Genetics exam 2

Mechanisms that make the bacterial chromosome more compact include the formation of micro- and macrodomains. DNA supercoiling. crossing over. both a and b.

both a and b.

How does exposing an E. coli cell to glucose affect the regulation of the lac operon via CAP? cAMP binds to CAP and transcription is increased. cAMP binds to CAP and transcription is decreased. cAMP does not bind to CAP and transcription is increased. cAMP does not bind to CAP and transcription is decreased.

cAMP does not bind to CAP and transcription is decreased.

The process in which completely unmethylated DNA becomes methylated is called maintenance methylation. de novo methylation. primary methylation. demethylation.

de novo methylation.

A chromatin-remodeling complex may change the locations of nucleosomes. evict nucleosomes from DNA. replace standard histones with histone variants. do all of the above.

do all of the above.

A double-stranded region of RNA forms a helical structure. obeys the AU/GC rule. may result in the formation of a structure such as a bulge loop or a stem-loop. does all of the above.

does all of the above.

Explain how the isolation of mutants was instrumental in our understanding of DNA replication.

ecause DNA replication is vital for cell division, mutations that block DNA replication prevent cell growth. For this reason, if researchers want to identify loss-of-function mutations in vital genes, they must screen for conditional mutants. One type of conditional mutant is a temperature-sensitive (ts) mutant. For example, a ts mutant might grow at 30°C (the permissive temperature) but fail to grow at a higher temperature, such as 42°C, which is chosen by the experimenter. The higher temperature at which the mutant strain fails to grow is called the nonpermissive temperature. The mutant strain fails to grow because the higher temperature inactivates the function of the protein encoded by the mutant gene.

CRISPR RNA (crRNA)

crRNA, found in prokaryotes, guides an endonuclease to foreign DNA, such as the DNA of a bacteriophage. Some CRISPR systems can target RNA.

Define nucleic acid.

DNA and its molecular cousin, RNA, are known as nucleic acids.

Each ribosomal subunit is composed of multiple proteins. rRNA. tRNA. both a and b.

both a and b.

To fulfill their role at the molecular level, chromosomal sequences facilitate four important processes:

(1) the synthesis of RNA and cellular proteins, (2) the replication of chromosomes, (3) the proper segregation of chromosomes, and (4) the compaction of chromosomes so they can fit within living cells.

Explain how the translation and degradation of an mRNA may be controlled by RNA-binding proteins.

(a) When the Fe3+ concentration is low, the binding of iron regulatory protein (IRP) to the iron response element (IRE) in the 5′-UTR of ferritin mRNA inhibits translation (left). When Fe3+ concentration is high and Fe3+ binds to IRP, the IRP is removed from the ferritin mRNA, so translation can proceed (right). (b) The binding of IRP to the IRE in the 3′-UTR of the transferrin receptor mRNA prevents the degradation of the mRNA. Therefore, the mRNA is available for the synthesis of transferrin receptor proteins when the Fe3+ concentration is low (left). When the Fe3+ concentration is high and iron binds to IRP, the IRP dissociates from the IRE, and the transferrin receptor mRNA is cleaved by endonucleases. This endonuclease cleavage removes the polyA tail, and then the mRNA is further degraded by an exonuclease (not shown) that degrades the rest of the transferrin receptor mRNA from the 3′ end. The ferritin mRNA has an IRE in its 5′-untranslated region (5′-UTR), which is the region between the 5′ end and the start codon. When iron levels in the cell are low, IRP binds to this IRE and the IRE/IRP complex prevents the translation of the ferritin mRNA. In other words, the IRE/IRP complex acts as a blocker of translation. However, when iron is abundant in the cytosol, the iron binds directly to the IRP and causes a conformational change that Page 381 prevents IRP from binding to the IRE. Under these conditions, the ferritin mRNA is translated to make more ferritin protein (Figure 15.19a, right), which stores the excess iron and thereby prevents its toxic buildup of iron within the cytosol.

A bidirectional enhancer has the following sequence: 5′-GTCA-3′ 3′-CAGT-5′ Which of the following sequences would also be a functional enhancer? 5′-ACTG-3′ 3′-TGAC-5′ 5′-TGAC-3′ 3′-ACTG-5′ 3′-GTCA-5′ 5′-CAGT-3′ 3′-TGAC-5′ 5′-ACTG-3′

5′-TGAC-3′ 3′-ACTG-5′

Describe the characteristics of a bacterial promoter.

A Promoter Is a Short Sequence of DNA That Is Necessary to Initiate Transcription Most of the promoter is located just ahead of, or upstream from, the site where transcription of a gene actually begins. By convention, the bases in a promoter sequence are numbered in relation to the transcriptional start site (Figure 12.4). This site is the first base used as a template for transcription and is denoted +1. The bases preceding this site are numbered in a negative direction. No base is numbered zero. two sequences, which are located at approximately the −35 and −10 sites in the promoter, are particularly important (see Figure 12.4). The sequence in the top DNA strand at the −35 site is 5′-TTGACA-3′, and the one at the −10 site is 5′-TATAAT-3′. The TATAAT sequence is called the Pribnow box

Chargaff's analysis of the base composition of DNA is consistent with base pairing between

A and T, and G and C.

Describe the structural features of a DNA strand.

A few structural features are worth noting. A phosphate group connects two sugar molecules via ester bonds. For this reason, the linkage in DNA (or RNA) strands is called a phosphodiester linkage. the 5′ carbons in every sugar molecule are above the 3′ carbons. Therefore, a strand has a directionality because all sugar molecules have the same orientation..

What is an operon? A site in the DNA where a regulatory protein binds A group of genes under the control of a single promoter An mRNA that encodes several genes All of the above

A group of genes under the control of a single promoter

Describe the structure of a nucleotide.

A nucleotide has three components: at least one phosphate group, a pentose sugar, and a nitrogenous base. As shown in Figure 9.4, nucleotides vary with regard to the sugar and the nitrogenous base. The two types of sugars are deoxyribose and ribose, which are found in DNA and RNA, respectively. The five different bases are subdivided into two categories: the purines and the pyrimidines. The purine bases, adenine (A) and guanine (G), contain a double-ring structure; the pyrimidine bases, thymine (T), cytosine (C), and uracil (U), contain a single-ring structure.

Describe the function of activators and repressors.

A repressor is a regulatory protein that binds to the DNA and inhibits transcription, whereas an activator is a regulatory protein that increases the rate of transcription. Transcriptional regulation by a repressor is termed negative control, and regulation by an activator is considered to be positive control.

Which of the following combinations will cause transcription to be activated? A repressor plus an inducer A repressor plus a corepressor An activator plus an inhibitor None of the above

A repressor plus an inducer

Which of the following statements about the spliceosome is false? A spliceosome splices pre-mRNA molecules. A spliceosome removes exons from RNA molecules. A spliceosome is composed of snRNPs. A spliceosome recognizes the exon-intron boundaries and the branch site.

A spliceosome removes exons from RNA molecules.

Describe the key structural features of a tRNA molecule.

A tRNA has three stem-loops, a few variable sites, and an acceptor stem with a 3′ single-stranded region (Figure 13.10). The acceptor stem is where an amino acid becomes attached to a tRNA (see the inset in the figure). All tRNA molecules have the sequence CCA at their 3′ ends. These three nucleotides are usually added enzymatically by the enzyme tRNA nucleotidyltransferase after the tRNA is made. A conventional numbering system for the nucleotides within a tRNA molecule begins at the 5′ end and proceeds toward the 3′ end.

List a few exceptions to the genetic code.

AUA Isoleucine Methionine in yeast and mammalian mitochondria UGA Stop Tryptophan in mammalian mitochondria CUU, CUC, CUA, CUG Leucine Threonine in yeast mitochondria AGA, AGG Arginine Stop codon in ciliated protozoa and in yeast and mammalian mitochondria UAA, UAG Stop Glutamine in ciliated protozoa UGA Stop Selenocysteine in certain genes found in bacteria, archaea, and eukaryotes UAG Stop Pyrrolysine in certain genes found in methane-producing archaea

What are the components of a single nucleosome? About 146 bp of DNA and four histone proteins About 146 bp of DNA and eight histone proteins About 200 bp of DNA and four histone proteins About 200 bp of DNA and eight histone proteins

About 146 bp of DNA and eight histone proteins

Compare and contrast two possible mechanisms for transcriptional termination in eukaryotes.

According to the allosteric model, RNA polymerase II becomes destabilized after it has transcribed the polyA signal sequence, and it eventually dissociates from the DNA. This destabilization may be caused by the release of proteins that function as elongation factors or by the binding of proteins that function as termination factors. A second model, called the torpedo model, suggests that RNA polymerase II is physically removed from the DNA. According to this model, the region of RNA that is still being transcribed and is downstream from the polyadenylation signal sequence is cleaved by an exonuclease that degrades the transcript in the 5′ to 3′ direction. When the exonuclease catches up to RNA polymerase II, this causes RNA polymerase II to dissociate from the DNA.

Outline the wobble rules.

According to the wobble rules, the first two positions pair strictly according to the AU/GC rule. However, the third position can tolerate certain types of mismatches (Figure 13.12). This proposal suggested that the base at the third position in the codon does not have to hydrogen bond as precisely with the corresponding base in the anticodon. When two or more tRNAs that differ at the wobble position are able to recognize the same codon, they are termed isoacceptor tRNAs. As an example, tRNAs with an anticodon of 3′-CCA-5′ or 3′-CCG-5′, which both carry glycine, can recognize a codon with the sequence of 5′-GGU-3′. In addition, the wobble rules enable a single type of tRNA to recognize more than one codon. For example, a tRNA with an anticodon sequence of 3′-AAG-5′, which carries phenylalanine, can recognize a 5′-UUC-3′ and a 5′-UUU-3′ codon.

enhancers

Activating sequences, known as enhancers, are needed to stimulate transcription. In the absence of enhancer sequences, most eukaryotic genes have very low levels of basal transcription.

Analyze the results of Hershey and Chase, and explain how they indicate that DNA is the genetic material.

After phages were given sufficient time to infect bacterial cells, the researchers separated the phage coats from the bacterial cells. They determined that most of the 32P had entered the bacterial cells whereas most of the 35S remained outside the cells. These results were consistent with the idea that the genetic material of bacteriophages is DNA not proteins

Which of the following are examples of RNA modification? Splicing Capping with 7-methylguanosine Adding a polyA tail All of the above

All of the above

Which of the following are examples of molecular changes that can have an epigenetic effect on gene expression? Chromatin remodeling Covalent histone modification Localization of histone variants DNA methylation Feedback loops All of the above

All of the above

Which of the following is a function of the 7-methylguanosine cap? Exit of mRNA from the nucleus Efficient splicing of pre-mRNA Initiation of translation All of the above

All of the above

Which of the following possibilities could explain how PcG complexes are able to silence genes? The compaction of nucleosomes The attachment of ubiquitin to histone proteins The direct inhibition of transcription factors, such as TFIID All of the above

All of the above

Which of the following is not a feature of the DNA double helix? It obeys the AT/GC rule. The DNA strands are antiparallel. The structure is stabilized by base stacking. All of the above are features of the DNA double helix.

All of the above are features of the DNA double helix.

How can methylation affect transcription? It may prevent the binding of regulatory transcription factors. It may enhance the binding of regulatory transcription factors. It may attract methyl-CpG-binding proteins, which inhibit transcription, to bind to a methylated sequence. All of the above are possible ways for methylation to affect transcription

All of the above are possible ways for methylation to affect transcription

During the initiation stage of translation in bacteria, which of the following events occur(s)? IF1 and IF3 bind to the 30S subunit. The mRNA binds to the 30S subunit and tRNAfMet binds to the start codon in the mRNA. IF2 hydrolyzes its GTP and is released; the 50S subunit binds to the 30S subunit. All of the above events occur.

All of the above events occur.

Which of the following statements regarding DNA polymerases in eukaryotes is not correct? DNA polymerase α synthesizes a short RNA-DNA primer. DNA polymerases ε and δ synthesize most of the leading and lagging strands, respectively. Lesion-replicating DNA polymerases can replicate over damaged DNA. All of the above statements are correct.

All of the above statements are correct.

Explain how an amino acid is attached to a tRNA via aminoacyl-tRNA synthetase.

Aminoacyl-tRNA synthetases catalyze a chemical reaction involving three different molecules: an amino acid, a tRNA molecule, and ATP. In the first step of the reaction, a synthetase recognizes a specific amino acid and also ATP (Figure 13.11). The Page 321 ATP is hydrolyzed, and AMP becomes attached to the amino acid; pyrophosphate is released. During the second step, the correct tRNA binds to the synthetase. The amino acid becomes covalently attached to the 3′ end of the tRNA molecule at the acceptor stem, and AMP is released. Finally, the tRNA with its attached amino acid is released from the enzyme. At this stage, the tRNA is called a charged tRNA or an aminoacyl-tRNA. In a charged tRNA molecule, the amino acid is attached to the 3′ end of the tRNA by a covalent bond

RNA component of signal recognition particle (SRP RNA)

An SRP is composed of one ncRNA and one or more proteins. In prokaryotes, SRP directs the synthesis of certain polypeptides to the plasma membrane. In eukaryotes, it directs them to the endoplasmic reticulum.

Antisense RNA

An antisense RNA is complementary to an mRNA. The binding of the antisense RNA to the mRNA blocks translation.

Describe how RNA primers are removed in eukaryotes.

Another key difference between bacterial and eukaryotic DNA replication is the way that RNA primers are removed. As discussed earlier in this chapter, bacterial RNA primers are removed by DNA polymerase I. By comparison, a DNA polymerase enzyme does not play this role in eukaryotes. Instead, an enzyme called flap endonuclease is primarily responsible for RNA primer removal. Flap endonuclease gets its name because it removes small RNA flaps that are generated by the action of DNA polymerase δ. In the diagram shown in Figure 11.21, DNA polymerase δ elongates the left Okazaki fragment until it runs into the RNA primer of the adjacent Okazaki fragment on the right. This causes a portion of the RNA primer to form a short flap, which is removed by flap endonuclease. As DNA polymerase δ continues to elongate the DNA, short flaps continue to be generated, which are sequentially removed by flap endonuclease. Eventually, all of the RNA primer is removed, and DNA ligase seals the DNA fragments together. hough flap endonuclease is thought to be the primary agent for RNA primer removal in eukaryotes, it cannot remove a flap that is too long. In such cases, the long flap is cleaved by an enzyme called Dna2 nuclease/helicase. This enzyme can cut a long flap, thereby generating a short flap. The short flap is then removed via flap endonuclease.

Describe how a regulatory transcription factor binds to a regulatory element.

As mentioned previously, when the binding of a regulatory transcription factor to a regulatory element increases transcription, the regulatory element is known as an enhancer. Such elements can stimulate transcription 10- to 1000-fold, a phenomenon known as up regulation. Alternatively, regulatory elements that serve to inhibit transcription are called silencers, and their action is called down regulation. Many regulatory elements are orientation-independent, or bidirectional. This means that the regulatory element functions in the forward or reverse direction. They can be located upstream, downstream, or in an intron

Analyze Noll's results and explain how they support the beads-on-a-string model.

As shown in the data, at a high DNase I concentration, the entire sample of chromosomal DNA was digested into fragments of approximately 200 bp in length. This result is predicted by the beads-on-a-string model. Furthermore, at a low or medium DNase I concentration, longer pieces were observed, and these were in multiples of 200 bp (400, 600, etc.). How do we explain these longer pieces? They occurred because occasional linker regions remained uncut at a low or medium DNase I concentration. For example, if one linker region was not cut, a DNA piece would contain two nucleosomes and be 400 bp in length. If two consecutive linker regions were not cut, this would produce a piece with three nucleosomes containing about 600 bp of DNA. Taken together, these results strongly supported the nucleosome model for chromatin structure.

Explain how eukaryotic mRNAs are modified to have a cap and a tail.

At their 5′ end, most mature mRNAs have a 7-methylguanosine covalently attached—an event known as capping. Capping occurs while the pre-mRNA is being made by RNA polymerase II, usually when the transcript is only 20-25 nucleotides in length. As shown in Figure 12.23, it is a three-step process. The nucleotide at the 5′ end of the transcript has three phosphate groups. First, an enzyme called RNA 5′-triphosphatase removes one of the phosphates, Page 298 and then a second enzyme, guanylyltransferase, hydrolyzes guanosine triphosphate (GTP) to attach a guanosine monophosphate (GMP) to the 5′ end. Finally, a methyltransferase attaches a methyl group to a nitrogen at position 7 in the base guanine. What are the functions of the 7-methylguanosine cap? The cap structure is recognized by cap-binding proteins, which perform various roles. For example, cap-binding proteins are required for the proper exit of most mRNAs from the nucleus. Also, the cap structure is recognized by initiation factors that are needed during the early stages of translation. Finally, the cap structure may be important in the efficient splicing of introns, particularly the first intron located nearest the 5′ end. o start the process, the pre-mRNA contains a polyadenylation signal sequence near its 3′ end. In mammals, the consensus sequence is AAUAAA. An endonuclease recognizes the signal sequence and cleaves the pre-mRNA at a location that is about 20 nucleotides beyond the 3′ end of the AAUAAA sequence. The fragment beyond the 3′ cut is degraded. Next, an enzyme known as polyA-polymerase attaches many adenine-containing nucleotides. The length of the polyA tail varies among different mRNAs; the maximum length is typically around 250 nucleotides. A long polyA tail facilitates mRNA export from the nucleus, stability of mRNA in the cytosol, and translation.

A key difference between B DNA and Z DNA is that

B DNA is right-handed, whereas Z DNA is left-handed.

Compare and contrast B DNA and Z DNA.

B DNA is the predominant form of DNA in living cells, though some is found in a Z DNA conformation. B DNA is a right-handed helix, whereas Z DNA is left-handed. In addition, the helical backbone in Z DNA appears to zigzag slightly as it winds itself around the double-helical structure. The numbers of base pairs per 360° turn are 10.0 in B and 12.0 in Z DNA. In B DNA, the bases tend to be centrally located, and the hydrogen bonds Page 221 between base pairs are oriented relatively perpendicular to the central axis. In contrast, the bases in Z DNA are substantially tilted relative to the central axis

Compare and contrast bacterial and eukaryotic translation.

Bacterial Eukaryotic Ribosome composition: 70S ribosomes:30S subunit— 21 proteins + 1 rRNA50S subunit— 34 proteins + 2 rRNAs 80S ribosomes:40S subunit— 33 proteins + 1 rRNA60S subunit— 49 proteins + 3 rRNAs Initiator tRNA: tRNAfmet tRNAMet Formation of the initiation complex: Requires IF1, IF2, and IF3 Requires more initiation factors compared to bacterial initiation Initial binding of mRNA to the ribosome: Requires a Shine-Dalgarno sequence Requires a 7-methylguanosine cap Selection of a start codon: AUG, GUG, or UUG located just downstream from the Shine-Dalgarno sequence According to Kozak's rules Elongation rate: Typically 15-20 amino acids per second Typically 2-6 amino acids per second Termination: Requires RF1, RF2, and RF3 Requires eRF1 and eRF3 Location of translation: Cytoplasm Cytosol Coupled to transcription: Yes No

Explain how RNA polymerase transcribes a bacterial gene.

Bacterial Transcription Is Initiated When RNA Polymerase Holoenzyme Binds at a Promoter The enzyme that catalyzes the synthesis of RNA is RNA polymerase. In E. coli, the core enzyme is composed of five subunits, α2ββ′ω. The association of a sixth subunit, sigma (σ) factor, with the core enzyme creates what is referred to as RNA polymerase holoenzyme. The holoenzyme is required to initiate transcription; the primary role of σ factor is to recognize the promoter. After RNA polymerase holoenzyme is assembled into its six subunits, it binds loosely to the DNA and then slides along the DNA, much as a train rolls down the tracks. How is a promoter identified? When the holoenzyme encounters a promoter, σ factor recognizes both the −35 and −10 sequences. The σ-factor protein contains a structure called a helix-turn-helix motif that can bind tightly to these sequences. Alpha (α) helices within the protein fit into the major groove of the DNA double helix and form hydrogen bonds with the bases. Hydrogen bonding occurs between nucleotides in the −35 and −10 sequences of the promoter and amino acid side chains in the helix-turn-helix structure of σ factor. the process of transcription Page 283 is initiated when σ factor within the holoenzyme has bound to the promoter to form a closed complex. For transcription to begin, the double-stranded DNA must then be unwound into an open complex. This unwinding first occurs at the TATAAT sequence in the −10 site A short strand of RNA is made within the open complex, and then σ factor is released from the core enzyme. The release of σ factor marks the transition to the elongation phase of transcription. The core enzyme may now slide down the DNA to synthesize a strand of RNA.

Describe the organization of sites along bacterial chromosomes.

Bacterial chromosomal DNA is usually a circular molecule, though some bacteria have linear chromosomes bacteria usually contain a single type of chromosome, more than one copy of that chromosome may be found within one bacterial cell. A typical chromosome is a few million base pairs (bp) in length. A bacterial chromosome commonly has a few thousand different genes, which are interspersed throughout the entire chromosome. Protein-encoding genes (also called structural genes) account for the majority of bacterial DNA. The nontranscribed regions of DNA located between adjacent genes are termed intergenic regions. bacterial chromosomes have one origin of replication (see Figure 10.1), a sequence that is a few hundred nucleotides Page 230 in length a variety of repetitive sequences have been identified in many bacterial species. These sequences are found in multiple copies and are usually interspersed within the intergenic regions throughout the bacterial chromosome. Repetitive sequences may play a role in a variety of genetic processes, including DNA folding, DNA replication, gene regulation, and genetic recombination.

Outline the key structural features of RNA.

Base pairing between A and U and between G and C may occur within one RNA molecule or between two separate RNA molecules. This base pairing causes short segments of RNA to form a double-stranded region that is helical. As shown in Figure 9.17, different arrangements of base pairing are possible, which result in structures called bulge loops, internal loops, multibranched junctions, and stem-loops (also called hairpins).

Compare and contrast the origins of replication in bacteria and eukaryotes.

Because eukaryotes have long, linear chromosomes, multiple origins of replication are needed so the DNA can be replicated within a reasonable length of time. The multiple replication forks eventually make contact with each other to complete the replication process. several replication origins have been identified and sequenced. They have been named ARS elements (for autonomously replicating sequence). ARS elements, which are about 50 bp in length, are necessary to initiate chromosome replication. ARS elements contain a high percentage of A and T bases and have a copy of the ARS consensus sequence, ATTTAT(A or G)TTTA, along with additional elements that enhance origin function. This arrangement is similar to bacterial origins of replication, which also have an AT-rich region and specific elements, such as DnaA boxes.

Explain how DNA replication occurs at the ends of eukaryotic chromosomes.

Binding of Telomerase Telomerase contains both protein subunits and RNA. The RNA part of telomerase, known as telomerase RNA component (TERC), contains a sequence complementary to that found in the telomeric repeat sequence. This allows telomerase to bind to the 3′ overhang region of the telomere. Polymerization Following binding, the RNA sequence beyond the binding site functions as a template for the synthesis of a six-nucleotide sequence at the end of the DNA strand. This synthesis is called polymerization, because it is analogous to the function of DNA polymerase. Telomere lengthening is catalyzed by two identical protein subunits of telomerase called telomerase reverse transcriptase (TERT). TERT's name indicates that it catalyzes the reverse of transcription; it uses an RNA template to synthesize DNA. Translocation Following polymerization, telomerase then moves—a process called translocation—to the new end of the DNA strand and attaches another six nucleotides to the end. This binding-polymerization-translocation cycle occurs many times in a row, thereby greatly lengthening the 3′ end of the DNA strand in the telomeric region. Page 272 The complementary strand is synthesized by primase, DNA polymerase, and DNA ligase, as described earlier in this chapter. The RNA primer is later removed, which leaves a 3′ overhang. DNA polymerase is unable to replicate the 3′ ends of DNA strands. DNA polymerase synthesizes DNA only in a 5′ to 3′ direction, and it cannot link together the first two individual nucleotides; it can elongate only preexisting strands. These two features of DNA polymerase function pose a problem at the 3′ ends of linear chromosomes. As shown in Figure 11.23, the 3′ end of a DNA strand cannot be replicated by DNA polymerase because a primer cannot be made upstream from this point.

To make a new DNA strand, which of the following is necessary? A template strand Nucleotides Heavy nitrogen Both a and b

Both a and b

Which of the following is a function of the piRISC? Inhibits transcription of TEs Causes the degradation of TE RNA Causes chromosome breakage Both a and b

Both a and b

To synthesize DNA, what does telomerase use as a template? It uses the DNA in the 3′ overhang region. It uses RNA that is a component of telomerase. No template is used. Both a and b are correct.

Both a and b are correct.

Which of the following functions does HOTAIR perform? Decoy Scaffold Guide Both b and c

Both b and c

COLDAIR

COLDAIR alters chromatin structure and thereby represses transcription by recognizing histone-modifying complexes, which modify the FLC gene in Arabidopsis. This modification allows flowering to occur.

Which of the following components are needed for the adaptation phase of the CRISPR-Cas system? crRNA and Cas1 crRNA and Cas2 crRNA and Cas9 Cas1 and Cas2

Cas1 and Cas2

Compare and contrast the functions of several types of proteins.

Cell shape and organization Tubulin: Forms cytoskeletal structures known as microtubules Transport Sodium channels: Transport sodium ions across the nerve cell membranes Hemoglobin: Transports oxygen in red blood cells Movement Myosin: Involved in muscle cell contraction Cell signaling Insulin: A hormone that influences cell metabolism and growth Insulin receptor: Recognizes insulin and initiates a cell response Cell surface recognition Integrins: Bind to large extracellular proteins Enzymes Hexokinase: Phosphorylates glucose during the first step in glycolysis β-Galactosidase: Cleaves lactose into glucose and galactose Glycogen synthetase: Uses glucose molecules as building blocks to synthesize a large carbohydrate known as glycogen RNA polymerase: Uses ribonucleotides as building blocks to synthesize RNA DNA polymerase: Uses deoxyribonucleotides as building blocks to synthesize DNA

Centromere and segragation

Centromeres are regions that play a role in the proper segregation of chromosomes during mitosis and meiosis. For most species, each eukaryotic chromosome contains a single centromere, which usually appears as a constricted region of a mitotic chromosome. Centromeres function as a site for the formation of kinetochores, which assemble just before and during the very early stages of mitosis and meiosis. The kinetochore is composed of a group of proteins that link the centromere to the spindle apparatus during mitosis and meiosis, ensuring the proper segregation of the chromosomes to each daughter cell.

Transcriptional activation of eukaryotic genes involves which of the following events? Changes in nucleosome locations Changes in histone composition within nucleosomes Changes in histone modifications All of the above

Changes in histone composition within nucleosomes

Outline the structures of radial loop domains

Chromosomes Are Further Compacted by Anchoring of the 30-nm Fiber into Radial Loop Domains Along the Nuclear Matrix The proteins of the nuclear matrix are involved in compacting the DNA into radial loop domains, similar to those described for the bacterial chromosome. During interphase, chromatin is organized into loops, often 25,000-200,000 bp in size, which are anchored to the nuclear matrix. The chromosomal DNA of eukaryotic species contains sequences called matrix-attachment regions (MARs) or scaffold-attachment regions (SARs), which are interspersed at regular intervals throughout the genome. The matrix-attachment regions (MARs), which contain a high percentage of A and T bases, bind to the nuclear matrix and create radial loops.

Compare and contrast the processes of transcription and RNA modification in bacteria and eukaryotes

Component Bacteria Eukaryotes Promoter Consists of −35 and −10 sequences For protein-encoding genes, the core promoter often consists of a TATA box and a transcriptional start site. RNA polymerase A single RNA polymerase Three types of RNA polymerases; RNA polymerase II transcribes protein-encoding genes. Initiation σ factor is needed for promoter recognition. Five general transcription factors assemble at the core promoter. Elongation Requires the release of σ factor Mediator controls the switch to the elongation phase. Termination ρ-dependent or ρ-independent According to the allosteric or torpedo model Splicing Very rare; self-splicing Commonly occurs in protein-encoding pre-mRNAs in complex eukaryotes via a spliceosome; self-splicing occurs rarely. Capping Does not occur Addition of 7-methylguanosine cap Tailing Added to 3′ end; promotes degradation Added to the 3′ end; promotes stability RNA editing Does not occur Occurs occasionally

Define DNA methylation, and explain how it affects transcription.

DNA Methylation Occurs on the Cytosine Base and Usually Inhibits Gene Transcription DNA structure can be modified by the covalent attachment of methyl groups, a mechanism called DNA methylation. Methylation can affect transcription in two ways. First, methylation of CpG islands may prevent or enhance the binding of regulatory transcription factors to the promoter region. For example, methylated CG sequences may prevent the binding of an activator protein to an enhancer element, presumably by the methyl group protruding into the major groove of the DNA (Figure 15.15a). The inability of an activator protein to bind to the DNA inhibits the initiation of transcription. However, CG methylation does not slow down the movement of RNA polymerase along a gene. In vertebrates and plants, coding regions downstream from the core promoter usually contain methylated CG sequences, but these do Page 377 not hinder the elongation phase of transcription. This suggests that methylation must occur in the vicinity of the promoter to have an effect on transcription. A second way that methylation inhibits transcription is via proteins known as methyl-CpG-binding proteins, which bind to methylated sequences (Figure 15.15b). These proteins contain a domain called the methyl-binding domain that specifically recognizes a methylated CpG island. Once bound to the DNA, the methyl-CpG-binding protein recruits other proteins to the region that inhibit transcription. For example, methyl-CpG-binding proteins may recruit histone deacetylase to a methylated CpG island near a promoter. Histone deacetylation removes acetyl groups from the histone proteins, which makes it more difficult for nucleosomes to be removed from the DNA. In this way, deacetylation tends to inhibit transcription

Describe how DNA gyrase causes DNA supercoiling.

DNA gyrase, also known as topoisomerase II. Page 232 This enzyme, which contains four subunits (two A and two B subunits), introduces negative supercoils (or relaxes positive supercoils) using energy from ATP To alter supercoiling, DNA gyrase has two sets of jaws that allow it to grab onto two regions of DNA. One of the DNA regions is grabbed by the lower jaws and then is wrapped in a right-handed direction around the two A subunits. The upper jaws then clamp onto another region of DNA. The DNA in the lower jaws is cut in both strands, and the other region of DNA is then released from the upper jaws and passed through this double-stranded break. To complete the process, the double-stranded break is ligated back together. The net result is that two negative supercoils have been introduced into the DNA molecule

A key difference between the nucleotides found in DNA and those in RNA is that

DNA has deoxyribose, but RNA has ribose. DNA has thymine, but RNA has uracil.

primosome

DNA helicase and primase are physically bound to each other to form a complex known as a primosome. This complex leads the way at the replication fork. The primosome tracks along the DNA, separating the parental strands and synthesizing RNA primers at regular intervals along the lagging strand. Being associated together within a complex allows the actions of DNA helicase and primase to be better coordinated.

Define processivity.

DNA polymerase III can catalyze the synthesis of the daughter strands so quickly because it is a processive enzyme. This means it does not dissociate from the growing strand after it has catalyzed the covalent joining of two nucleotides. Rather, as depicted in Figure 11.8a, it remains clamped to the DNA template strand and slides along the template as it Page 267 catalyzes the synthesis of the daughter strand.

Compare and contrast the synthesis of the leading and lagging strands.

DNA polymerase III catalyzes the attachment of nucleotides to the 3′ end of each primer, in a 5′ to 3′ direction. In the leading strand, one RNA primer is made at the origin, and then DNA polymerase III attaches nucleotides in a 5′ to 3′ direction as it slides toward the opening of the replication fork. The synthesis of the leading strand is continuous. In the lagging strand, the synthesis of DNA also proceeds in a 5′ to 3′ manner, but it does so in the direction away from the replication fork. In the lagging strand, RNA primers repeatedly initiate the synthesis of short segments of DNA; the synthesis is discontinuous. The length of these fragments in bacteria is typically 1000-2000 nucleotides. In eukaryotes, the fragments are shorter: 100-200 nucleotides. Each fragment contains a short RNA primer at the 5′ end, which is made by primase. The remainder of the fragment is a strand of DNA made by DNA polymerase. The DNA fragments made in this manner are known as Okazaki fragments, To complete the synthesis of Okazaki fragments along the lagging strand, three additional events must occur: removal of the RNA primers, synthesis of DNA in the area where the primers have been removed, and the covalent attachment of adjacent fragments of DNA

Explain the proofreading function of DNA polymerase.

DNA polymerase can identify a mismatched nucleotide and remove it from the daughter strand. This occurs by exonuclease cleavage of the bonds between adjacent nucleotides at the 3′ end of the newly made strand. The ability to remove mismatched bases by this mechanism is called the proofreading function of DNA polymerase. Proofreading occurs by the removal of nucleotides in the 3′ to 5′ direction at the 3′ exonuclease site. After the mismatched nucleotide is removed, DNA polymerase resumes DNA synthesis in the 5′ to 3′ direction.

Describe how nucleotides are connected to a growing DNA strand.

DNA polymerase catalyzes the covalent attachment between the phosphate in one nucleotide and the sugar in the previous nucleotide. The formation of this covalent bond requires an input of energy. Prior to bond formation, the nucleotide about to be attached to the growing strand is a deoxyribonucleoside triphosphate (dNTP). It contains three phosphate groups attached at the 5′ carbon (C) atom of deoxyribose. The dNTP first enters the catalytic site of DNA polymerase and binds to the template strand according to the AT/GC rule. Next, the 3′ hydroxyl (—OH) group on the previous nucleotide reacts with the phosphate group (PO42−) adjacent to the sugar on the incoming nucleotide. This reaction is highly exergonic and results in a covalent bond between the sugar at the 3′ end of the DNA strand and the PO42− of the incoming nucleotide. The formation of this covalent bond causes the newly made strand to grow in the 5′ to 3′ direction. As shown in Figure 11.16, pyrophosphate (PPi) is released and broken down into 2 phosphates.

Explain how DnaA protein initiates DNA replication.

DNA replication begins with Page 258 the binding of DnaA proteins to sequences within the origin of replication known as DnaA boxes. The DnaA boxes serve as recognition sites for the binding of the DnaA proteins. When DnaA proteins are in their ATP-bound form, they bind to the five DnaA boxes in oriC to initiate DNA replication. DnaA proteins also bind to each other to form a complex. Other DNA-binding proteins, such as HU and IHF, cause the DNA to bend around the complex of DnaA proteins, which results in the separation of the strands at the AT-rich region. Following separation of the AT-rich region, the DnaA proteins, with the help of DnaC proteins, recruit the enzyme DNA helicase to this site. When a DNA helicase encounters a double-stranded region, it breaks the hydrogen bonds between the two strands, thereby generating two single strands. Two Page 259 DNA helicases begin strand separation within the oriC region and continue to separate the DNA strands beyond the origin. These proteins use the energy from ATP hydrolysis to catalyze the separation of the double-stranded parental DNA. In E. coli, DNA helicases bind to single-stranded DNA and travel along the DNA in a 5′ to 3′ direction to keep the replication fork moving. As shown in Figure 11.6, the action of DNA helicases promotes the movement of two replication forks outward from oriC in opposite directions. This initiates the replication of the bacterial chromosome in both directions, an event termed bidirectional replication.

For the Igf2 gene, where do de novo methylation and maintenance methylation occur? De novo methylation occurs in sperm, and maintenance methylation occurs in egg cells. De novo methylation occurs in egg cells, and maintenance methylation occurs in sperm cells. De novo methylation occurs in sperm, and maintenance methylation occurs in somatic cells of offspring. De novo methylation occurs in egg cells, and maintenance methylation occurs in somatic cells of offspring.

De novo methylation occurs in sperm, and maintenance methylation occurs in somatic cells of offspring.

Which of the following could be the components of a single nucleotide found in DNA?

Deoxyribose, phosphate, and thymine

Explain how epigenetic modifications are involved in developmental changes that lead to the formation of specific cell types.

During embryonic development, many genes undergo epigenetic changes that enable them to be transcribed or cause them to be permanently repressed. Such epigenetic changes are then transmitted during subsequent cell divisions. For example, an embryonic cell that will eventually give rise to muscle tissue is programmed to undergo epigenetic modifications that will enable the transcription of muscle-specific genes and repress the transcription of genes that should not be expressed in muscle cells. esearchers have discovered that two competing groups of protein complexes—the trithorax group (TrxG) and the polycomb group (PcG)—are key regulators of epigenetic changes that are programmed during development. TrxG and PcG complexes regulate many different genes, particularly those that encode transcription factors that control developmental changes and cell differentiation.

Outline the three stages of translation.

During initiation, the ribosomal subunits, mRNA, and the first tRNA assemble to form a complex. After the initiation complex is formed, the ribosome slides along the mRNA in the 5′ to 3′ direction, moving over the codons. This is the elongation stage of translation. As the ribosome moves, tRNA molecules sequentially bind to the mRNA at the A site in the ribosome, bringing with them the appropriate amino acids. Therefore, amino acids are linked in the order dictated by the codon sequence in the mRNA. Finally, a stop codon is reached, signaling the termination of translation. At this point, disassembly occurs, and the newly made polypeptide is released.

Describe the specificity between the amino acid carried by a tRNA and a codon in mRNA.

During mRNA-tRNA recognition, the anticodon in a tRNA molecule binds to a codon in mRNA in an antiparallel manner and according to the AU/GC rule (Figure 13.9). For example, if the anticodon in the tRNA is 3′- AAG-5′, it will bind to a 5′-UUC-3′ codon.

Describe the mechanism of genomic imprinting of the Igf2 gene in mammals.

During oogenesis, the lack of methylation allows CTCFs to promote the formation of a loop, which inhibits the expression of Igf2. During spermatogenesis, methylation prevents CTCFs from binding, thereby preventing the formation of this loop. When a loop does not form, the enhancer can activate the expression of Igf2.

Describe the structural features of DNA that enable it to be replicated.

During the replication process, the two complementary strands of DNA come apart and serve as template strands, or parental strands, for the synthesis of two new strands of DNA. After the double helix has separated, individual nucleotides have access to the template strands. Hydrogen bonding between individual nucleotides and the template strands must obey the AT/GC rule. To complete the replication process, a covalent bond is formed between the phosphate of one nucleotide and the sugar of the previous nucleotide.

Describe the structure of a chromosome territory.

Each chromosome in the cell nucleus is located in a discrete chromosome territory. As seen here, each chromosome occupies its own distinct territory. The binding of each chromosome to the nuclear matrix is thought to play a key role in forming these chromosome territories.

Define epigenetics and epigenetics inheritance.

Epigenetics is the study of mechanisms that lead to changes in gene expression that can be passed from cell to cell and are reversible, but do not involve a change in the sequence of DNA. This type of change may also be called an epimutation—a heritable change in gene expression that does not alter the sequence of DNA. In multicellular species that reproduce via gametes (i.e., sperm and egg cells), an epigenetic change that is passed from parent to offspring is called epigenetic inheritance, or transgenerational epigenetic inheritance. For example, as we learned in Chapter 5, genomic imprinting is an epigenetic change that is passed from parent to offspring. However, not all epigenetic changes fall into this category. For example, an organism may be exposed to an environmental agent in cigarette smoke that causes an epigenetic change in a lung cell that is subsequently transmitted from cell to cell and promotes lung cancer. Such a change would not be transmitted to offspring.

Explain how histone modifications affect transcription.

First, they may directly influence interactions within nucleosomes. For example, positively charged lysines within the core histone proteins can be acetylated by a type of enzyme called histone acetyltransferase. The attachment of the acetyl group (—COCH3) eliminates the positive charge on the lysine side chain, thereby disrupting the electrostatic attraction between the histone protein and the negatively charged DNA backbone The pattern of covalent modifications to the amino-terminal tails provides binding sites for proteins that subsequently affect the degree of transcription. One pattern of histone modification may attract proteins that inhibit transcription, which would silence the transcription of genes in the region. A different combination of histone modifications may attract proteins, such as chromatin-remodeling complexes, which would serve to alter the positions of nucleosomes in a way that promotes gene transcription. For example, the acetylation of histones attracts certain chromatin remodelers that shift nucleosomes or evict histone octamers, thereby aiding in the transcription of genes.

Compare and contrast different mechanisms of RNA splicing.

Group I and II introns are self-splicing. (a) The splicing of group I introns involves the binding of a free guanosine to a site within the intron, leading to the cleavage of RNA at the 3′ end of exon 1. The bond between a different nucleotide in the intron strand (in this case, guanine) and the 5′ end of exon 2 is cleaved. The 3′ end of exon 1 then forms a covalent bond with the 5′ end of exon 2. (b) In group II introns, a similar self-splicing mechanism occurs, except that the 2′ —OH group on an adenine nucleotide (already within the intron) begins the catalytic process. (c) Pre-mRNA splicing requires the aid of a multicomponent structure known as the spliceosome. Type of Intron Mechanism of Removal Occurrence Group I Self-splicing Found in rRNA genes within the nucleus of Tetrahymena and other simple eukaryotes. Found in a few protein-encoding, tRNA, and rRNA genes within mitochondrial DNA (in fungi and plants) and in chloroplast DNA. Found very rarely in tRNA genes within bacteria. Group II Self-splicing Found in a few protein-encoding, tRNA, and rRNA genes within mitochondrial DNA (in fungi and plants) and in chloroplast DNA. Also found rarely in bacterial genes. Pre-mRNA Spliceosome Very commonly found in protein-encoding genes within the nucleus of eukaryotic cells.

Explain how the ncRNA known as HOTAIR plays a role in gene repression.

HOTAIR acts as a scaffold for the binding of two histone-modifying complexes known as Polycomb Repressive Complex 2 (PRC2) and Lysine-Specific Demethylase 1 (LSD1). PRC2 binds to the 5′ end of HOTAIR, and LSD1 binds to the 3′ end. HOTAIR then guides PRC2 and LSD1 to a target gene by binding to a region near the gene that contain many purines, which is called a GA-rich region. For example, HOTAIR binds to a GA-rich region that is next to a specific HoxD gene on human chromosome 2. A portion of HOTAIR is complementary to this GA-rich region. he next event involves histone modifications. As described in Chapter 16, PRC2 functions as a histone methylase that trimethylates lysine 27 on histone H3. LSD1 forms a complex with other proteins and demethylates mono- and dimethylated lysines. In particular, it removes methyl groups from lysine 4 on histone H3. Although the mechanism of gene repression is not well understood, these histone modifications (H3K27 trimethylation and H3K4 demethylation) may inhibit transcription in two ways: The modifications may directly inhibit the ability of RNA polymerase to transcribe the target gene. For example, these histone modifications may prevent RNA polymerase from forming a preinitiation complex. Page 417 Rather than directly affecting transcription, the histone modifications carried out by PRC2 and LSD1 may attract other chromatin-modifying enzymes to the target gene. For example, they may attract a histone deacetylase to the target gene, which would foster a closed chromatin conformation.

Hox transcript antisense intergenic RNA (HOTAIR)

HOTAIR alters chromatin structure and thereby represses transcription by guiding histone-modifying complexes to target genes.

Describe how helicase, topoisomerase, and single-strand binding protein are important for the unwinding of the DNA double helix.

Helicase breaks the hydrogen bonds between the bases of the two strands topoisomeras II relieves tension from positive supercoiling single-strand binding proteins keep thee parental strands separate

Distinguish between cis- and trans-epigenetic mechanisms that maintain epigenetic changes.

In a cis-epigenetic mechanism, the epigenetic change at a given site is maintained only at that site; it does not affect the expression of the same gene located elsewhere in the cell nucleus By comparison, some epigenetic phenomena are explained by trans-epigenetic mechanisms that are maintained by diffusible molecules, such as proteins or ncRNAs. An example of a trans-epigenetic mechanism is a feedback loop. In this mechanism, an epigenetic change is established by activating a gene that encodes a transcription factor. After the transcription factor is initially made, it stimulates its own expression. Experimentally, researchers can distinguish between cis- and trans-epigenetic mechanisms by conducting cell-fusion experiments. In the example shown in Figure 16.3, one cell has gene B epigenetically modified so it is transcriptionally activated, whereas this same gene in another cell is inactive. When the cells are fused, two different outcomes are possible. According to a cis-epigenetic mechanism, the epigenetic modification will be maintained only for the copy of gene B that was originally modified. The other copy will remain inactive. This pattern will be maintained following cell division. By comparison, if a trans-epigenetic mechanism is operating, both copies of gene B will be expressed because the fused cell will contain enough of the transcription factor protein to stimulate both copies. This pattern will also be maintained following cell division.

Describe the organization of the lac operon.

In bacteria, it is common for a few genes to be arranged together in an operon—a group of two or more genes that are transcribed from a single promoter. An operon encodes a polycistronic mRNA, an RNA that contains the sequences of two or more genes. To facilitate transcription, an operon is flanked by a promoter that signals the beginning of transcription and a terminator that specifies the end of transcription. Two or more genes are found between these two sequences. One of these units, known as the lac operon, contains a CAP site; lac promoter (lacP); operator site (lacO); three protein-encoding genes, lacZ, lacY, and lacA; and a terminator. The lacZ gene encodes the enzyme β-galactosidase, an enzyme that cleaves lactose into galactose and glucose. As a side reaction, β-galactosidase also converts a small percentage of lactose into allolactose, a structurally similar sugar (Figure 14.3b). As we will see later, allolactose acts as a small effector molecule for regulating the lac operon. The lacY gene encodes lactose permease, a membrane protein required for the active transport of lactose into the cytoplasm of the bacterium. The lacA gene encodes galactoside transacetylase, an enzyme that covalently modifies lactose and lactose analogs by the attachment of hydrophobic acetyl groups.

Outline the structures the 30 nm fiber

In eukaryotic chromatin, nucleosomes associate with each other to form a more compact structure that is 30 nm in diameter. The 30-nm fiber shortens the total length of DNA another sevenfold. The structure of the 30-nm fiber has proven difficult to determine because the conformation of the DNA may be substantially altered when it is extracted from living cells. Most models for the structure of the 30-nm fiber fall into two main classes. The solenoid model suggests a helical structure in which contact between nucleosomes Page 241 produces a symmetrically compact structure within the 30-nm fiber (Figure 10.13b). This type of model is still favored by some researchers in the field. However, some experimental data suggest that the 30-nm fiber may not form such a regular structure. Instead, an alternative zigzag model, advocated by Rachel Horowitz, Christopher Woodcock, and others, is based on techniques such as cryoelectron microscopy (electron microscopy at low temperature). According to the zigzag model, linker regions within the 30-nm structure are variably bent and twisted, and little face-to-face contact occurs between nucleosomes (Figure 10.13c). At this level of compaction, the overall picture of chromatin that emerges is an irregular, fluctuating, three-dimensional zigzag structure with stable nucleosome units connected by deformable linker regions. In 2005, Timothy Richmond and colleagues were the first to solve the crystal structure of a segment of DNA containing multiple nucleosomes, in this case four. The structure with four nucleosomes revealed that the linker DNA zigzags back and forth between each nucleosome, a feature consistent with the zigzag model.

Describe the variation in size of eukaryotic genomes.

In many cases, this variation is not related to the complexity of the species. For example, two closely related species of salamander, Plethodon richmondi and Plethodon larselli, differ considerably in genome size

Analyze the results of Jacob, Monod, and Pardee, and explain how they indicated that the lacI gene encodes a diffusible repressor protein.

In the absence of lactose, the lac operons were repressed—even the operon on the bacterial chromosome. How do we explain these results? Because the normal lacI gene on the F′ factor was not physically located next to the chromosomal lac operon, this result is consistent with the idea that the lacI gene codes for a repressor protein that can diffuse throughout the cell and bind to any lac operon. The hypothesis that the lacI− mutation resulted in the synthesis of an internal activator was rejected. If that hypothesis had been correct, the merozygote strain would have still made an internal inducer, and the lac operons in the merozygote would have been expressed in the absence of lactose. This result was not obtained. The interactions between regulatory proteins and DNA sequences illustrated in this experiment led to the definition of two genetic terms. A trans-effect is a form of genetic regulation that can occur even though two DNA segments are not physically adjacent. The action of lac repressor on the lac operon is a trans-effect. A regulatory protein, such as lac repressor, is called a trans-acting factor.

Describe the four criteria that the genetic material must meet.

Information: The genetic material must contain the information necessary to construct an entire organism. In other words, it must provide the blueprint for determining the inherited traits of an organism. Transmission: During reproduction, the genetic material must be passed from parents to offspring. Replication: Because the genetic material is passed from parents to offspring, and from mother cell to daughter cells during cell division, it must be copied. Variation: Within any species, a significant amount of phenotypic variability occurs. For example, Mendel studied several characteristics in pea plants that varied among different strains. These included height (tall versus dwarf) and seed color (yellow versus green). Therefore, the genetic material must also vary in ways that can account for the known phenotypic differences within each species.

Sigma (σ) factor is needed during which stage(s) of transcription? Initiation Elongation Termination All of the above

Initiation

Outline the structures of nucleosomes, the 30-nm fiber, and radial loop domains.

Linear DNA Wraps Around Histone Proteins to Form Nucleosomes, the Repeating Structural Unit of Chromatin A nucleosome consists of 146 or 147 bp of DNA wrapped around an octamer of core histone proteinsEach octamer contains eight histone subunits: two copies each of four different histone proteins. The octamer of histones contains two molecules each of four different histone proteins: H2A, H2B, H3, and H4. These are called the core histone proteins The DNA is negatively supercoiled over the surface of this octamer; it makes 1.65 negative superhelical turns around a histone octamer. The amount of DNA required to wrap around the histone octamer is 146 or 147 bp. At its widest point, a single nucleosome is about 11 nm in diameter. Another histone, H1, is found in most eukaryotic cells and is called the linker histone. It binds to the DNA in the linker region between nucleosomes and may help to compact adjacent nucleosomes

Describe how DNA methylation is heritable.

Methylated DNA sequences are inherited during cell division. Experimentally, if fully methylated DNA is introduced into a plant or vertebrate cell, the DNA will remain fully methylated even in subsequently produced daughter cells. However, if the same sequence of nonmethylated DNA is introduced into a cell, it will remain nonmethylated in the daughter cells. These observations indicate that the pattern of methylation is retained following DNA replication and, therefore, is inherited in future daughter cells.

Define histone variant, and explain why histone variants are functionally important.

Most of the histone genes encode standard histone proteins. However, a few have accumulated mutations that change the amino acid sequence of the histone proteins. These altered histones are called histone variants. Among eukaryotic species, histone variants have been identified for H1, H2A, H2B, and H3, but not for H4. Certain histone variants play specialized roles in chromatin structure and function. In all eukaryotes, histone variants are incorporated into a subset of nucleosomes to create functionally specialized regions of chromatin. In most cases, the standard histones are incorporated into the nucleosomes while new DNA is synthesized during S phase of the cell cycle. Later, some of the standard histones are replaced by histone variants via chromatin-remodeling complexes.

On its chromosome, an E. coli cell has a genotype of lacI− lacZ+ lacY+ lacA+. It has an F′ factor with a genotype of lacI+ lacZ+ lacY+ lacA+. What is the expected level of expression of the lac operon genes (lacZ+ lacY+ lacA+) in the absence of lactose? Both lac operons will be expressed. Neither lac operon will be expressed. Only the chromosomal lac operon will be expressed. Only the lac operon on the F′ factor will be expressed.

Neither lac operon will be expressed.

Going from simple to complex, which of the following is the proper order for the structure of DNA?

Nucleotide, DNA strand, double helix, chromosome

Describe the four levels of complexity of DNA.

Nucleotides form the repeating structural unit of nucleic acids. Nucleotides are linked together in a linear manner to form a strand of DNA or RNA. Two strands of DNA (or sometimes strands of RNA) interact with each other to form a double helix. The three-dimensional structure of DNA results from the folding and bending of the double helix. Within living cells, DNA is associated with a wide variety of proteins that influence its structure. Chapter 10 examines the roles of these proteins in creating the three-dimensional structure of DNA found within chromosomes.

Outline the steps whereby glucocorticoid receptors and CREB proteins regulate genes.

Once inside the cell, the glucocorticoid hormone binds to the glucocorticoid receptor, which then releases a heat shock protein known as HSP90. This exposes a nuclear localization signal (NLS). Two glucocorticoid receptors then form a homodimer and travel into the nucleus, where the dimer binds to a glucocorticoid response element (GRE) that is next to a particular gene. The binding of the glucocorticoid receptors to the GRE activates the transcription of the adjacent target gene. An extracellular signaling molecule binds to a receptor in the plasma membrane, thereby activating a G protein, which then activates adenylyl cyclase, leading to the synthesis of cAMP. Next, cAMP binds to protein kinase A, which activates it. Protein kinase A then travels into the nucleus and phosphorylates the CREB protein. Once phosphorylated, CREB protein acts as a transcriptional activator by promoting the binding of CBP, which is a coactivator.

List the factors that contribute to combinatorial control.

One or more activator proteins may stimulate the ability of RNA polymerase to initiate transcription. One or more repressor proteins may inhibit the ability of RNA polymerase to initiate transcription. The function of activators and repressors may be modulated in a variety of ways, including the binding of small effector molecules, protein-protein interactions, and covalent modifications. Regulatory proteins may alter the composition or arrangements of nucleosomes in the vicinity of a promoter, thereby affecting transcription. DNA methylation may inhibit transcription, either by preventing the binding of an activator protein or by recruiting proteins that change the structure of chromatin in a way that inhibits transcription.

During the peptidyl transfer reaction, the polypeptide, which is attached to a tRNA in the _______, becomes bound via _______to an amino acid attached to a tRNA in the _______. A site, several hydrogen bonds, P site A site, a peptide bond, P site P site, a peptide bond, A site P site, several hydrogen bonds, A site

P site, a peptide bond, A site

Outline the functions of different DNA polymerases in eukaryotes.

Polymerase Types* Function α Initiates DNA replication in conjunction with primase ε Replication of the leading strand δ Replication of the lagging strand γ Replication of mitochondrial DNA η, κ, ι, ξ (lesion-replicating polymerases) Replication of damaged DNA α, β, δ, ε, σ, λ, μ, φ, θ, η DNA repair or other functions†

Outline how primase, DNA polymerase, and DNA ligase are needed to make strands of DNA at the replication fork.

Primase synthesizes an RNA primer DNA polymerase III synthesizes a daughter strand of DNA α Synthesizes DNA ε 3′ to 5′ proofreading (removes mismatched nucleotides) θ Accessory protein that stimulates the proofreading function β Clamp protein, which allows DNA polymerase to slide along the DNA without falling off τ, γ, δ, δ′, ψ, and χ Clamp loader complex, involved with helping the clamp protein bind to the DNA DNA polymerase I exises the RNA primers and fills it in with DNA DNA Ligase covalently links okazaki fragments together DNA polymerases II, IV, and V play a role in DNA repair and the replication of damaged DNA.

Analyze the experimental evidence that double-stranded RNA is more potent than antisense RNA at inhibiting mRNA

Prior to this work, researchers had often introduced antisense RNA into cells as a way to inhibit mRNA translation. Because antisense RNA is complementary to mRNA, the antisense RNA binds to the mRNA, thereby preventing translation. Oddly, in other experiments, researchers introduced sense RNA (RNA with the same sequence as mRNA) into cells instead, and, unexpectedly, this also inhibited mRNA expression. Another curious observation was that the effects of antisense RNA often persisted for a very long time, much longer than would have been predicted by the relatively short half-lives of most RNA molecules in the cell. These results indicated that the control embryos contained a high amount of mex-3 mRNA, which was known from previous research. In the embryos that had received antisense RNA, mex-3 mRNA levels were decreased, but detectable, as shown by light-green labeling. Remarkably, Page 419 in embryos that received double-stranded RNA, no mex-3 mRNA was detected! These results indicated that double-stranded RNA is more potent at silencing mRNA than is antisense RNA. In this case, the double-stranded RNA caused the mex-3 mRNA to be degraded. Fire and Mello used the term RNA interference (RNAi) to describe the phenomenon in which double-stranded RNA causes the silencing of mRNA.

Compare and contrast epigenetic changes that are programmed during development versus those that are caused by environmental agents.

Programmed Changes During Development Genomic imprinting Certain genes, such as the Igf2 gene discussed in Chapter 5, undergo different patterns of DNA methylation during oogenesis and spermatogenesis. Such patterns affect whether the maternal or paternal allele is expressed in offspring. X-chromosome inactivation As described in Chapter 5 and later in this chapter, X-chromosome inactivation occurs during embryogenesis in female mammals. Cell differentiation The differentiation of cells into particular cell types involves epigenetic changes such as DNA methylation and covalent histone modification. Environmental Agents Temperature In some species of flowering plants, cold winter temperatures cause specific types of covalent histone modifications that affect the expression of specific genes the following spring. This process may be necessary for seed germination or flowering in the spring. Diet The different diets of queen and worker bees alter DNA methylation patterns, which affects the expression of many genes. Such effects may underlie the different body types of queen and worker bees. Toxins Cigarette smoke contains a variety of toxins that affect DNA methylation and covalent histone modifications in lung cells. These epigenetic changes may play a role in the development of lung cancer. In addition, metals, such as cadmium and nickel, and certain chemicals found in pesticides and herbicides, cause epigenetic changes that can affect gene expression.

Which of the following base sequences is used during transcription? Promoter and terminator Start and stop codons Ribosome-binding site Both a and b

Promoter and terminator

Define repetitive sequence and explain how this type of sequence affects genome sizes.

Protein-encoding genes are typically unique sequences of DNA In the case of humans, unique sequences make up roughly 41% of the entire genome (Figure 10.9). These unique sequences include the protein-encoding regions of genes (2%), introns (24%), and unique regions that are not found within genes (15%). Moderately repetitive sequences are found a few hundred to several thousand times in a genome. By comparison, some moderately repetitive sequences do not play a functional role and are derived from transposable elements (TEs)—short segments of DNA that have the ability to move within a genome. This category of repetitive sequences is discussed in greater detail in Chapter 20. Highly repetitive sequences are found tens of thousands or even millions of times throughout a genome.Since that time, this gene has become a type of TE called a retroelement, which is transcribed into RNA, copied into DNA, and inserted into the genome. Over the past 65 million years, the Alu sequence has been copied and inserted into the human genome many times and is now present in about 1,000,000 copies.

List the functions of the three types of RNA polymerases in eukaryotes.

RNA polymerase I: transcribes all of the genes for ribosomal RNA (rRNA) except for the 5S rRNA. RNA polymerase II: transcribes all protein-encoding genes. Therefore, it is responsible for the synthesis of all mRNAs. It also transcribes the genes for most snRNAs which are needed for RNA splicing (discussed later in this chapter). In addition, it transcribes several types of genes that produce other non-coding RNAs (described in Chapter 17), including most long non-coding RNAs, microRNAs, and snoRNAs. RNA polymerase III: transcribes Page 287 all tRNA genes and the 5S rRNA gene. To a much lesser extent than RNA polymerase II, it also transcribes a few genes that produce other non-coding RNAs, such as snRNAs, long non-coding RNAs, microRNAs, and snoRNAs.

Which RNA polymerase in eukaryotes is responsible for the transcription of genes that encode proteins? RNA polymerase I RNA polymerase II RNA polymerase III All of the above

RNA polymerase II

RNaseP RNA

RNaseP is involved in the processing of tRNA molecules. The RNaseP RNA is the catalytic component.

Explain how small effector molecules affect the function of activators and repressors.

Rather, an effector molecule exerts its effects by binding to a repressor or activator. The binding of the effector molecule causes a conformational change in the regulatory protein and thereby influences whether or not the protein can bind to the DNA. Genetic regulatory proteins that respond to small effector molecules typically have two binding sites. One site is where the protein binds to the DNA; the other is the binding site for the effector molecule. An inducer is a small effector molecule that causes transcription to increase. An inducer may accomplish this in two ways: it can bind to a repressor protein and prevent it from binding to the DNA, or it can bind to an activator protein and cause it to bind to the DNA. In either case, the transcription rate is increased. Genes that are regulated in this manner are called inducible genes. A corepressor is a small molecule that binds to a repressor protein, thereby causing the protein to bind to the DNA. An inhibitor binds to an activator protein and prevents it from binding to the DNA. Both corepressors and inhibitors act to reduce the rate of transcription. Therefore, the genes they regulate are termed repressible genes.

topoisomerase I

Relaxes negative supercoils. This enzyme can bind to a negatively supercoiled region and introduce a break in one of the DNA strands. After one DNA strand has been broken, the DNA molecule rotates to relieve the tension that is caused by negative supercoiling. This rotation relaxes negative supercoiling. The broken strand is then repaired. The competing actions of DNA gyrase and topoisomerase I govern the overall supercoiling of the bacterial DNA.

Describe how DNA replication is terminated.

Replication Is Terminated When the Replication Forks Meet at the Termination Sequences On the opposite side of the E. coli chromosome from oriC is a pair of termination sequences, known as ter sequences. A protein known as the termination utilization Page 264 substance (Tus) binds to the ter sequences and stops the movement of the replication forks. one of the ter sequences, designated T1, prevents the advancement of the fork that is moving left to right, but allows the movement of the other fork (see the inset in Figure 11.13). Alternatively, T2 prevents the advancement of the fork that is moving right to left, but allows the advancement of the other fork. In any given cell, only one ter sequence is required to stop the advancement of one replication fork, and then the other fork ends its synthesis of DNA when it reaches the halted fork. In other words, DNA replication ends when oppositely advancing forks meet, usually at T1 or T2. Finally, DNA ligase covalently links the two daughter strands, creating two circular, double-stranded molecules.

Outline the general functions of ncRNAs.

Scaffold Some ncRNAs contain binding sites for multiple components, such as a group of different proteins. In this way, an ncRNA can act as a scaffold for the formation of a complex, as shown in Figure 17.1b. Guide Another function of having multiple binding sites is that some ncRNAs can guide one molecule to a specific location in a cell. For example, an ncRNA may bind to a protein and guide it to a specific site in the cell's DNA that is part of a particular gene. This specificity occurs because the ncRNA has a binding site for the protein and another binding site for the DNA. Alteration of Protein Function or Stability When an ncRNA binds to a protein, it can alter that protein's structure, which in turn, can have a variety of effects. The binding of the ncRNA may affect: the ability of the protein to act as a catalyst the ability of the protein to bind to another molecule, such as another protein, DNA, or RNA the stability of the protein Ribozyme Another interesting feature of some ncRNAs is that they function as ribozymes, which are RNA molecules with catalytic function. For example, in Chapter 12, we considered RNaseP (refer back to Figure 12.17). RNaseP recognizes tRNA molecules. The RNA component of RNaseP functions as an endonuclease that cleaves a tRNA, reducing its size. Blocker An ncRNA may physically prevent or block a cellular process from happening. For example, as described in Chapter 14, an antisense RNA is a type of ncRNA that is complementary to an mRNA. When an antisense RNA binds to an mRNA in the vicinity of the start codon, it blocks the ability of the mRNA to be translated Decoy Some ncRNAs recognize other ncRNAs and sequester them, thereby preventing them from working. For example, a decoy ncRNA may bind to a different ncRNA called a microRNA (miRNA), which is described later in this chapter. The function of an miRNA is to inhibit the translation of a particular mRNA. However, as shown below, if a decoy RNA binds to an miRNA, it is unable to carry out its function. When multiple decoy ncRNAs are found in a cell, they collectively act like a sponge by binding to the miRNAs and preventing them from functioning.

Describe the ability of ncRNAs to bind to other molecules and macromolecules.

Some ncRNAs bind to DNA or to another RNA through complementary base pairing. This binding allows the ncRNAs to affect processes such as transcription and translation. In addition, ncRNAs can bind to proteins or small molecules.

The Fidelity of DNA Replication Is Ensured by Three Mechanisms

Stability of Base Pairing The hydrogen bonding between G and C or between A and T is much more stable than that between mismatched pairs of bases. Due to this stability alone, only 1 mistake per 1000 nucleotides would be made. Structure of the Active Site of DNA Polymerase The active site of DNA polymerase preferentially catalyzes the attachment of nucleotides when the correct bases are located in opposite strands. Helix distortions caused by mispairing usually prevent an incorrect nucleotide from properly occupying the active site of DNA polymerase. By comparison, the correct nucleotide occupies the active site with precision and promotes induced fit, which is a conformational change in the enzyme that is necessary for catalysis. The inability of incorrect nucleotides to promote induced fit decreases the error rate to a range from 1 in 100,000 to 1 in 1 million. Proofreading DNA polymerase decreases the error rate even further by the enzymatic removal of mismatched nucleotides. As shown in Figure 11.17, DNA polymerase can identify a mismatched nucleotide and remove it from the daughter strand. This occurs by exonuclease cleavage of the bonds between adjacent nucleotides at the 3′ end of the newly made strand. The ability to remove mismatched bases by this mechanism is called the proofreading function of DNA polymerase. Proofreading occurs by the removal of nucleotides in the 3′ to 5′ direction at the 3′ exonuclease site. After the mismatched nucleotide is removed, DNA polymerase resumes DNA synthesis in the 5′ to 3′ direction.

Telomerase RNA component (TERC)

TERC facilitates the binding of telomerase to the telomere and acts as a template for DNA replication.

Explain how general transcription factors and RNA polymerase assemble at the promoter and form an open complex.

TFIID binds to the TATA box. TFIID is a complex of proteins that includes the TATA binding protein (TBP) and several TBP associated factors (TAFs). TFIIB and TFIID. TFIIB promotes the binding of RNA polymerase II to the core promoter TFIIF is bound to RNA polymerase II. TFIIE and TFIIH bind to RNA polymerase II to form a preinitiation or closed complex. TFIIH acts as a helicase to form an open complex and phosphorylate the CTD of RNA polymerase II. CTD phosphorylation breaks the contact between TFIIB and RNA polymerase II. TFIIB, TFIIE, and TFIIH are released.

Explain how a regulatory transcription factor exerts its effects via TFIID or mediator.

TFIID is a general transcription factor that binds to the TATA box and is needed to recruit RNA polymerase II to the core promoter. Activator proteins can enhance the ability of TFIID to initiate transcription. One possibility is that activator proteins might help TFIID bind to the TATA box, or they might enhance the function of TFIID in a way that facilitates its ability to recruit RNA polymerase II. In some cases, activator proteins exert their effects by interacting with coactivators—proteins that increase the rate of transcription but do not directly bind to the DNA itself. This type of activation is shown in Figure 15.4a. Coactivators typically contain a transactivation domain that promotes the activation of RNA polymerase, often by interacting with general transcription factors. mediator refers to the observation that this complex mediates the interaction between RNA polymerase II and regulatory transcription factors. As discussed in Chapter 12, mediator controls the ability of RNA polymerase II to progress to the elongation stage of transcription via phosphorylation of the carboxyl-terminal domain (CTD). Transcriptional activators stimulate the ability of mediator to cause the phosphorylation of the carboxyl-terminal domain, thereby facilitating the switch between the initiation and elongation stages. In contrast, repressors have the opposite effect. In the example shown in Figure 15.5a, an activator binds to a distant enhancer element. The activator protein and mediator are brought together by the formation of a loop within the intervening DNA. Alternatively, a repressor protein may prevent mediator from allowing RNA polymerase to proceed to the elongation phase of transcription

Telomeres

Telomeres serve several important functions in the replication and stability of the chromosome. As discussed in Chapter 8, telomeres prevent chromosomal rearrangements such as translocations. In addition, they prevent chromosome shortening in two ways. First, the telomeres protect chromosomes from digestion via enzymes called exonucleases that recognize the ends of DNA. Second, an unusual form of DNA replication occurs at the telomere to ensure that eukaryotic chromosomes do not become shortened with each round of DNA replication

Describe the key features of a bacterial origin of replication.

The GATC methylation sites within oriC are involved with regulating DNA replication. These sites are methylated by an enzyme known as DNA adenine methyltransferase (Dam). Prior to DNA replication, the GATC sites are methylated in both strands. This full methylation facilitates the initiation of DNA replication at the origin. Following DNA replication, the newly made strands are not methylated, because adenine rather than methyladenine is incorporated into the daughter strands. The initiation of DNA replication at the origin does not readily occur until after it has become fully methylated. Because it takes several minutes for Dam to methylate the GATC sites within this region, DNA replication does not occur again too quickly.

Summarize the steps that occur for transcriptional activation of a eukaryotic gene.

The NFR at the transcriptional start site is flanked by two well-positioned nucleosomes that are termed the −1 and +1 nucleosomes. In yeast, the transcriptional start site (TSS) is usually at the boundary between the NFR and the +1 nucleosome. However, in animals, the TSS is about 60 bp farther upstream and within the NFR. The +1 nucleosome typically contains histone variants H2A.Z and H3.3. The nucleosomes downstream from the +1 nucleosome tend to be evenly spaced near the beginning of a eukaryotic gene, but their spacing becomes less regular farther downstream. The ends of many eukaryotic genes appear to have a well-positioned nucleosome that is followed by an NFR. This arrangement at the ends of genes may be important for transcriptional termination. a transcriptional activator binds to an enhancer in the NFR. The activator then recruits a chromatin-remodeling complex and a histone-modifying enzyme to this region. The chromatin remodelers may shift nucleosomes or temporarily evict histone octamers from the promoter region. Further changes in chromatin structure are necessary for elongation to occur. RNA polymerase II cannot transcribe DNA that is tightly wrapped in nucleosomes. For transcription to occur, histone octamers are evicted or destabilized so that RNA polymerase II can move along the DNA. Histone-modifying enzymes also play a key role in histone removal and replacement during the elongation phase of transcription. Histone-modifying enzymes have been found to travel with RNA polymerase II during the elongation phase of transcription.

Outline the steps of RNA interference.

The RNAs that promote interference are of two types: microRNAs and small-interfering RNAs . MicroRNAs (miRNAs) are ncRNAs that are transcribed from endogenous eukaryotic genes—genes that are normally found in the genome. They play key roles in regulating gene expression, particularly during embryonic development in animals and plants. Most commonly, a single type of miRNA inhibits the translation of several different mRNAs. An miRNA and an mRNA bind to each other because they have sequences that are partially complementary. In humans, nearly 2000 genes encode miRNAs, though that number may be an underestimate. Research suggests that 60% of human protein-encoding genes are regulated by miRNAs. By comparison, small-interfering RNAs (siRNAs) are ncRNAs that usually originate from sources that are exogenous, which means they are not normally made by cells. The sources of siRNAs can be viruses that infect a cell, or researchers can make siRNAs to study gene function experimentally, as in Experiment 17A. In most cases, an siRNA is a perfect match to a single type of mRNA. The functioning of siRNAs is thought to play a key role in preventing certain types of viral infections. In addition, siRNAs have become important experimental tools in molecular biology.

Outline the three stages of transcription.

The Three Stages of Transcription Are Initiation, Elongation, and Termination Initiation: the promotor acts as a recognition site for transcription factors that enable RNA polymerase to bind the promoter. Then DNA is denatured to form a bubble complex Elongation: RNA polymerase slides along the DNA to synthesize RNA Termination: A terminator is reached that causes the RNA polymerrase and RNA transcript to dissociate

Outline how the information within DNA is used to make mRNA and a polypeptide.

The ability of mRNA to be translated into a specific sequence of amino acids relies on the genetic code. The sequence of bases within an mRNA molecule provides coded information that is read in groups of three nucleotides known as codons (see Figure 13.3). The sequence of three bases in most codons specifies a particular amino acid. These codons are termed sense codons.The codon AUG, which specifies methionine, is used as a start codon; it is usually the first codon that begins a polypeptide sequence. The AUG codon can also be used to specify additional methionines within the coding sequence. Finally, three codons, UAA, UAG, and UGA, which are known as stop codons, are used to end the process of translation. Stop codons are also known as termination codons or nonsense codons. The codons in mRNA are recognized by the anticodons in transfer RNA (tRNA) molecules (see Figure 13.3). Anticodons are three-nucleotide sequences that are complementary to codons in mRNA. The tRNA molecules carry the amino acids that are specified by the Page 310 codons in the mRNA. In this way, the order of codons in mRNA dictates the order of amino acids within a polypeptide.

Describe the characteristics of a eukaryotic promoter for a protein-encoding gene.

The core promoter is a relatively short DNA sequence that is necessary for transcription to take place. It typically consists of a TATAAA sequence called the TATA box and the transcriptional start site, where transcription begins. The TATA box, which is usually about 25 bp upstream from a transcriptional start site, is important in determining the precise starting point for transcription. If it is missing from the core promoter, the transcription start site becomes undefined, and transcription may start at a variety of different locations. The core promoter, by itself, produces a low level of transcription. This is termed basal transcription.

Which of the following characteristics is typical of a eukaryotic gene that can be transcribed? The core promoter is wrapped around a nucleosome. The core promoter is found in a nucleosome-free region. The terminator is wrapped around a nucleosome. None of the above characteristics is typical of such a gene.

The core promoter is found in a nucleosome-free region.

The fourth codon in an mRNA sequence is GGG, which specifies glycine. If we assume that no amino acids are removed from the polypeptide, which of the following statements is correct? The third amino acid from the N-terminus is glycine. The fourth amino acid from the N-terminus is glycine. The third amino acid from the C-terminus is glycine. The fourth amino acid from the C-terminus is glycine.

The fourth amino acid from the N-terminus is glycine.

trans-acting factors

The transcription factors that control the expression of a gene are themselves encoded by genes; regulatory genes that encode transcription factors may be far away from the genes they control, even on a different chromosome. When a gene encoding a trans-acting factor is expressed, the transcription factor protein can diffuse into the cell nucleus and bind to its appropriate cis-acting element.

Describe the processing of ribosomal RNAs and tRNAs.

The large ribosomal RNA gene is transcribed into a long 45S rRNA primary transcript. This transcript is cleaved to produce 18S, 5.8S, and 28S rRNA molecules, which become associated with protein subunits in the ribosome. This processing occurs within the nucleolus of the cell. The production of tRNA molecules requires processing via exonucleases and endonucleases. An exonuclease cleaves a covalent bond between two nucleotides at one end of a strand. Starting at one end, an exonuclease digests a strand, one nucleotide at a time. Some exonucleases begin this digestion only from the 3′ end, traveling in the 3′ to 5′ direction, whereas others begin only at the 5′ end and digest in the 5′ to 3′ direction. By comparison, an endonuclease cleaves the bond between two adjacent nucleotides within a strand.

Outline the processes that make a bacterial chromosome more compact.

The loops that emanate from the core, which are called microdomains, are typically 10,000 base pairs (10 kbp) in length. An E. coli chromosome is expected to have about 400 to 500 hundred of them. The lengths and boundaries of these microdomains are thought to be dynamic, changing in response to environmental conditions. In E. coli, many adjacent microdomains are further organized into macrodomains that are about 800 to 1000 kbp in length; each macrodomain contains about 80 to 100 microdomains. The macrodomains are not evident in Figure 10.3. To form micro- and macrodomains, bacteria use a set of DNA-binding proteins called nucleoid-associated proteins (NAPs) that facilitate chromosome compaction and organization. These proteins either bend the DNA or act as bridges that cause different regions of DNA to bind to each other.

Outline the procedure of chromatin immunoprecipitation sequencing, and explain how it was used to identify nucleosome-free regions that flank eukaryotic genes.

The mapping has allowed the determination of (1) where nucleosomes are located, (2) where histone variants are found, and (3) where covalent modifications of histones occur. This amazing achievement has been possible in large part by using an approach called chromatin immunoprecipitation sequencing (ChIP-Seq). As shown in Figure 15.11, ChIP-Seq begins with living cells. The cells are treated with formaldehyde, which covalently links proteins to the DNA. Such a linkage is called a crosslink. Next, the cells are broken open, and the chromatin is exposed to a high concentration of micrococcal nuclease (MNase), which is an enzyme that digests DNA. MNase digests the linker regions between nucleosomes, but it is unable to cleave DNA that is attached to the core histone proteins. Following this digestion, what remains is a collection of millions of nucleosomes that have DNA attached to them. The nucleosomes can be removed from this mixture using antibodies that specifically recognize histone proteins. All nucleosomes are removed if the antibody recognizes a histone such as H4, which is found in all nucleosomes and does not exist as a histone variant. Alternatively, specific types of nucleosomes can be removed from this mixture using antibodies that recognize particular histones. The antibodies, which are attached to heavy beads, are added, and they recognize specific nucleosomes. The material is then centrifuged and the DNA and proteins attached to the beads form a pellet at the bottom of the centrifuge tube. This step is called immunoprecipitation. Next, the crosslinks between the histones and DNA are broken, and the addition of proteases partially digests the core histones, thereby removing the histone octamers from DNA. The DNA is then subjected to gel electrophoresis. Only those DNA fragments that are about 150 bp in length are saved. This is the length of DNA that wraps around one nucleosome.

Outline the types of molecular changes that underlie epigenetic gene regulation.

The most common types of molecular changes that underlie epigenetic control are DNA methylation, chromatin remodeling, covalent histone modification, the localization of histone variants, and feedback loops DNA methylation Methyl groups may be attached to cytosine bases in DNA. When methylation occurs near promoters, transcription is usually inhibited. Chromatin remodeling Nucleosomes may be moved to new locations or evicted. When such changes occur in the vicinity of promoters, the level of transcription may be altered. Also, larger-scale changes in chromatin structure may occur, such as those that happen during X-chromosome inactivation in female mammals. Covalent histone modification Specific amino acid side chains found in the amino-terminal tails of histones can be covalently modified. For example, they can be acetylated or phosphorylated. Such modifications may enhance or inhibit transcription. Localization of histone variants Histone variants may become localized to specific locations, such as near the promoters of genes, and affect transcription. Feedback loop The activation of a gene that encodes a transcription factor may result in a feedback loop in which that transcription factor continues to stimulate its own expression.

Describe the four levels of protein structure.

The primary structure of a polypeptide is its amino acid sequence. (b) Certain regions of a primary structure will fold into a secondary structure; the two types of secondary structures are called α helices and β sheets. (c) Regions of secondary structure are connected by irregularly shaped regions to form the tertiary structure of a polypeptide. (d) Some polypeptides associate with each other to form a protein with a quaternary structure.

List the components of the replisome.

The primosome is physically associated with two DNA polymerase holoenzymes to form a replisome. DNA polymerase III proteins act in concert to replicate the leading and lagging strands. The term dimeric DNA polymerase is used to describe two DNA polymerase holoenzymes that move as a unit during DNA replication. For this to occur, the lagging strand is looped out with respect to the DNA polymerase that synthesizes the lagging strand. This loop allows the lagging-strand polymerase to make DNA in a 5′ to 3′ direction yet move as a unit with the leading-strand polymerase. Interestingly, when the lagging-strand polymerase reaches the end of an Okazaki fragment, it must be released from the template DNA and "hop" to the RNA primer that is closest to the fork. The clamp loader complex (see Table 11.2), which is part of DNA polymerase holoenzyme, then reloads the holoenzyme at the site where the next RNA primer has been made. Similarly, after primase synthesizes an RNA primer in the 5′ to 3′ direction, it must hop over the primer and synthesize the next primer closer to the replication fork.

What is the genetic code? The relationship between a three-base codon sequence and an amino acid or the end of translation The entire base sequence of an mRNA molecule The entire sequence from the promoter to the terminator of a gene The binding of tRNA to mRNA

The relationship between a three-base codon sequence and an amino acid or the end of translation

Outline the structural features of ribosomes.

The ribosome can be thought of as the macromolecular arena where translation takes place. Bacterial cells have one type of ribosome that is found within the cytoplasm. Eukaryotic cells contain biochemically distinct ribosomes in different cellular locations Unless otherwise noted, the term eukaryotic ribosome refers to ribosomes in the cytosol, not to those found within organelles. Likewise, the description of eukaryotic translation refers to translation via cytosolic ribosomes. Each ribosome is composed of structures called the large and small subunits. This term is perhaps misleading because each ribosomal subunit itself is formed from the assembly of many different proteins and RNA molecules called ribosomal RNA, or rRNA. Ribosomes contain discrete sites where tRNAs bind and the polypeptide is synthesized. In 1964, James Watson was the first to propose a two-site model for tRNA binding to the ribosome. These sites are known as the peptidyl site (P site) and aminoacyl site (A site).

Describe the evidence that the development of queen honeybees is due to exposure to royal jelly.

The striking differences between queens and worker bees are largely caused by differences in their diets. Certain worker bees, called nurse bees, produce a secretion called royal jelly from glands in their mouths. All female larvae are initially fed royal jelly, but those that are bathed in royal jelly throughout their entire larval development and fed it into adulthood become queens. In contrast, female larvae that are weaned at an early stage of development and switched to a diet of pollen and nectar become worker bees. Bee larvae were fed a diet that should produce worker bees. These larvae were injected with a substance that inhibits DNA methyltransferase. The result was that most of them became queen bees with fully developed ovaries! Page 403 While other factors may contribute to the development of queens, these results are consistent with the hypothesis that royal jelly may also contain a substance that inhibits DNA methylation. Such inhibition is thought to allow the expression of genes that contribute to the development of traits that are observed in queen bees.

Compare and contrast the structures of nucleotides found in DNA and in RNA.

The sugar in DNA is always deoxyribose. In RNA, the sugar is always ribose. Also, the base thymine is not found in RNA, which contains the base uracil instead of thymine. Adenine, guanine, and cytosine occur in both DNA and RNA. Nucleosides composed of ribose and guanine, cytosine, or uracil are named guanosine, cytidine, and uridine, respectively. Nucleosides made of deoxyribose and adenine, guanine, thymine or cytosine are called deoxyadenosine, deoxyguanosine, deoxythymidine, and deoxycytidine, respectively.

Describe the process of RNA editing.

The term RNA editing refers to the process of making a change Page 299 in the nucleotide sequence of an RNA molecule that involves additions or deletions of particular nucleotides or a conversion of one type of base to another, such as a cytosine to a uracil. In the case of mRNAs, editing can have various effects, such as generating start codons, generating stop codons, and changing the coding sequence for a polypeptide.

cis-acting elements.

The term cis comes from chemistry nomenclature meaning "next to." Cis-acting elements, though possibly far away from the core promoter, are always found within the same chromosome as the genes they regulate.

Analyze the results of Griffith and explain how they indicate that DNA is the genetic material.

The transformed bacteria acquired the information to make a capsule. Among different strains, variation exists in the ability to create a capsule and to cause mortality in mice. The genetic material that is necessary to create a capsule must be replicated so it can be transmitted from mother to daughter cells during cell division. Taken together, these observations are consistent with the idea that the formation of a capsule is governed by the bacteria's genetic material, meeting the four criteria described previously. Griffith's experiments showed that some genetic material from the dead bacteria had been transferred to the living bacteria and provided them with a new trait.

How do PRC1 complexes silence gene expression?

Three mechanisms have been proposed, which are not mutually exclusive. The first mechanism involves chromatin compaction. PRC1 can catalyze the aggregation of nucleosomes into a more compact knot-like structure, which would silence gene expression (see Figure 16.6). A second mechanism involves another covalent modification—the attachment of a ubiquitin molecule to histone H2A. Though this covalent modification is associated with the silencing of many genes, the molecular mechanism by which it represses transcription is not understood. Finally, a third possibility is a direct interaction with a transcription factor. In the model shown in Figure 16.6, PRC1 interacts with TFIID (described in Chapter 15), thereby inhibiting transcription.

Describe how chromatin-remodeling complexes alter nucleosomes.

Tne result of ATP-dependent chromatin remodeling is a change in the positions of nucleosomes (Figure 15.9a). This may involve shifts in nucleosomes to new locations or changes in the relative spacing of nucleosomes over a long stretch of DNA. A second effect is that remodelers may evict histone octamers from the DNA, thereby creating gaps where nucleosomes are not found (Figure 15.9b). A third possibility is that remodelers may change the composition of nucleosomes by removing standard histones and replacing them with histone variants

Compare and contrast two mechanisms for transcriptional termination in bacteria.

Transcription Is Terminated by Either an RNA-Binding Protein or an Intrinsic Terminator In ρ-dependent termination, the termination process requires two components. First, a site in the DNA, called the rut site (for rho utilization site), encodes a sequence in the RNA that acts as a recognition site for the binding of the ρ protein The ρ protein functions as a helicase, an enzyme that can separate RNA-DNA hybrid regions. After the rut site is synthesized in the RNA, ρ protein binds to the RNA and moves in the direction of RNA polymerase. The second component of ρ-dependent termination is the site where termination actually takes place. At this terminator site, the DNA encodes an RNA sequence containing several GC base pairs that form a stem-loop structure. RNA synthesis terminates several nucleotides beyond this stem-loop.his binding results in a conformational change that causes RNA polymerase to pause in its synthesis of RNA. The pause allows ρ protein to catch up to the stem-loop, pass through it, and break the hydrogen bonds between the DNA and RNA within the open complex. When this occurs, the completed RNA strand is separated from the DNA along with RNA polymerase. The process of ρ-independent termination does not require the ρ protein. In this case, the terminator involves two adjacent nucleotide sequences (Figure 12.11). One sequence promotes the formation of a stem-loop. The second sequence, which is downstream from the stem-loop, is a uracil-rich sequence located at the 3′ end of the RNA. As shown in Figure 12.11, the formation of the stem-loop causes RNA polymerase to pause in its synthesis of RNA. This pausing is stabilized by other proteins that bind to RNA polymerase. For example, a protein called NusA binds to RNA polymerase and promotes pausing at stem-loop sequences. At the precise time that RNA polymerase pauses, the uracil-rich sequence in the RNA transcript is bound to the DNA template strand. As previously mentioned, the hydrogen bonding of RNA to DNA keeps RNA polymerase clamped onto the DNA. However, the binding of this Page 285 uracil-rich sequence to the DNA template strand is relatively weak, causing the RNA transcript to spontaneously dissociate from the DNA and cease further transcription. Because this process does not require the ρ protein to physically remove the RNA transcript from the DNA, it is also referred to as intrinsic termination.

Explain the functions of condensin and cohesin.

Two multiprotein complexes called condensin and cohesin, which play a critical role in chromosomal condensation and sister chromatid alignment, respectively. oth contain a category of proteins called SMC proteins. SMC stands for structural maintenance of chromosomes. These proteins use energy from ATP to catalyze changes in chromosome structure. Together with topoisomerases, SMC complexes have been shown to promote major changes in DNA structure. An emerging theme is that SMC complexes actively fold, tether, and manipulate DNA strands. The monomers, which are connected at a hinge region, have Page 246 two long coiled arms with a head region that binds ATP (Figure 10.19). The length of each monomer is about 50 nm, which is equivalent to approximately 150 bp of DNA. As M phase begins, the nuclear envelope breaks apart and condensin is observed to coat the individual chromatids as the chromosomes become highly compacted. Although the mechanism of compaction is not entirely understood, recent evidence suggests that condensin proteins form a ring around the DNA. Compaction occurs because multiple condensin proteins bring chromatin loops closer together and hold them in place the function of cohesin is to promote the binding (i.e., cohesion) between sister chromatids. After S phase and until the middle of prophase, sister chromatids remain attached to each other along their length.

silencers

Under certain conditions, it may be necessary to prevent transcription of a given gene. This occurs via silencers—DNA sequences that are recognized Page 288 by transcription factors that inhibit transcription.

If a tRNA has an anticodon with the sequence 5′-CAG-3′, which amino acid does it carry? Aspartic acid Valine Leucine Glutamine

Valine

Explain how coat color in mice is epigenetically modified by dietary factors.

When DNA is methylated, DNA methyltransferase removes a methyl group from S-adenosyl methionine and transfers it to a cytosine base in DNA. A variety of nutrients can increase the synthesis of S-adenosyl methionine in cells. These include folic acid, vitamin B12, betaine, and choline chloride. Waterland and Jirtle divided female mice into a control group that was fed a normal diet and an experimental group that was fed a diet supplemented with folic acid, vitamin B12, betaine, and choline chloride. Both groups were fed their respective diets before and during pregnancy and up to the stage of weaning. Offspring carrying the Avy allele were then analyzed with regard to their coat color and levels of DNA methylation. As expected, a range of coat colors was observed among the offspring (Figure 16.8c). However, the offspring of females that had been fed a supplemented diet tended to have darker coats. For example, over 25% of the offspring with heavily mottled coats had mothers that were fed a supplemented diet (red bars), whereas less than 10% had mothers that were given a normal diet (blue bars).

Outline the key experiments that led to the discovery of the DNA double helix.

When a purified substance, such as DNA, is subjected to X-rays, it produces a well-defined diffraction pattern if the molecular structure has a regular pattern. An interpretation of the diffraction pattern (using mathematical theory) can ultimately provide information concerning the structure of the molecule.

Analyze the results of Avery, MacLeod, and McCarty and explain how they indicate that DNA is the genetic material.

When the DNA extracts were treated with RNase or protease, they still converted type R bacteria into type S. These results indicated that any RNA or protein in the extract was not acting as the genetic material. However, when the extract was treated with DNase, it lost its ability to convert type R into type S bacteria. These results indicated that the degradation of the DNA in the extract by DNase prevented conversion of type R to type S. This interpretation is consistent with the hypothesis that DNA is the genetic material. A more elegant way of saying this is that "DNA is the transforming principle."

Outline the process of X-chromosome inactivation.

XCI is an epigenetic phenomenon. During early embryonic development in female mammals, one of the X chromosomes in each somatic cell is randomly chosen for inactivation. After this occurs, the same X chromosome is maintained in an inactivated state during subsequent cell divisions (refer back to Figure 5.4). A region of the X chromosome called the X-inactivation center (Xic) plays a key role in this process (Figure 16.5a). Within the Xic are two genes called Xist, for X inactive-specific transcript, and Tsix. Xist is expressed from the inactivated X chromosome, whereas Tsix is expressed from the active X chromosome. The two genes are transcribed in opposite directions Xist RNA recruits protein complexes to Xi that cause covalent modifications of specific sites in histone tails. For example, Xist RNA forms a complex with PRC2, which is described later in this chapter (look ahead to Figure 16.6). The Xist-PRC2 complexes cover the whole X chromosome. PRC2 modifies histones, and these histone changes are recognized by other proteins that promote changes in chromatin structure. Collectively, covalent histone Page 395 modifications, the incorporation of macroH2A into nucleosomes, and the methylation of many CpG islands are Page 396 thought to play key roles in the silencing of genes on Xi and its compaction into a Barr body. These epigenetic changes in Xi are then maintained in subsequent cell divisions.

X inactive-specific transcript (Xist RNA)

Xist RNA coats one of the X chromosomes in female mammals and plays a role in its compaction and resulting inactivation.

For XCI to occur, where are the Xist and Tsix genes expressed? Xist is expressed only on Xa, and Tsix is expressed only on Xi. Xist is expressed only on Xi, and Tsix is expressed only on Xa. Xist is expressed only on Xa, and Tsix is expressed only on Xa. Xist is expressed only on Xi, and Tsix is expressed only on Xi.

Xist is expressed only on Xi, and Tsix is expressed only on Xa.

In a DNA strand, a phosphate connects a 3′ carbon atom in one deoxyribose to

a 5′ carbon in an adjacent deoxyribose.

Explain how the CRISPR-Cas system defends bacteria against bacteriophages.

a bacteriophage infects a bacteria cell. The Cas11 and Cas2 genes are expressed and form a complex taht cleaves the bacteriophage into small pieces. a piece is inserted into the Crispr gene Expression If a bacterial cell has already been adapted to a bacteriophage, a subsequent bacteriophage infection will result in the expression phage, in which the system gets ready for action by expressing the Crispr, tracr, and Cas9 genes (Figure 17.7c). The Crispr gene is transcribed from a single promoter and produces a long ncRNA called pre-crRNA. The gene encoding the tracrRNA is also transcribed, which produces many molecules of tracrRNA, another kind of ncRNA. A region of the tracrRNA is complementary to the repeat sequences of the pre-crRNA. Several molecules of tracrRNA base-pair with the pre-crRNA. The pre-crRNA is then cleaved into many small molecules, now called crRNAs. Each crRNA is attached to a tracrRNA. A region of the tracrRNA is Page 425 recognized by the Cas9 protein. The tracrRNA acts as a guide that causes the tracrRNA-crRNA complex to bind to a Cas9 protein. Interference After the tracrRNA-crRNA-Cas9 complex is formed, the bacterial cell is ready to destroy the bacteriophage DNA. This phase is called interference, because it resembles the process of RNA interference described earlier in this chapter (refer back to Figure 17.4). Each spacer within a crRNA is complementary to one Page 426 of the strands of the bacteriophage DNA. Therefore, the crRNA acts as a guide that causes the tracrRNA-crRNA-Cas9 complex to bind to that strand (Figure 17.7d). After binding, the Cas9 protein functions as an endonuclease that makes double-strand breaks in the bacteriophage DNA. This cleavage inactivates the phage, and thereby prevents its proliferation.

When mice carrying the Avy allele exhibit a darker coat, this phenotype is thought to be caused by dietary factors that result in a greater level of DNA methylation and a decrease in the expression of the Agouti gene. a lower level of DNA methylation and a decrease in the expression of the Agouti gene. a greater level of DNA methylation and the overexpression of the Agouti gene. a lower level of DNA methylation and the overexpression of the Agouti gene.

a greater level of DNA methylation and a decrease in the expression of the Agouti gene.

A release factor is referred to as a "molecular mimic" because its structure is similar to a ribosome. an mRNA. a tRNA. an elongation factor.

a tRNA.

In one cell, gene C is expressed, whereas in another cell, gene C is inactive. After the cells are fused experimentally, both copies of gene C are expressed. This observation could be explained by a cis-epigenetic mechanism. a trans-epigenetic mechanism. DNA methylation. both a and b.

a trans-epigenetic mechanism.

tandem arrays

also known as tandem repeats. In a tandem array, a very short nucleotide sequence is repeated many times in a row. In Drosophila, for example, 19% of the chromosomal DNA consists of highly repetitive sequences found in tandem arrays

Analyze the experiment of Meselson and Stahl and explain how the results were consistent with the semiconservative model of DNA replication.

after one round Page 256 of DNA replication (i.e., after one generation), all of the DNA sedimented at a density that was half-heavy. After approximately two rounds of DNA replication (i.e., after 1.9 generations), a mixture of light DNA and half-heavy DNA was observed. This result was consistent with the semiconservative model of DNA replication, because some DNA molecules should contain all light DNA, and other molecules should be half-heavy (see Figure 11.2b). The dispersive model predicts that after two generations, the heavy nitrogen would be evenly dispersed among four strands, each strand containing 1/4 heavy nitrogen and 3/4 light nitrogen (see Figure 11.2c). However, this result was not obtained. Instead, the results of the Meselson and Stahl experiment provided compelling evidence in favor of only the semiconservative model for DNA replication.

Which of the following can bind to ncRNAs? DNA RNA Proteins Small molecules All of the above

all

A bacterial chromosome typically contains a few thousand genes. one origin of replication. some repetitive sequences. all of the above.

all of the above

DNA gyrase promotes negative supercoiling. relaxes positive supercoils. cuts DNA strands as part of its function. does all of the above.

all of the above

The basal transcription apparatus is composed of five general transcription factors. RNA polymerase II. a DNA sequence containing a TATA box and transcriptional start site. all of the above.

all of the above

The chromosomes of eukaryotes typically contain a few hundred to several thousand different genes. multiple origins of replication. a centromere. telomeres at their ends. all of the above.

all of the above

The compaction leading to a metaphase chromosome involves which of the following? The formation of nucleosomes The formation of the 30-nm fiber Anchoring and further compaction of the radial loops All of the above

all of the above

Which of the following is not a key difference between bacteria and eukaryotes? Initiation of transcription requires more proteins in eukaryotes. A 7-methylguanosine cap is added only to eukaryotic mRNAs. Splicing is common in complex eukaryotes but not in bacteria. All of the above are key differences.

all of the above

Regulatory transcription factors can be modulated by the binding of small effector molecules. protein-protein interactions. covalent modifications. all of the above.

all of the above.

The origin of replication in E. coli contains an AT-rich region. DnaA box sequences. GATC methylation sites. all of the above.

all of the above.

The site(s) on a ribosome where tRNA molecules may be located include the A site. the P site. the E site. all of the above.

all of the above.

The binding of _______ to lac repressor causes lac repressor to _______ to the operator site, thereby _______ transcription. glucose, bind, inhibiting allolactose, bind, inhibiting glucose, not bind, increasing allolactose, not bind, increasing

allolactose, not bind, increasing

In the experiment of Avery, McLeod, and McCarty, the addition of RNase or protease to a DNA extract

allowed the conversion of type R bacteria into type S bacteria.

Outline how alternative splicing occurs, and describe its benefits.

alternative splicing, which refers to the phenomenon that a pre-mRNA can be spliced in more than one way. Alternative splicing produces two or more polypeptides from the same gene that have differences in their amino acid sequences, leading to possible changes in their functions. In most cases, the alternative versions of the protein have similar functions, because most of their amino acid sequences are identical to each other. Nevertheless, alternative splicing produces differences in amino acid sequences that Page 296 provide each polypeptide with its own unique characteristics. Because alternative splicing allows two or more different polypeptide sequences to be derived from a single gene, some geneticists have speculated that an important advantage of this process is that it allows an organism to carry fewer genes in its genome.

An enzyme known as _______attaches an amino acid to the _______of a tRNA, thereby producing _______. aminoacyl-tRNA synthetase, anticodon, a charged tRNA aminoacyl-tRNA synthetase, 3′ single-stranded region of the acceptor stem, a charged tRNA polynucleotide phosphorylase, anticodon, a charged tRNA polynucleotide phosphorylase, anticodon, an aminoacyl-tRNA

aminoacyl-tRNA synthetase, 3′ single-stranded region of the acceptor stem, a charged tRNA

Explain how riboswitches can regulate transcription and translation.

an RNA molecule can exist in two different secondary conformations. The conversion from one conformation to the other is due to the binding of a small molecule. As described in Table 14.2, a riboswitch can regulate transcription, translation, RNA stability, and splicing. Transcription The 5´ region of an mRNA may exist in one conformation that forms a ρ-independent terminator, which causes attenuation of transcription. The other conformation does not form a terminator and is completely transcribed. Translation The 5´ region of an mRNA may exist in one conformation in which the Shine-Dalgarno sequence cannot be recognized by the ribosome, whereas the other conformation has an accessible Shine-Dalgarno sequence that allows the mRNA to be translated. RNA stability One mRNA conformation may be stable, whereas the other conformation acts as ribozyme that causes self-degradation. Splicing In eukaryotes, one pre-mRNA conformation may be spliced in one way, whereas another conformation is spliced in a different way.

A ribozyme is a complex between RNA and a protein. an RNA that encodes a protein that functions as an enzyme. an RNA molecule with catalytic function. a protein that degrades RNA molecules.

an RNA molecule with catalytic function.

A site in a chromosome where DNA replication begins is a promoter. an origin of replication. an operator. a replication fork.

an origin of replication.

A regulatory transcription factor protein typically contains _________ that binds to the ________ of the DNA. an α helix, backbone an α helix, major groove a β sheet, backbone a β sheet, major groove

an α helix, major groove

An epigenetic modification to a specific gene may initially be established by a transcription factor. a non-coding RNA. both a and b. none of the above.

both a and b.

In the lagging strand, DNA is made in the direction _________ the replication fork and is made as __________. toward, one continuous strand away from, one continuous strand toward, Okazaki fragments away from, Okazaki fragments

away from, Okazaki fragments

For the following five sequences, what is the consensus sequence? 5′-GGGAGCG-3′ 5′-GAGAGCG-3′ 5′-GAGTGCG-3′ 5′-GAGAACG-3′ 5′-GAGAGCA-3′ a 5′-GGGAGCG-3′ b 5′-GAGAGCG-3′ c 5′-GAGTGCG-3′ d 5′-GAGAACG-3′

b 5′-GAGAGCG-3′

The proofreading function of DNA polymerase involves the recognition of a ________ and the removal of a short segment of DNA in the __________ direction. missing base, 5′ to 3′ base pair mismatch, 5′ to 3′ missing base, 3′ to 5′ base pair mismatch, 3′ to 5′

base pair mismatch, 3′ to 5′

Describe the organization of a protein-encoding gene and its mRNA transcript.

base sequences within an mRNA are used during the translation process. In bacteria, a short sequence within the mRNA, the ribosome-binding site (also known as the Shine-Dalgarno sequence), provides a location for a ribosome to bind and begin translation. The bacterial ribosome recognizes this site because it is complementary to a sequence in ribosomal RNA. In addition, mRNA contains a series of codons, read as groups of three nucleotides, which contain the information for a polypeptide's sequence. The first codon, which is very close to the ribosome-binding site, is the start codon. This is followed by many more codons that dictate the sequence of amino acids within the synthesized polypeptide. Finally, a stop codon signals the end of translation.

Epigenetic changes may be programmed during development. be caused by environmental changes. involve changes in the DNA sequence of a gene. be both a and b.

be both a and b.

Describe three ways that the function of a regulatory transcription factor can be modulated.

binding of a small effector molecule (hormone) protein-protein interaction covalent modifictation (phosphorylation) Usually, one or more of these modulating effects are important in determining whether a transcription factor binds to the DNA or influences transcription by RNA polymerase.

The process of RNA interference may lead to the degradation of an mRNA. the inhibition of translation of an mRNA. the synthesis of an mRNA. both a and b

both a and b

With regard to the 30-nm fiber, a key difference between the solenoid and zigzag models is the solenoid model suggests a helical structure. the zigzag model suggests a more irregular pattern of nucleosomes. the zigzag model does not include nucleosomes. both a and b are correct.

both a and b are correct.

A type of secondary structure found in proteins is an α helix. a β sheet. both a and b. none of the above.

both a and b.

An enhancer is a _____________ that ___________ the rate of transcription. trans-acting factor, increases trans-acting factor, decreases cis-acting element, increases cis-acting element, decreases

cis-acting element, increases

In the CRISPR-Cas system, what does the tracrRNA bind to? crRNA and Cas1 crRNA and Cas2 crRNA and Cas9 Cas1 and Cas2

crRNA and Cas9

In eukaryotes, RNA primers are primarily removed by DNA polymerase I. DNA polymerase α. flap endonuclease. helicase.

flap endonuclease

euchromatin

he less compacted regions, known as euchromatin, usually are capable of gene transcription. In euchromatin, the 30-nm fiber forms radial loop domains.

A chromosome territory is a region along a chromosome where many genes are clustered. along a chromosome where the nucleosomes are close together. in a cell nucleus where a single chromosome is located. in a cell nucleus where multiple chromosomes are located.

in a cell nucleus where a single chromosome is located.

Distinguish between general and regulatory transcription factors.

general transcription factors, which are required for the binding of RNA polymerase to the core promoter and its progression to the elongation stage. General transcription factors are necessary for any transcription to occur. In addition, eukaryotic cells possess a diverse array of regulatory transcription factors that serve to regulate the rate of transcription of target genes.

For a riboswitch that controls transcription, the binding of a small molecule such as TPP controls whether the RNA has an antiterminator or terminator stem-loop. has a Shine-Dalgarno antisequestor or the Shine-Dalgarno sequence within a stem-loop. is degraded from its 5′ end. has an antiterminator or terminator stem-loop and has a Shine-Dalgarno antisequestor or the Shine-Dalgarno sequence within a stem-loop.

has an antiterminator or terminator stem-loop.

Describe the levels of compaction that lead to a metaphase chromosome.

he first level involves the wrapping of DNA around histone octamers to form nucleosomes. The next level is the formation of a 30-nm fiber in which nucleosomes form a zigzag or solenoid structure via binding of histone H1 and nonhistone proteins to the DNA (Figure 10.17b). The 30-nm fiber then forms radial loop domains by anchoring to the protein filaments of the nuclear matrix. The average distance that the loops radiate from the protein filaments is approximately 300 nm. When cells prepare to divide, the protein filaments come closer together and form a more compact scaffold for anchoring the radial loops (Figure 10.17d). This additional level of compaction greatly shortens Page 245 the overall length of a chromosome and produces a diameter of approximately 700 nm. By the end of prophase, sister chromatids are highly compacted. Two parallel chromatids have a diameter of approximately 1400 nm and a much shorter length than interphase chromosomes. The overall size of a metaphase chromosome is much smaller than a chromosome territory

The enzyme known as ________ uses ________ and separates the DNA strands at the replication fork. helicase, ATP helicase, GTP gyrase, ATP gyrase, GTP

helicase, ATP

Describe how triplex DNA is formed.

his structure can form in vitro when natural double-stranded DNA and a third short strand that is synthetically made are mixed. The synthetic strand binds into the major groove of the naturally occurring double-stranded DNA n other words, the synthetic third strand incorporates itself into the triple helix due to specific interactions between its bases and those of the biological DNA. The pairing rules are that a thymine in the synthetic DNA hydrogen bonds at an AT pair in the biological DNA and a cytosine in the synthetic DNA hydrogen bonds at a GC pair.

The role of cohesin is to make chromosomes more compact. allow for the replication of chromosomes. hold sister chromatids together. promote the separation of sister chromatids.

hold sister chromatids together.

According to the histone code hypothesis, the pattern of histone modifications acts like a language that influences chromatin structure. promotes transcriptional termination. inhibits the elongation of RNA polymerase. does all of the above.

influences chromatin structure.

The binding of iron regulatory protein (IRP) to the iron response element (IRE) in the 5′ region of the ferritin mRNA results in the inhibition of translation of the ferritin mRNA. stimulation of translation of the ferritin mRNA. degradation of the ferritin mRNA. both a and c.

inhibition of translation of the ferritin mRNA.

The three stages of transcription are initiation, ribosome binding, and termination. elongation, ribosome binding, and termination. initiation, elongation, and termination. initiation, regulation, and termination.

initiation, elongation, and termination.

Describe the steps that occur during the initiation, elongation, and termination stages of translation.

initiation: he mRNA, tRNAfMet, and ribosomal subunits associate with each other to form an initiation complex. The formation of this complex requires the participation of three initiation factors: IF1, IF2, and IF3. First, IF1 and IF3 bind to the 30S ribosomal subunit, which prevents the association of the 50S subunit. Next, the mRNA binds to the 30S subunit. This binding is facilitated by a nine-nucleotide sequence within the bacterial mRNA called the Shine-Dalgarno sequence. The location of this sequence is shown in Figure 13.15 and in more detail in Figure 13.16. How does the Shine-Dalgarno sequence facilitate the binding of mRNA to the ribosome? The Shine-Dalgarno sequence is complementary to a short sequence within the 16S rRNA, which promotes the hydrogen bonding of the mRNA to the 30S subunit. elongation: Binding of a Charged tRNA to the A Site To begin elongation, a charged tRNA brings a new amino acid to the ribosome so it can be attached to the end of the growing polypeptide., The next step of elongation is a reaction called peptidyl transfer—the polypeptide is removed from the tRNA in the P site and transferred to the amino acid at the A site. After the peptidyl transfer reaction is complete, the ribosome moves, or translocates, to the next codon in the mRNA. This moves the tRNAs at the P and A sites to the E and P sites, respectively. Finally, the uncharged tRNA exits the E site. termination: At the top of this figure, the completed polypeptide is attached to a tRNA in the P site. A stop codon is located at the A site. In the first step, RF1 or RF2 binds to the stop codon at the A site and RF3 (not shown) binds at a different location on the ribosome. After RF1 (or RF2) and RF3 have bound, the bond between the polypeptide and the tRNA is hydrolyzed. The polypeptide and tRNA are then released from the ribosome. The final step in translational termination is the disassembly of ribosomal subunits, mRNA, and the release factors.

DNA polymerase III is a processive enzyme, which means that it does not dissociate from the growing strand after it has attached a nucleotide to the 3′ end. it makes a new strand very quickly. it proceeds toward the opening of the replication fork. it copies DNA with relatively few errors.

it does not dissociate from the growing strand after it has attached a nucleotide to the 3′ end.

Explain how the lac operon is regulated by lac repressor and by catabolite activator protein.

lac repressor protein, which binds to the sequence of nucleotides found within the lac operator site. Once bound, lac repressor prevents RNA polymerase from transcribing the lacZ, lacY, and lacA genes (Figure 14.4a). The binding of the repressor to the operator site is a reversible process. In the absence of allolactose, lac repressor is bound to the operator site most of the time. When four molecules of allolactose bind to the repressor, a conformational change occurs that prevents lac repressor from binding to the operator site. Under these conditions, RNA polymerase is free to transcribe the operon (Figure 14.4b). In genetic terms, we say that the operon has been induced. This form of transcriptional regulation is influenced by the presence of glucose, which is a catabolite—a substance that is broken down inside the cell. The presence of glucose ultimately leads to repression of the lac operon. When exposed to both glucose and lactose, E. coli cells first use glucose, and catabolite repression prevents the use of lactose. Glucose, however, is not itself the small effector molecule that binds directly to a genetic regulatory protein. Instead, this form of regulation involves a small effector molecule, cyclic-AMP (cAMP), which is produced from ATP via an enzyme known as adenylyl cyclase. When a bacterium is exposed to glucose, the transport of glucose into the cell stimulates a signaling pathway that causes the intracellular concentration of cAMP to decrease because the pathway inhibits adenylyl cyclase, the enzyme needed for cAMP synthesis. The effect of cAMP on the lac operon is mediated by the activator protein called catabolite activator protein (CAP). CAP is composed of two subunits, each of which binds one molecule of cAMP.

Describe the organization of sites along a eukaryotic chromosome.

linear occur in sets (usually diploid) tens of millions to hundreds of millions bp a few hundred to several thousand genes origin of replication every ~100,000 bp a centromere that forms the recognition site for kinetochore proteins repetitive sequences are usually near centromeric and telomeric regions three types of regions that are required for chromosomal replication and segregation: origins of replication, centromeres, and telomeres.

With regard to a promoter, a transcriptional start site is located at the −35 sequence and is recognized by σ factor. located at the −35 sequence and is where the first base is used as a template for transcription. located at the +1 site and is recognized by σ factor. located at the +1 site and is where the first base is used as a template for transcription.

located at the +1 site and is recognized by σ factor.

The anticodon of a tRNA is located in the 3′ single-stranded region of the acceptor stem. loop of the first stem-loop. loop of the second stem-loop. loop of the third stem-loop.

loop of the second stem-loop.

MicroRNA (miRNA), small-interfering RNA (siRNA)

miRNAs and siRNAs regulate the expression and degradation of mRNAs.

In the Hershey and Chase experiment involving T2 phage,

most of the 32P entered the bacterial cells whereas most of the 35S did not

PIWI-interacting RNA (piRNA)

pi-RNA associates with PIWI proteins and prevents the movement of transposable elements.

List the different types of RNA modifications.

processing splicing 5' capping 3' polyA tailing RNA editing base modification

Negative supercoiling may enhance activities like transcription and DNA replication because it allows the binding of proteins to the major groove. promotes DNA strand separation. makes the DNA more compact. causes all of the above.

promotes DNA strand separation.

drugs that specifically block DNA Gyrase

quinolones and coumarins—inhibit gyrase and other bacterial topoisomerases, thereby blocking bacterial cell growth. These drugs do not inhibit eukaryotic topoisomerases, which are structurally different from their bacterial counterparts. This finding has been the basis for the production of many drugs with important antibacterial applications. An example is ciprofloxacin (known also by the brand name Cipro), which is used to treat a wide spectrum of bacterial diseases, including anthrax.

Which of the following is an example of a moderately repetitive sequence? rRNA genes Most protein-encoding genes Both a and b None of the above

rRNA genes

Ribosomal RNA (rRNA)

rRNAs are components of ribosomes, which are the site of polypeptide synthesis.

When an ncRNA functions as a decoy, it contains binding sites for many different proteins, thereby promoting the formation of a large complex. recognizes other ncRNAs and sequesters them, thereby preventing them from working. can alter its conformation, allowing it to switch between active and inactivate conformations. may physically prevent or block a cellular process from happening.

recognizes other ncRNAs and sequesters them, thereby preventing them from working.

A repressor is a __________ that _________ transcription. small effector molecule, inhibits small effector molecule, enhances regulatory protein, inhibits regulatory protein, enhances

regulatory protein, inhibits

In Noll's experiment to test the beads-on-a-string model, exposure of nuclei to a low concentration of DNase I resulted in a single band of DNA with a size of approximately 200 bp. several bands of DNA in multiples of 200 bp. a single band of DNA with a size of 100 bp. several bands of DNA in multiples of 100 bp.

several bands of DNA in multiples of 200 bp

Small nuclear RNA (snRNA)

snRNAs associate with proteins to form subunits of the spliceosome, which splices pre-mRNAs in eukaryotes.

The reading frame begins with a _______and is read _______. promoter, one base at a time promoter, in groups of three bases start codon, one base at a time start codon, in groups of three bases

start codon, in groups of three bases

Transfer RNA (tRNA)

tRNA molecules recognize mRNA codons during translation and carry the appropriate amino acid.

The complementarity of DNA strands is based on the chemical properties of a phosphodiester linkage. the binding of proteins to the DNA. the AT/GC rule. none of the above.

the AT/GC rule.

The Kozak rules determine the choice of the start codon in complex eukaryotes. the choice of the start codon in bacteria. the site in the mRNA where translation ends. how fast the mRNA is translated.

the choice of the start codon in complex eukaryotes.

Combinatorial control refers to the phenomenon that transcription factors always combine with each other when regulating genes. the combination of many factors determines the expression of any given gene. small effector molecules and regulatory transcription factors are found in many different combinations. genes and regulatory transcription factors must combine with each other during gene regulation

the combination of many factors determines the expression of any given gene.

Define chromatin.

the complex of DNA and proteins that is found within eukaryotic chromosomes.

Evidence or approaches that led to the discovery of the DNA double helix include

the determination of structures using ball-and-stick models. the X-ray diffraction data of Franklin. the base composition data of Chargaff.

A groove in the DNA refers to

the indentations where the bases are in contact with the surrounding water.

In eukaryotes, DNA replication is initiated at an origin of replication by DnaA proteins. the origin recognition complex. DNA polymerase δ. MCM helicase.

the origin recognition complex.

The model that correctly describes the process of DNA replication is the conservative model. the semiconservative model. the dispersive model. all of the above.

the semiconservative model.

heterochromatin

tightly compacted regions of chromosomes. These regions of the chromosome are usually transcriptionally inactive. Heterochromatin is most abundant in the centromeric regions of the chromosome and, to a lesser extent, in the telomeric regions. The term constitutive heterochromatin refers to chromosomal regions that are always heterochromatic and permanently inactive with regard to transcription. Constitutive heterochromatin usually contains highly repetitive DNA sequences, such as tandem repeats, rather than gene sequences. Facultative heterochromatin refers to chromatin that can occasionally interconvert between heterochromatin and euchromatin. An example of facultative heterochromatin occurs in female mammals when one of the two X chromosomes is converted to a heterochromatic Barr body.

With regard to transcriptional termination in eukaryotes, which model suggests that RNA polymerase is physically removed from the DNA? Allosteric model Torpedo model Both models Neither model

torpedo model

Outline the key structural features of the DNA double helix.

two strands of dna form a right-handed double helix the bases in opposite strands hydrogen bond according to the AT/GC rule the two strands run antiparallel with regard to their directionality about 10 base pairs reults in a 360 degree turn

Define ribozyme.

which are RNA molecules with catalytic function

A uracil-rich sequence occurs at the end of the RNA in ρ-dependent termination. ρ-independent termination. both a and b. none of the above.

ρ-independent termination.