Nucleic Acids HW

Describe three properties common to the reactions catalyzed by DNA polymerase, RNA polymerase, reverse transcriptase, and RNA replicase. How is the enzyme polynucleotide phosphorylase similar to and different from these four enzymes?

(1) Use of a template strand of nucleic acid; (2) synthesis in the 5'-3' direction; (3) use of nucleoside triphosphate substrates, with formation of a phosphodiester bond and displacement of PPi. Polynucleotide phosphorylase forms phosphodiester bonds but differs in all other listed properties.

One strand of a double-helical DNA has the sequence (5')GCGCAATATTTCTCAAAATATTGCGC(3'). Write the base sequence of the complementary strand. What special type of sequence is contained in this DNA segment? Does the double-stranded DNA have the potential to form any alternative structures?

(59)GCGCAATATTTTGAGAAATATTGCGC(39); it contains a palindrome. The individual strands can form hairpin structures; the two strands can form a cruciform.

Calculate the weight in grams of a double-helical DNA molecule stretching from the Earth to the moon (~320,000 km). The DNA double helix weighs about 1 x 1018 g per 1,000 nucleotide pairs; each base pair extends 3.4 Å. For an interesting comparison, your body contains about 0.5 g of DNA!

9.4x10^-4 g

The template strand of a segment of double-helical DNA contains the sequence (5')CTTAACACCCCTGACTTCGCGCCGTCG(3') (a) What is the base sequence of the mRNA that can be transcribed from this strand? (b) What amino acid sequence could be coded by the mRNA in (a), starting from the 5' end? (c) If the complementary (nontemplate) strand of this DNA were transcribed and translated, would the resulting amino acid sequence be the same as in (b)? Explain the biological significance of your answer.

(a) (5')CGACGGCGCGAAGUCAGGGGUGUUAAG(3') (b) Arg-Arg-Arg-Glu-Val-Arg-Gly-Val-Lys (c) No. The complementary antiparallel strands in double- helical DNA do not have the same base sequence in the 5'-3' direction. RNA is transcribed from only one specific strand of duplex DNA. The RNA polymerase must therefore recognize and bind to the correct strand.

Human pancreatic ribonuclease has 128 amino acid residues. (a) What is the minimum number of nucleotide pairs required to code for this protein? (b) The mRNA expressed in human pancreatic cells was copied with reverse transcriptase to create a "library" of human DNA. The sequence of the mRNA coding for human pancreatic ribonuclease was determined by sequencing the complementary DNA (cDNA) from this library that included an open reading frame for the protein. Use the Entrez database system (www.ncbi.nlm.nih.gov/Entrez) to find the published sequence of this mRNA (search the nucleotide database for accession number D26129). What is the length of this mRNA? (c) How can you account for the discrepancy between the size you calculated in (a) and the actual length of the mRNA?

(a) 384 nucleotide pairs (b) 1,620 nucleotide pairs (c) Most of the nucleotides are untranslated regions at the 3' and 5' ends of the mRNA. Also, most mRNAs code for a signal sequence (Chapter 27) in their protein products, which is eventually cleaved off to produce the mature and functional protein.

Replication of the E. coli Chromosome The E. coli chromosome contains 4,639,221 bp. (a) How many turns of the double helix must be unwound during replication of the E. coli chromosome? (b) From the data in this chapter, how long would it take to replicate the E. coli chromosome at 37 _C if two replication forks proceeded from the origin? Assume replication occurs at a rate of 1,000 bp/s. Under some conditions E. coli cells can divide every 20 min. How might this be possible? (c) In the replication of the E. coli chromosome, about how many Okazaki fragments would be formed? What factors guarantee that the numerous Okazaki fragments are assembled in the correct order in the new DNA?

(a) 4.42 10^5 turns (b) ~40 min. In cells dividing every 20 min, a replicative cycle is initiated every 20 min, each cycle beginning before the prior one is complete. (c) ~2,000 to ~5,000 Okazaki fragments. The fragments are 1,000 to 2,000 nucleotides long and are firmly bound to the template strand by base pairing. Each fragment is quickly joined to the lagging strand, thus preserving the correct order of the fragments. Answer C relies on the use of 5'-3' exonuclease activity (nick translation) and ligation of the nick after the RNA primer is removed, thus the joining.

(a) How long would it take for the E. coli RNA polymerase to synthesize the primary transcript for the E. coli genes encoding the enzymes for lactose metabolism (the 5,300 bp lac operon, considered in Chapter 28)? (b) How far along the DNA would the transcription "bubble" formed by RNA polymerase move in 10 seconds?

(a) 60 to 100 s (b) 500 to 900 nucleotides

Describe the probable effects on gene expression in the lac operon of a mutation in (a) the lac operator that deletes most of O1; (b) the lacI gene that inactivates the repressor; and (c) the promoter that alters the region around position _10.

(a) Constitutive, low-level expression of the operon; most mutations in the operator would make the repressor less likely to bind. (b) Either constitutive expression, as in (a), or constant repression, if the mutation destroyed the capability to bind to lactose and related compounds and hence the response to inducers. (c) Either increased or decreased expression of the operon (under conditions in which it is induced), depending on whether the mutation made the promoter more or less similar, respectively, to the consensus E. coli promoter.

To examine the roles of hydrogen bonds and hydrophobic interactions between transcription factors and DNA, go to FirstGlance in Jmol at http://firstglance.jmol.org and enter the PDB ID 1TGH. This file models the interactions between a human TATA-binding protein and a segment of double-stranded DNA. Once the structure loads, click the "Spin" button to stop the molecule from rotating. When the molecule has reloaded, click the "Contacts" link. With the radio button for "Chains" selected, click on any part of the protein (Chain A, displays in blue) to select it as the target. Click "Show Atoms Contacting Target" and, in the list of contacts to display, check only "Show putatively hydrogen-bonded non-water" to display hydrogen bonds between the protein and the TATA box DNA. Then click on the rightmost option for viewing images (Maximum detail: Target & Contacts Balls and Sticks, Colored by Element). With this view you should be able to use the zoom and rotate controls and mouse clicks to answer the following questions. (a) Which of the base pairs in the DNA form hydrogen bonds with the protein? Which of these contribute to the specific recognition of the TATA box by this protein? (Hydrogen-bond length between hydrogen donor and hydrogen acceptor ranges from 2.5 to 3.3 Å.) (b) Which amino acid residues in the protein interact with these base pairs? (c) What is the sequence of the DNA in this model and which portions of the sequence are recognized by the TATA-binding protein? (d) Examine the hydrophobic interactions in this complex. Are they rare or numerous? To answer this question, click on "Return to contacts" and check the option to "Show hydrophobic (apolar van der Waals) interactions."

(a) For 10% expression (90% repression), 10% of the repressor has bound inducer and 90% is free and available to bind the operator. The calculation uses Eqn 5-8 (p. 161), with 5 0.1 andKd5104M. [IPTG] [IPTG] 5[IPTG]1K 5[IPTG]11024M 0.1 5 [IPTG] so 0.9 [IPTG] 5 105 or [IPTG] 5 1.1 3 105 M [IPTG] 1 1024 M For 90% expression, 90% of the repressor has bound inducer, so 5 0.9. Entering the values for and Kd in Eqn 5-8 gives [IPTG] 5 9 3 104 M. Thus, gene expression varies 10-fold over a roughly 10-fold [IPTG] range. (b) You would expect the protein levels to be low before induction, rise during induction, and then decay as synthesis stops and the proteins are degraded. (c) As shown in (a), the lac operon has more levels of expression than just on or off; thus it does not have characteristic A. As shown in (b), expression of the lac operon subsides once the inducer is removed; thus it lacks characteristic B. (d) GFP-on: repts and GFP are expressed at high levels; repts represses OP, so no LacI protein is produced. GFP-off: LacI is expressed at a high level; LacI represses OPlac, so repts and GFP are not produced. (e) IPTG treatment switches the system from GFP-off to GFP-on. IPTG has an effect only when LacI is present, so affects only the GFP-off state. Adding IPTG relieves the repression of OPlac, allowing high-level expression of repts, which turns off expression of LacI, and high-level expression of GFP. (f) Heat treatment switches the system from GFP-on to GFP-off. Heat has an effect only when repts is present, so affects only the GFP-on state. Heat inactivates repts and relieves the repression of OP, allowing high- level expression of LacI. LacI then acts at OPlac to repress synthesis of repts and GFP. (g) Characteristic A: The system is not stable in the intermediate state. At some point, one repressor will act more strongly than the other due to chance fluctuations in expression; this shuts off expression of the other repressor and locks the system in one state. Characteristic B: Once one repressor is expressed, it prevents the synthesis of the other; thus the system remains in one state even after the switching stimulus has been removed. (h) At no time does any cell express an intermediate level of GFP—this is a confirmation of characteristic A. At the intermediate concentration (X) of inducer, some cells have switched to GFP-on while others have not yet made the switch and remain in the GFP-off state; none are in between. The bimodal distribution of expression levels at [IPTG] 5 X is caused by the mixed population of GFP-on and GFP-off cells.

Predict the amino acid sequences of peptides formed by ribosomes in response to the following mRNA sequences, assuming that the reading frame begins with the first three bases in each sequence. (a) GGUCAGUCGCUCCUGAUU (b) UUGGAUGCGCCAUAAUUUGCU (c) CAUGAUGCCUGUUGCUAC (d) AUGGACGAA

(a) Gly-Gln-Ser-Leu-Leu-Ile (b) Leu-Asp-Ala-Pro (c) His-Asp-Ala-Cys-Cys-Tyr (d) Met-Asp-Glu in eukaryotes; fMet-Asp-Glu in bacteria

DNA Replication Kornberg and his colleagues incubated soluble extracts of E. coli with a mixture of dATP, dTTP, dGTP, and dCTP, all labeled with 32P in the a-phosphate group. After a time, the incubation mixture was treated with trichloroacetic acid, which precipitates the DNA but not the nucleotide precursors. The precipitate was collected, and the extent of precursor incorporation into DNA was determined from the amount of radioactivity present in the precipitate. (a) If any one of the four nucleotide precursors were omitted from the incubation mixture, would radioactivity be found in the precipitate? Explain. (b) Would 32P be incorporated into the DNA if only dTTP were labeled? Explain. (c) Would radioactivity be found in the precipitate if 32P labeled the b or g phosphate rather than the a phosphate of the deoxyribonucleotides? Explain.

(a) No. Incorporation of 32P into DNA results from the synthesis of new DNA, which requires the presence of all four nucleotide precursors. (b) Yes. Although all four nucleotide precursors must be present for DNA synthesis, only one of them has to be radioactive in order for radioactivity to appear in the new DNA. (c) No. Radioactivity is incorporated only if the 32P label is in the phosphate; DNA polymerase cleaves off pyrophosphate—i.e., the - and -phosphate groups.

Answer

(a) Points Y. (b) Points X.

Elucidation of the three-dimensional structure of DNA helped researchers understand how this molecule conveys information that can be faithfully replicated from one generation to the next. To see the secondary structure of double-stranded DNA, go to the Protein Data Bank website (www.pdb.org). Use the PDB identifiers listed below to retrieve the structure summaries for the two forms of DNA. Open the structures using Jmol, and use the controls in the Jmol menu (accessed with a control-click or by clicking on the Jmol logo in the lower right corner of the image screen) to complete the following exercises. Refer to the Jmol help links as needed. (a) Obtain the file for 141D, a highly conserved, repeated DNA sequence from the end of the HIV-1 (the virus that causes AIDS) genome. Display the molecule as a ball-and-stick structure and color by element. Identify the sugar-phosphate backbone for each strand of the DNA duplex. Locate and identify individual bases. Identify the 5' end of each strand. Locate the major and minor grooves. Is this a right- or left-handed helix?(b) Obtain the file for 145D, a DNA with the Z conformation. Display the molecule as a ball-and-stick structure and color by element. Identify the sugar-phosphate backbone for each strand of the DNA duplex. Is this a right- or left-handed helix? (c) To fully appreciate the secondary structure of DNA, view the molecules in stereo. On the control menu, Select > All, then Style > Stereographic > Cross-eyed viewing or Wall-eyed viewing. You will see two images of the DNA molecule. Sit with your nose approximately 10 inches from the monitor and focus on the tip of your nose (cross-eyed) or the opposite edges of the screen (wall-eyed). In the background you should see three images of the DNA helix. Shift your focus to the middle image, which should appear three-dimensional. (Note that only one of the two authors can make this work.)

(a) Right-handed. The base at one 59 end is adenine; at the other 59 end, cytidine. (b) Left-handed (c) If you cannot see the structures in stereo, see additional tips in the expanded solutions manual, or use a search engine to find tips online.

Some aminoacyl-tRNA synthetases do not recognize and bind the anticodon of their cognate tRNAs but instead use other structural features of the tRNAs to impart binding specificity. The tRNAs for alanine apparently fall into this category. (a) What features of tRNAAla are recognized by Ala-tRNA synthetase? (b) Describe the consequences of a C G mutation in the third position of the anticodon of tRNAAla. (c) What other kinds of mutations might have similar effects? (d) Mutations of these types are never found in natural populations of organisms. Why? (Hint: Consider what might happen both to individual proteins and to the organism as a whole.)

(a) The Ala-tRNA synthetase recognizes the G3-U70 base pair in the amino acid arm of tRNAAla. (b) The mutant tRNAAla would insert Ala residues at codons encoding Pro. (c) A mutation that might have similar effects is an alteration in tRNAPro that allowed it to be recognized and aminoacylated by Ala-tRNA synthetase. (d) Most of the proteins in the cell would be inactivated, so these would be lethal mutations and hence never observed. This represents a powerful selective pressure for maintaining the genetic code.

Bacterial endospores form when the environment is no longer conducive to active cell metabolism. The soil bacterium Bacillus subtilis, for example, begins the process of sporulation when one or more nutrients are depleted. The end product is a small, metabolically dormant structure that can survive almost indefinitely with no detectable metabolism. Spores have mechanisms to prevent accumulation of potentially lethal mutations in their DNA over periods of dormancy that can exceed 1,000 years. B. subtilis spores are much more resistant than are the organism's growing cells to heat, UV radiation, and oxidizing agents, all of which promote mutations. (a) One factor that prevents potential DNA damage in spores is their greatly decreased water content. How would this affect some types of mutations? (b) Endospores have a category of proteins called small acid-soluble proteins (SASPs) that bind to their DNA, preventing formation of cyclobutane-type dimers. What causes cyclobutane dimers, and why do bacterial endospores need mechanisms to prevent their formation?

(a) Water is a participant in most biological reactions, including those that cause mutations. The low water content in endospores reduces the activity of mutation-causing enzymes and slows the rate of nonenzymatic depurination reactions, which are hydrolysis reactions. (b) UV light induces formation of cyclobutane pyrimidine dimers. Because B. subtilis is a soil organism, spores can be lofted to the top of the soil or into the air, where they may be subject to prolonged UV exposure.

Much important confirmatory evidence on the genetic code has come from assessing changes in the amino acid sequence of mutant proteins after a single base has been changed in the gene that encodes the protein. Which of the following amino acid replacements would be consistent with the genetic code if the replacements were caused by a single base change? Which cannot be the result of a single-base mutation? Why? (a) Phe Leu (b) Lys Ala (c) Ala Thr (d) Phe Lys

(a), (c), (e), and (g) only; (b), (d), and (f) cannot be the result of single-base mutations: (b) and (f) would require substitutions of two bases, and (d) would require substitutions of all three bases.

DNA unwinding, such as that occurring in replication, affects the superhelical density of DNA. In the absence of topoisomerases, the DNA would become overwound ahead of a replication fork as the DNA is unwound behind it. A bacterial replication fork will stall when the superhelical density (j) of the DNA ahead of the fork reaches +0.14 (see Chapter 24). Bidirectional replication is initiated at the origin of a 6,000 bp plasmid in vitro, in the absence of topoisomerases. The plasmid initially has a j of -0.06. How many base pairs will be unwound and replicated by each replication fork before the forks stall? Assume that each fork travels at the same rate and that each includes all components necessary for elongation except topoisomerase.

1,200 bp (600 in each direction) Super helical density = delta Lk/Initial Lk Delta Lk = Lkf-Lki Delta Lk = delta Twist + Delta Writhe Initial Lk = twist Twist = 6000bp/10.5bp per twist

A new RNA polymerase activity is discovered in crude extracts of cells derived from an exotic fungus. The RNA polymerase initiates transcription only from a single, highly specialized promoter. As the polymerase is purified, its activity declines, and the purified enzyme is completely inactive unless crude extract is added to the reaction mixture. Suggest an explanation for these observations.

A dissociable factor necessary for activity (e.g., a specificity factor similar to the s subunit of the E. coli enzyme) may have been lost during purification of the polymerase.

DNA polymerases are capable of editing and error correction, whereas the capacity for error correction in RNA polymerases seems to be quite limited. Given that a single base error in either replication or transcription can lead to an error in protein synthesis, suggest a possible biological explanation for this difference.

A single base error in DNA replication, if not corrected, would cause one of the two daughter cells, and all its progeny, to have a mutated chromosome. A single base error in RNA transcription would not affect the chromosome; it would lead to formation of some defective copies of one protein, but because mRNAs turn over rapidly, most copies of the protein would not be defective. The progeny of this cell would be normal.

The gene encoding the E.coli enzyme β-galactosidase begins with the sequence ATGACCATGATTACG. What is the sequence of the RNA transcript specified by this part of the gene?

AUGACCAUGAUUACG

Explain why the absorption of UV light by double-stranded DNA increases (the hyperchromic effect) when the DNA is denatured.

Base stacking in nucleic acids tends to reduce the absorption of UV light. Denaturation involves loss of base stacking, and UV absorption increases.

The RNA genome of phage Qβ is the nontemplate strand, or coding strand, and when introduced into the cell it functions as an mRNA. Suppose the RNA replicase of phage Qβ synthesized primarily template-strand RNA and uniquely incorporated this, rather than nontemplate strands, into the viral particles. What would be the fate of the template strands when they entered a new cell? What enzyme would have to be included in the viral particles for successful invasion of a host cell?

Because the template-strand RNA does not encode the enzymes needed to initiate viral infection, it would probably be inert or simply degraded by cellular ribonucleases. Replication of the template-strand RNA and propagation of the virus could occur only if intact RNA replicase (RNA-dependent RNA polymerase) were introduced into the cell along with the template strand.

A part of a sequenced chromosome has the sequence (on one strand) ATTGCATCCGCGCGTGCGCGCGCGATCCCGTTACTTTCCG. Which part of this sequence is most likely to take up the Z conformation?

CGCGCGTGCGCGCGCG

In the scheme of Figure 8-35, each new base to be added to the growing oligonucleotide is modified so that its 3' hydroxyl is activated and the 5' hydroxyl has a dimethoxytrityl (DMT) group attached. What is the function of the DMT group on the incoming base?

DMT is a blocking group that prevents reaction of the incoming base with itself.

The 5,386 bp genome of bacteriophage φX174 includes genes for 10 proteins, designated A to K, with sizes given in the table below. How much DNA would be required to encode these 10 proteins? How can you reconcile the size of the φX174 genome with its protein-coding capacity?

DNA with a minimum of 5,784 bp; some of the coding sequences must be nested or overlapping.

E. coli cells are growing in a medium containing lactose but no glucose. Indicate whether each of the following changes or conditions would increase, decrease, or not change the expression of the lac operon. It may be helpful to draw a model depicting what is happening in each situation. (a) Addition of a high concentration of glucose (b) A mutation that prevents dissociation of the Lac repressor from the operator (c) A mutation that completely inactivates b-galactosidase (d) A mutation that completely inactivates galactoside permease (e) A mutation that prevents binding of CRP to its binding site near the lac promoter

Each condition decreases expression of lac operon genes.

A biochemist replaces the DNA-binding domain of the yeast Gal4 protein with the DNA-binding domain from the Lac repressor, and finds that the engineered protein no longer regulates transcription of the GAL genes in yeast. Draw a diagram of the different functional domains you would expect to find in the Gal4 protein and in the engineered protein. Why does the engineered protein no longer regulate transcription of the GAL genes? What might be done to the DNA-binding site recognized by this chimeric protein to make it functional in activating transcription of GAL genes?

Gal4 DNA-binding domain Gal4 activator domain Lac repressor DNA-binding domain Gal4 activator domain The engineered protein cannot bind to the Gal4 binding site in the GAL gene (UASG) because it lacks the Gal4 DNA-binding domain. Modify the Gal4p DNA binding site to give it the nucleotide sequence to which the Lac repressor normally binds (using methods described in Chapter 9).

What is the minimum number of transesterification reactions needed to splice an intron from an mRNA transcript? Explain.

Generally two: one to cleave the phosphodiester bond at one intron-exon junction, the other to link the resulting free exon end to the exon at the other end of the intron. If the nucleophile in the first step were water, this step would be a hydrolytic event and only one transesterification step would be required to complete the splicing process.

Most amino acids have more than one codon and attach to more than one tRNA, each with a different anticodon. Write all possible anticodons for the four codons of glycine: (5')GGU, GGC, GGA, and GGG. (a) From your answer, which of the positions in the anticodons are primary determinants of their codon specificity in the case of glycine? (b) Which of these anticodon-codon pairings has/have a wobbly base pair? (c) In which of the anticodon-codon pairings do all three positions exhibit strong Watson-Crick hydrogen bonding?

Glycine codons (5')GGU (5')GGC (5')GGA (5')GGG Anticodons (5')ACC, GCC, ICC (5')GCC, ICC (5')UCC, ICC (5')CCC, UCC (a) The 3' and middle position (b) Pairings with anticodons (5')GCC, ICC, and UCC (c) Pairings with anticodons (5')ACC and CCC

A duplex DNA oligonucleotide in which one of the strands has the sequence TAATACGACTCACTATAGGG has a melting temperature (tm) of 59 °C. If an RNA duplex oligonucleotide of identical sequence (substituting U for T) is constructed, will its melting temperature be higher or lower?

Higher

A researcher isolates mutant variants of the bacterial translation factors IF-2, EF-Tu, and EF-G. In each case, the mutation allows proper folding of the protein and the binding of GTP but does not allow GTP hydrolysis. At what stage would translation be blocked by each mutant protein?

IF-2: The 70S ribosome would form, but initiation factors would not be released and elongation could not start. EF-Tu: The second aminoacyl-tRNA would bind to the ribosomal A site, but no peptide bond would form. EF-G: The first peptide bond would form, but the ribosome would not move along the mRNA to vacate the A site for binding of a new EF-Tu-tRNA.

The cells of many eukaryotic organisms have highly specialized systems that specifically repair G-T mismatches in DNA. The mismatch is repaired to form a G≡C (not A=T) base pair. This G-T mismatch repair mechanism occurs in addition to a more general system that repairs virtually all mismatches. Suggest why cells might require a specialized system to repair G-T mismatches.

In eukaryotic DNA, about 5% of C residues are methylated. 5-Methylcytosine can spontaneously deaminate to form thymine; the resulting G-T pair is one of the most common mismatches in eukaryotic cells.

Conclusions from the Meselson-Stahl Experiment The Meselson-Stahl experiment proved that DNA undergoes semiconservative replication in E. coli. In the "dispersive" model of DNA replication, the parent DNA strands are cleaved into pieces of random size, then joined with pieces of newly replicated DNA to yield daughter duplexes. Explain how the results of Meselson and Stahl's experiment ruled out such a model.

In random, dispersive replication, in the second generation, all the DNAs would have the same density and would appear as a single band, not the two bands observed in the Meselson-Stahl experiment. This is because the experiment used cells grown in N15 isotopic environment and switched them to N14 prior to replication. Due to the semiconservative method of replication, there are varying densities of DNA strands on account of the isotopic N.

Heavy Isotope Analysis of DNA Replication A culture of E. coli growing in a medium containing 15NH4Cl is switched to a medium containing 14NH4Cl for three generations (an eightfold increase in population). What is the molar ratio of hybrid DNA (15N-14N) to light DNA (14N-14N) at this point?

In this extension of the Meselson-Stahl experiment, after three generations the molar ratio of 15N-14N DNA to 14N-14N DNA is 2/6 0.33. This answer relies on the comparison of hybrid to light-only DNA, and not a percentage of hybrid to overall DNA.

The isoleucyl-tRNA synthetase has a proofreading function that ensures the fidelity of the aminoacylation reaction, but the histidyl-tRNA synthetase lacks such a proofreading function. Explain.

Isoleucine is similar in structure to several other amino acids, particularly valine. Distinguishing between valine and isoleucine in the aminoacylation process requires the second filter of a proofreading function. Histidine has a structure unlike that of any other amino acid, and this structure provides opportunities for binding specificity adequate to ensure accurate aminoacylation of the cognate tRNA.

Prepare a table that lists the names and compares the functions of the precursors, enzymes, and other proteins needed to make the leading strand versus the lagging strand during DNA replication in E. coli.

Leading strand: Precursors: dATP, dGTP, dCTP, dTTP (also needs a template DNA strand and DNA primer); enzymes and other proteins: DNA gyrase, helicase, single-stranded DNA- binding protein, DNA polymerase III, topoisomerases, and pyrophosphatase. Lagging strand: Precursors: ATP, GTP, CTP, UTP, dATP, dGTP, dCTP, dTTP (also needs an RNA primer); enzymes and other proteins: DNA gyrase, helicase, single-stranded DNA-binding protein, primase, DNA polymerase III, DNA polymerase I, DNA ligase, topoisomerases, and pyrophosphatase. NAD is also required as a cofactor for DNA ligase.

How would transcription of the E. coli trp operon be affected by the following manipulations of the leader region of the trp mRNA? (a) Increasing the distance (number of bases) between the leader peptide gene and sequence 2 (b) Increasing the distance between sequences 2 and 3 (c) Removing sequence 4 (d) Changing the two Trp codons in the leader peptide gene to His codons (e) Eliminating the ribosome-binding site for the gene that encodes the leader peptide (f) Changing several nucleotides in sequence 3 so that it can base-pair with sequence 4 but not with sequence 2

Less attenuation of transcription. The ribosome completing the translation of sequence 1 would no longer overlap and block sequence 2; sequence 2 would always be available to pair with sequence 3, preventing formation of the attenuator structure. (b) More attenuation of transcription. Sequence 2 would pair less efficiently with sequence 3; the attenuator structure would be formed more often, even when sequence 2 was not blocked by a ribosome. (c) No attenuation of transcription. The only regulation would be that afforded by the Trp repressor. (d) Attenuation loses its sensitivity to Trp tRNA. It might become sensitive to His tRNA. (e) Attenuation would rarely, if ever, occur. Sequences 2 and 3 always block formation of the attenuator. (f) Constant attenuation of transcription. Attenuator always forms, regardless of the availability of tryptophan.

Which positions in the purine ring of a purine nucleotide in DNA have the potential to form hydrogen bonds but are not involved in Watson-Crick base pairing?

N-3 and N-7

6. The Chemistry of DNA Replication All DNA polymerases synthesize new DNA strands in the 5' 3' direction. In some respects, replication of the antiparallel strands of duplex DNA would be simpler if there were also a second type of polymerase, one that synthesized DNA in the 3' 5' direction. The two types of polymerase could, in principle, coordinate DNA synthesis without the complicated mechanics required for lagging strand replication. However, no such 3' 5'-synthesizing enzyme has been found. Suggest two possible mechanisms for 3' 5' DNA synthesis. Pyrophosphate should be one product of both proposed reactions. Could one or both mechanisms be supported in a cell? Why or why not? (Hint: You may suggest the use of DNA precursors not actually present in extant cells.)

Mechanism 1: 3'-OH of an incoming dNTP attacks the phosphate of the triphosphate at the 5' end of the growing DNA strand, displacing pyrophosphate. This mechanism uses normal dNTPs, and the growing end of the DNA always has a triphosphate on the 5' end. Mechanism 2: This uses a new type of precursor, nucleotide 3'- triphosphates. The growing end of the DNA strand has a 5'-OH, which attacks the phosphate of an incoming deoxynucleotide 3'-triphosphate, displacing pyrophosphate. Note that this mechanism would require the evolution of new metabolic pathways to supply the needed deoxynucleotide 3'-triphosphates.

To prepare genomic regions for transcription, certain histones in the resident nucleosomes are acetylated and methylated at specific locations. Once transcription is no longer needed, these modifications need to be reversed. In mammals, the methylation of Arg residues in histones is reversed by peptidylarginine deiminases (PADIs). The reaction promoted by these enzymes does not yield unmethylated arginine. Instead, it produces citrulline residues in the histone. What is the other product of the reaction? Suggest a mechanism for this reaction.

Methylamine. The reaction proceeds with attack of water on the guanidinium carbon of the modified arginine.

Some E. coli mutants contain defective DNA ligase. When these mutants are exposed to 3H-labeled thymine and the DNA produced is sedimented on an alkaline sucrose density gradient, two radioactive bands appear. One corresponds to a high molecular weight fraction, the other to a low molecular weight fraction. Explain.

Mutants with defective DNA ligase produce a DNA duplex in which one of the strands remains in pieces (as Okazaki fragments). When this duplex is denatured, sedimentation results in one fraction containing the intact single strand (the high molecular weight band) and one fraction containing the unspliced fragments (the low molecular weight band).

A given sequence of bases in an mRNA will code for one and only one sequence of amino acids in a polypeptide, if the reading frame is specified. From a given sequence of amino acid residues in a protein such as cytochrome c, can we predict the base sequence of the unique mRNA that coded it? Give reasons for your answer.

No. Because nearly all the amino acids have more than one codon (e.g., Leu has six), any given polypeptide can be coded for by a number of different base sequences. However, some amino acids are encoded by only one codon and those with multiple codons often share the same nucleotide at two of the three positions, so certain parts of the mRNA sequence encoding a protein of known amino acid sequence can be predicted with high certainty.

Predict the likely effects of a mutation in the sequence (5')AAUAAA in a eukaryotic mRNA transcript.

Normal posttranscriptional processing at the 3' end (cleavage and polyadenylation) would be inhibited or blocked.

The plasmid cloning vector pBR322 (see Fig. 9-3) is cleaved with the restriction endonuclease PstI. An isolated DNA fragment from a eukaryotic genome (also produced by PstI cleavage) is added to the prepared vector and ligated. The mixture of ligated DNAs is then used to transform bacteria, and plasmid-containing bacteria are selected by growth in the presence of tetracycline. (a) In addition to the desired recombinant plasmid, what other types of plasmids might be found among the transformed bacteria that are tetracycline resistant? How can the types be distinguished? (b) The cloned DNA fragment is 1,000 bp long and has an EcoRI site 250 bp from one end. Three different recombinant plasmids are cleaved with EcoRI and analyzed by gel electrophoresis, giving the patterns shown below. What does each pattern say about the cloned DNA? Note that in pBR322, the PstI and EcoRI restriction sites are about 750 bp apart. The entire plasmid with no cloned insert is 4,361 bp. Size markers in lane 4 have the number of nucleotides noted

Plasmids in which the original pBR322 was regenerated without insertion of a foreign DNA fragment; these would retain resistance to ampicillin. Also, two or more molecules of pBR322 might be ligated together with or without insertion of foreign DNA. (b) The clones in lanes 1 and 2 each have one DNA fragment inserted in different orientations. The clone in lane 3 has two DNA fragments, ligated such that the EcoRI proximal ends are joined.

Vertebrate and plant cells often methylate cytosine in DNA to form 5-methylcytosine. In these same cells, a specialized repair system recognizes G-T mismatches and repairs them to G≡C base pairs. How might this repair system be advantageous to the cell? (Explain in terms of the presence of 5-methylcytosine in the DNA.)

Spontaneous deamination of 5-methylcytosine (see Fig. 8-30) produces thymine, and thus a G-T mismatched pair. These are among the most common mismatches in the DNA of eukaryotes. The specialized repair system restores the GqC pair.

A researcher engineers a lac operon on a plasmid but inactivates all parts of the lac operator (lacO) and the lac promoter, replacing them with the binding site for the LexA repressor (which acts in the SOS response) and a promoter regulated by LexA. The plasmid is introduced into E. coli cells that have a lac operon with an inactive lacI gene. Under what conditions will these transformed cells produce b-galactosidase?

The E. coli cells will produce -galactosidase when they are subjected to high levels of a DNA-damaging agent, such as UV light. Under such conditions, RecA binds to single-stranded chromosomal DNA and facilitates the autocatalytic cleavage of the LexA repressor, releasing LexA from its binding site and allowing transcription of downstream genes.

The chemical mechanisms used to avoid errors in protein synthesis are different from those used during DNA replication. DNA polymerases use a 3' 5' exonuclease proofreading activity to remove mispaired nucleotides incorrectly inserted into a growing DNA strand. There is no analogous proofreading function on ribosomes and, in fact, the identity of an amino acid attached to an incoming tRNA and added to the growing polypeptide is never checked. A proofreading step that hydrolyzed the previously formed peptide bond after an incorrect amino acid had been inserted into a growing polypeptide (analogous to the proofreading step of DNA polymerases) would be impractical. Why? (Hint: Consider how the link between the growing polypeptide and the mRNA is maintained during elongation; see Figs. 27-29 and 27-30.)

The amino acid most recently added to a growing polypeptide chain is the only one covalently attached to a tRNA and thus is the only link between the polypeptide and the mRNA encoding it. A proofreading activity would sever this link, halting synthesis of the polypeptide and releasing it from the mRNA.

A Drosophila egg that is bcd_/bcd_ may develop normally, but the adult fruit fly will not be able to produce viable offspring. Explain.

The bcd mRNA needed for development is contributed to the egg by the mother. The egg develops normally even if its genotype is bcd/bcd, as long as the mother has one normal bcd gene and the bcd allele is recessive. However, the adult bcd/bcd female will be sterile because she has no normal bcd mRNA to contribute to her own eggs.

The RNA viruses have relatively small genomes. For example, the single-stranded RNAs of retroviruses have about 10,000 nucleotides and the Qb RNA is only 4,220 nucleotides long. Given the properties of reverse transcriptase and RNA replicase described in this chapter, can you suggest a reason for the small size of these viral genomes?

These enzymes lack a 3'-5' proofreading exonuclease and have a high error rate; the likelihood of a replication error that would inactivate the virus is much less in a small genome than in a large one.

The gene for a eukaryotic polypeptide 300 amino acid residues long is altered so that a signal sequence recognized by SRP occurs at the polypeptide's amino terminus and a nuclear localization signal (NLS) occurs internally, beginning at residue 150. Where is the protein likely to be found in the cell?

The protein would be directed into the ER, and from there the targeting would depend on additional signals. SRP binds the amino-terminal signal early in protein synthesis and directs the nascent polypeptide and ribosome to receptors in the ER. Because the protein is translocated into the lumen of the ER as it is synthesized, the NLS is never accessible to the proteins involved in nuclear targeting.

Sickle-cell hemoglobin has a Val residue at position 6 of the β-globin chain, instead of the Glu residue found in normal hemoglobin A. Can you predict what change took place in the DNA codon for glutamate to account for replacement of the Glu residue by Val?

The two DNA codons for Glu are GAA and GAG, and the four DNA codons for Val are GTT, GTC, GTA, and GTG. A single base change in GAA to form GTA or in GAG to form GTG could account for the Glu S Val replacement in sickle-cell hemoglobin. Much less likely are two-base changes, from GAA to GTG, GTT, or GTC; and from GAG to GTA, GTT, or GTC.

Methionine is one of two amino acids with only one codon. How does the single codon for methionine specify the initiating residue and interior Met residues of polypeptides synthesized by E. coli?

There are two tRNAs for methionine: tRNAfMet, which is the initiating tRNA, and tRNAMet, which can insert a Met residue in interior positions in a polypeptide. Only fMet-tRNAfMet is recognized by the initiation factor IF-2 and is aligned with the initiating AUG positioned at the ribosomal P site in the initiation complex. AUG codons in the interior of the mRNA can bind and incorporate only Met-tRNAMet.

The death cap mushroom, Amanita phalloides, contains several dangerous substances, including the lethal α-amanitin. This toxin blocks RNA elongation in consumers of the mushroom by binding to eukaryotic RNA polymerase II with very high affinity; it is deadly in concentrations as low as 10-8 M. The initial reaction to ingestion of the mushroom is gastrointestinal distress (caused by some of the other toxins). These symptoms disappear, but about 48 hours later, the mushroom-eater dies, usually from liver dysfunction. Speculate on why it takes this long for α-amanitin to kill.

Though RNA synthesis is quickly halted by -amanitin toxin, it takes several days for the critical mRNAs and the proteins in the liver to degrade, causing liver dysfunction and death. Simply, the functional components of the cell last for a day or two, after which the body is out of functional parts.

The secreted bacterial protein OmpA has a precursor, ProOmpA, which has the amino-terminal signal sequence required for secretion. If purified ProOmpA is denatured with 8 M urea and the urea is then removed (such as by running the protein solution rapidly through a gel filtration column) the protein can be translocated across isolated bacterial inner membranes in vitro. However, translocation becomes impossible if ProOmpA is first allowed to incubate for a few hours in the absence of urea. Furthermore, the capacity for translocation is maintained for an extended period if ProOmpA is first incubated in the presence of another bacterial protein called trigger factor. Describe the probable function of this factor.

Trigger factor is a molecular chaperone that stabilizes an unfolded and translocation-competent conformation of ProOmpA.

Write all the possible mRNA sequences that can code for the simple tripeptide segment Leu-Met-Tyr. Your answer will give you some idea about the number of possible mRNAs that can code for one polypeptide.

UUAAUGUAU UUGAUGUAU CUUAUGUAU CUCAUGUAU CUAAUGUAU CUGAUGUAU UUAAUGUAC UUGAUGUAC CUUAUGUAC CUCAUGUAC CUAAUGUAC CUGAUGUAC

What factors promote the fidelity of replication during the synthesis of the leading strand of DNA? Would you expect the lagging strand to be made with the same fidelity? Give reasons for your answers.

Watson-Crick base pairing between template and leading strand; proofreading and removal of wrongly inserted nucleotides by the 3'-exonuclease activity of DNA polymerase III. Yes—perhaps. Because the factors ensuring fidelity of replication are operative in both the leading and the lagging strands, the lagging strand would probably be made with the same fidelity. However, the greater number of distinct chemical operations involved in making the lagging strand might provide a greater opportunity for errors to arise.

Hydrolysis of the N-glycosyl bond between deoxyribose and a purine in DNA creates an AP site. An AP site generates a thermodynamic destabilization greater than that created by any DNA mismatched base pair. This effect is not completely understood. Examine the structure of an AP site (see Fig. 8-30b) and describe some chemical consequences of base loss.

Without the base, the ribose ring can be opened to generate the noncyclic aldehyde form. This, and the loss of base-stacking interactions, could contribute significant flexibility to the DNA backbone.