been isolated that produce a defective gene product with little or no 5 3 polymerase activity, but normal 5 3 exonuclease activity. However, no polA mutant has been identified that is completely deficient in the 5 3 exonuclease activity, while retaining 5 3 polymerase activity, of DNA polymerase I. How can these results be explained?

ANS: The 5 3 exonuclease activity of DNA polymerase I is essential to the survival of the bacterium, whereas the 5 3 polymerase activity of the enzyme is not essential.

9.17 Compare and contrast the structures of the A, B, and Z forms of DNA.

ANS: The B form of DNA helix is that proposed by Watson and Crick and is the conformation that DNA takes under physiological conditions. It is a right-handed double helical coil with 10 bases per turn of the helix and a diameter of 1.9 nm. It has a major and a minor groove. Z-DNA is left-handed, has 12 bases per turn, a single deep groove, and is 1.8 nm in diameter. Its sugar-phosphate backbone takes a zigzagged path, and it is G:C rich. A-DNA is a right-handed helix with 11 base pairs per turn. It is a shorter, thicker double helix with a diameter of 0.23 nm and has a narrow, deep major groove and a broad, shallow minor groove. A-DNA forms in vitro under high salt concentrations or in a partially dehydrated state.

8.11 You have identified three mutations —a, b, and c—in Streptococcus pneumoniae. All three are recessive to their wild-type alleles a+, b+, and c+. You prepare DNA from a wild-type donor strain and use it to transform a strain with genotype a b c. You observe a+b+ transformants and a+c+ transformants, but no b+c+ transformants. Are these mutations closely linked? If so, what is their order on the Streptococcus chromosome?

ANS: The a, b, and c mutations are closely linked and in the order b — a — c on the chromosome.

10.31 In the yeast S. cerevisiae, haploid cells carrying a mutation called est1 (for ever-shorter telomeres) lose distal telomere sequences during each cell division. Predict the ultimate phenotypic effect of this mutation on the progeny of these cells.

ANS: The chromosomes of haploid yeast cells that carry the est1 mutation become shorter during each cell division. Eventually, chromosome instability results from the complete loss of telomeres, and cell death occurs because of the deletion of essential genes near the ends of chromosomes.

10.10 Arrange the following enzymes in the order of their action during DNA replication in E. coli: (1) DNA polymerase I, (2) DNA polymerase III, (3) DNA primase, (4) DNA gyrase, and (5) DNA helicase.

ANS: The correct sequence of action is 4, 5, 3, 2, 1.

11.8. Many eukaryotic genes contain noncoding introns that separate the coding sequences or exons of these genes. At what stage during the expression of these split genes are the noncoding intron sequences removed?

ANS: The entire nucleotide-pair sequences—including the introns—of the genes are transcribed by RNA polymerase to produce primary transcripts that still contain the intron sequences. The intron sequences are then spliced out of the primary transcripts to produce the mature, functional RNA molecules. In the case of protein-encoding nuclear genes of higher eukaryotes, the introns are spliced out by complex macromolecular structures called spliceosomes (see Figure 11.27).

11.4. On the basis of what evidence was the messenger RNA hypothesis established?

ANS: The genetic information of cells is stored in DNA, which is located predominantly in the chromosomes. The gene products (polypeptides) are synthesized primarily in the cytoplasm on ribosomes. Some intermediate must therefore carry the genetic information from the chromosomes to the ribosomes. RNA molecules (mRNAs) were shown to perform this function by means of RNA pulse-labeling and pulse chase experiments combined with autoradiography (see Figure 11.6). The enzyme RNA polymerase was subsequently shown to catalyze the synthesis of mRNA using chromosomal DNA as a template. Finally, the mRNA molecules synthesized by RNA polymerase were shown to faithfully direct the synthesis of specific polypeptides when used in in vitro protein synthesis systems.

8.5 In what way does the integration of the chromosome into the host chromosome during a lysogenic infection differ from crossing over between homologous chromosomes?

ANS: The insertion of the phage chromosome into the host chromosome is a site-specific recombination process catalyzed by an enzyme that recognizes specific sequences in the and E. coli chromosomes. Crossing over between homologous chromosomes is not sequence-specific. It can occur at many sites along the two chromosomes.

8.9 The wild-type phage T4 chromosome is terminally redundant; that is, it contains the same genes at both ends. In the case of wild-type T4, this terminal redundancy includes about 3% of the entire chromosome. Some mutant phage are missing the entire rII locus (see Chapter 14), which is over 3,000 nucleotide pairs in length and represents almost 2% of the chromosome. Will these rII mutations have any effect on the terminally redundant regions of the chromosomes of the phage? If so, what effect?

ANS: The length of the terminal redundancy will be larger in rII deletion mutants. If 2 percent of the chromosome has been deleted, the length of the terminally redundant region will be 5 percent (3 percent in wild-type phage plus 2 percent as a result of the deletion).

8.8 If a single T4 phage—with its linear chromosome—infects an E. coli cell and goes through the lytic cycle, the chromosomes of the progeny phage will be circularly permuted with different terminally redundant ends. How are these circularly permuted, terminally redundant molecules produced?

ANS: The linear chromosome replicates to produce two or more linear molecules, which undergo recombination within the terminally redundant regions to produce long DNA molecules—concatamers—many chromosomes in length (see Figure 8.12). When the concatameric DNA is packaged into phage heads, each head holds one complete chromosome plus a little bit extra (the terminally redundant region). Therefore, as each head is filled and the DNA cleaved, a population of circularly permutated and terminally redundant chromosomes is produced (see Figure 8.13).

9.31 When total DNA from the kangaroo rat (Dipodomys ordii) is analyzed by equilibium density-gradient centrifugation, a fraction of the DNA forms a distinct satellite band in the gradient. This satellite DNA contains a highly repeated DNA sequence that can be isolated from the gradient free of other DNA sequences. Many such highly repetitive DNA sequences are located in regions of heterochromatin adjacent to the centromeres of chromosomes ("centromeric heterochromatin"). How could a researcher determine where the satellite DNA sequence is located in the chromosomes of the kangaroo rat?

ANS: The locations of the satellite DNA sequence in the chromosomes can be determined by in situ hybridization (see Focus on In Situ Hybridization).

8.4 How does the structure of the prophage differ from the structure of the chromosome packaged in the virion?

ANS: The mature (packaged) lambda chromosome and the lambda prophage are circular permutations of one another (see Fig. 8.6).

10.26 What is a primosome, and what are its functions? What essential enzymes are present in the primosome? What are the major components of the E. coli replisome? How can geneticists determine whether these components are required for DNA replication?

ANS: The primosome is a protein complex that initiates the synthesis of Okazaki fragments during lagging strand synthesis. The major components of the E. coli DNA primosome are DNA primase and DNA helicase. Geneticists have been able to show that both DNA primase and DNA helicase are required for DNA replication by demonstrating that mutations in the genes encoding these enzymes result in the arrest of DNA synthesis in mutant cells under conditions where the altered proteins are inactive.

10.28 Two mutant strains of E. coli each have a temperature-sensitive mutation in a gene that encodes a product required for chromosome duplication. Both strains replicate their DNA and divide normally at 25°C, but are unable to replicate their DNA or divide at 42°C. When cells of one strain are shifted from growth at 25°C to growth at 42°C, DNA synthesis stops immediately. When cells of the other strain are subjected to the same temperature shift, DNA synthesis continues, albeit at a decreasing rate, for about a half hour. What can you conclude about the functions of the products of these two genes?

ANS: The product of the first gene is required for DNA chain extension, whereas the product of the second gene is only required for the initiation of DNA synthesis.

9.23 The satellite DNAs of Drosophila virilis can be isolated, essentially free of main-band DNA, by density-gradient centrifugation. If these satellite DNAs are sheared into approximately 40-nucleotide-pair-long fragments and are analyzed in denaturation-renaturation experiments, how would you expect their hybridization kinetics to compare with the renaturation kinetics observed using similarly sheared main-band DNA under the same conditions? Why?

ANS: The satellite DNA fragments would renature much more rapidly than the main-band DNA fragments. In D. virilus satellite DNAs, all three have repeating heptanucleotide-pair sequences. Thus essentially every 40 nucleotide-long (average) single-stranded fragment from one strand will have a sequence complementary (in part) with every single-stranded fragment from the complementary strand. Many of the nucleotide-pair sequences in main-band DNA will be unique sequences (present only once in the genome).

9.18 The temperature at which one-half of a double-stranded DNA molecule has been denatured is called the melting temperature, Tm. Why does Tm depend directly on the GC content of the DNA?

ANS: The value of Tm increases with the GC content because GC base pairs, connected by three hydrogen bonds, are stronger than AT base pairs connected by two hydrogen bonds.

8.12 A nutritionally defective E. coli strain grows only on a medium containing thymine, whereas another nutritionally defective strain grows only on medium containing leucine. When these two strains were grown together, a few progeny were able to grow on a minimal medium with neither thymine or leucine. How can this result be explained?

ANS: There are two possible explanations. One possibility is that a spontaneous mutation caused reversion of either auxotrophic strain to the prototrophic condition. Because this requires only one mutation in one cell, this is a possibility, although rare. Another, more likely, possibility is that conjugation occurred between the E. coli parental auxotrophic strains. During conjugation, genes from the parental strains recombined. Because each parent had a wildtype gene copy for either thymine or leucine, recombinant progeny containing the wildtype copy of each gene would be able to synthesize both nutrients and grow on minimal medium.

8.10 Geneticists have used mutations that cause altered phenotypes such as white eyes in Drosophila, white flowers and wrinkled seeds in peas, and altered coat color in rabbits to determine the locations of genes on the chromosomes of these eukaryotes. What kinds of mutant phenotypes have been used map genes in bacteria?

ANS: Three main types of bacterial mutants have been used to map genes in bacteria; these include: mutants unable to utilize specific sugars as energy sources (such as lactose), mutants unable to synthesize essential metabolites (these are called auxotrophs), and mutants resistant to drugs and antibiotics. Whereas wild-type bacteria can use almost any sugar as an energy source, can grow on minimal media, and are killed by antibiotics, mutants in genes controlling these processes result in different growth characteristics. These growth phenotypes can be used to map genes in bacteria.

8.1. By what criteria are viruses living? nonliving?

ANS: Viruses reproduce and transmit their genes to progeny viruses. They utilize energy provided by host cells and respond to environmental and cellular signals like other living organisms. However, viruses are obligate parasites; they can reproduce only in appropriate host cells.

9.27 Of what special interest is the biophysical technique of viscoelastometry to geneticists?

ANS: Viscoelastometry is a procedure used to measure the viscosity of molecules in solution. In addition, viscoelastometric methods can be used to estimate the sizes of the largest DNA molecules present in aqueous solutions. By using viscoelastometry, scientists have obtained evidence which indicates that all of the DNA present in chromosomes of eukaryotes exists as giant, "chromosome-size" DNA molecules (one huge DNA molecule per chromosome). These data eliminated early models of chromosome structure with multiple DNA molecules joined end-to-end by protein or RNA "linkers."

9.6 How did the reconstitution experiment of Fraenkel-Conrat and colleagues show that the genetic information of tobacco mosaic virus (TMV) is stored in its RNA rather than its protein?

ANS: When tobacco leaves were infected with reconstituted virus particles containing RNA from type A viruses and protein from type B viruses, the progeny viruses were type A, showing that RNA, not protein, carries the genetic information in TMV.

10.32 Assume that the sequence of a double-stranded DNA shown below is present at one end of a large DNA molecule in a eukaryotic chromosome. 5′ (centromere sequence) -gattccccgggaagcttggggggcccatcttcgtacgtctttgca-3′ 3′-(centromere sequence) -ctaaggggcccttcgaaccccccgggtagaagcatgcagaaacgt-5′ You have reconstituted a eukaryotic replisome that is active in vitro. However, it lacks telomerase activity. If you isolate the DNA molecule shown above and replicate it in your in vitro system, what products would you expect?

ANS: Without telomerase, the 5′ end of the newly replicated strand will be missing some bases. The exact number of missing bases doesn't matter: 5′-GATTCCCCGGGAAGCTTGGGGGGCCCATCT T CGT ACGTCTTTGCA-3′ 3′-CTAAGGGGCCCTTCGAACCCCCCGCGGGTAGAAGCAT G-5′ 5′-AAGCTT GGGGGGCCCATCT T CGT ACGTCT T T G CA-3′ 3′-CTAAGGGGCCCTTCGAA CCCCCCGGGTAGAAGCATG CAGAAACG T-5′

9.12 DNA was extracted from cells of Staphylococcus afermentans and analyzed for base composition. It was found that 37 percent of the bases are cytosine. With this information, is it possible to predict what percentage of the bases are adenine? If so, what percentage? If not, why not?

ANS: Yes. Because DNA in bacteria is double-stranded, the 1:1 base-pair stoichiometry applies. Therefore, if 37% of the bases are cytosine, then 37% are guanine. This means that the remaining 26% of the bases are adenine and thymine. Thus, 26%/2 = 13% of the bases are adenine.

To determine the location of the his- mutation on the E. coli chromosome, you perform interrupted mating experiments with 5 different Hfr strains. The following chart shows the time of entry (minutes, in parentheses) of the wild-type alleles of the first 5 markers (mutant genes) into the His- strain. (BOOK)

Look at answer sheet

An F+ strain, marked at 10 loci, gives rise spontaneously to Hfr progeny whenever the F factor becomes incorporated into the chromosome of the F+ strain. The F factor can integrate into the circular chromosome at many points, so that the resulting Hfr strains transfer the genetic markers in different orders. For any Hfr strain, the order of markers entering a recipient cell can be determined by interrupted mating experiments. From the following data for several Hfr strains derived from the same F+, determine the order of markers in the F+ strain. (BOOK)

Z at top of circle--> to the right--> HERSKOWITZ

8.26 Mutations nrd 11 (gene nrd B, encoding the beta subunit of the enzyme ribonucleotide reductase), am M69 (gene 63, encoding a protein that aids tail-fiber attachment), and nd 28 (denA, encoding the enzyme endonuclease II) are known to be located between gene 31 and gene 32 on the bacteriophage T4 chromosome. Mutations am N54 and am A453 are located in genes 31 and 32, respectively. Given the three-factor cross data in the following table, what is the linear order of the five mutant sites? (BOOK)

amA453—nrd11—nd¬28—amM69—amN54.

8.22 The data in the following table were obtained from three-point transduction tests made to determine the order of mutant sites in the A gene encoding the subunit of tryptophan synthetase in E. coli. Anth is a linked, unselected marker. In each cross, trp+ recombinants were selected and then scored for the anth marker (anth+ or anth). What is the linear order of anth and the three mutant alleles of the A gene indicated by the data in the table? (BOOK)

anth—A34—A223—A46.

8.24 Two additional mutations in the trp A gene of E. coli, trp A58 and trp A487, were ordered relative to trp A223 and the outside marker anth by three-factor transduction crosses as described in Problem 8.22. The results of these crosses are summarized in the following table. What is the linear order of anth and the three mutant sites in the trp A gene? (BOOK)

anth—A487—A223—A58.

In E. coli, the ability to utilize lactose as a carbon source requires the presence of the enzymes -galactosidase and -galactoside permease. These enzymes are encoded by two closely linked genes, lacZ and lacY, respectively. Another gene, proC, controls, in part, the ability of E. coli cells to synthesize the amino acid proline. The alleles strr and strs, respectively, control resistance and sensitivity to streptomycin. Hfr H is known to transfer the two lac genes, proC, and str, in that order, during conjugation. A cross was made between Hfr H of genotype lacZ lacY+ proC+ strs and an F strain of genotype lacZ+ lacY proC strr. After about 2 hours, the mixture was diluted and plated out on medium containing streptomycin but no proline. When the resulting proC+ strr recombinant colonies were checked for their ability to grow on medium containing lactose as the sole carbon source, very few of them were capable of fermenting lactose. When the reciprocal cross (Hfr H lacZ+ lacY proC+ strs X F lacZ lacY+ proC strr) was done, many of the proC+ strr recombinants were able to grow on medium containing lactose as the sole carbon source. What is the order of the lacZ and lacY genes relative to proC?

lacY—lacZ—proC.

8.23 Bacteriophage P1 mediates generalized transduction in E. coli. A P1 transducing lysate was prepared by growing P1 phage on pur+ pro- his- bacteria. Genes pur, pro, and his encode enzymes required for the synthesis or purines, proline, and histidine, respectively. The phage and transducing particles in this lysate were then allowed to infect pur- pro+ his+ cells. After incubating the infected bacteria for a period of time sufficient to allow transduction to occur, they were plated on minimal medium supplemented with proline and histidine, but no purines to select for pur+ transductants. The pur+ colonies were then transferred to minimal medium with and without proline and with and without histine to determine the frequencies of each of the outside markers. Given the following results, what is the order of the three genes on the E. coli chromosome? (BOOK)

pro — pur — his.