MCB4403 Test 2 Smartworks (Ch 6, 10, 12, 13, 14, 15, & 17)

The pyruvate dehydrogenase complex plays an important role in cellular metabolism by connecting _________ to _________.

- glycolysis - the TCA cycle Explanation: Pyruvate dehydrogenase functions in respiring organisms by converting pyruvate from glycolysis into acetyl-CoA. Oxaloacetate then condenses with acetyl-CoA in the first reaction of the TCA cycle to produce citrate.

Both DNA sequencing and polymerase chain reactions (PCR) require __________ in order to synthesize a new DNA molecule. A. restriction endonucleases B. primers C. dideoxynucleotides (ddNTPs) D. reverse transcriptase E. fluorescent nucleotides

B. primers Explanation: Short primers are required in the reaction mixtures, along with dNTPs and DNA polymerase, for both PCR and sequencing reactions in order to synthesize new DNA molecules.

Label the steps in which clustered regularly interspaced short palindromic repeats (CRISPR) are used to protect a bacterial cell. Note that the diagram should be labeled clockwise from the phage entry. A. Viral DNA forms new spacer. B. Viral DNA inactivated. C. crRNAs act as a homing beacon for viral DNA. D. Phage dsDNA enters. E. Viral DNA transcribed to RNA. F. Cleaved phage RNA joins Cas complex.

Explanation: A piece of phage DNA is copied as a "spacer" into the host genome. If the bacteria survives the infection, upon reinfection with the same kind of phage the spacers are transcribed into CRISPR RNA. A processed spacer (crRNA) joins the CAS complex to recognize and cleave phage DNA.

Place the steps in the order which they occur during induction of bioluminescence in Allivibrio fischeri (referred to as Vibrio fischeri in the video; it has since been reclassified).

Explanation: Autoinducer (AI) is produced constitutively at low levels by A. fischeri. As in other Gram-negative bacteria, the AI (an acyl homoserine lactone) diffuses freely across the cytoplasmic membrane. Under conditions of high cell density, AI accumulates and eventually diffuses back into the cells to combine with the transcriptional regulator, LuxR. The LuxR-AI complex activates transcription of the lux operon, which encodes genes for the light-producing enzyme, luciferase, and other proteins required to supply and regenerate reaction substrates.

Quorum sensing represents a method by which bacteria can measure what? - the size of a prospective host - external pH - the size of their populations - osmolarity - temperature

the size of their populations Explanation: The legal term "quorum" refers to the minimum number of members required for a group to conduct official business or proceedings. Certain small molecules (often referred to as autoinducers) that are produced and released by bacteria allow the cells to monitor the size of their populations. This method of gene regulation is especially useful for cooperative behaviors such as bioluminescence, virulence, and horizontal gene transfer.

Which of the following are retroviral sequences that have contributed to the human genome? A. Alu sequence B. VSV C. LTR D. HERV-K

A. Alu sequence D. HERV-K Explanation: Both HERV-K and the Alu sequence are retroelements found in the human genome. HERV-K is an endogenous retrovirus that activates during normal development of human embryos. The Alu sequence is a retrotransposon that can disrupt human genes. HERV-K is an endogenous retrovirus that activates during normal development of human embryos. The Alu sequence is a retrotransposon that can disrupt human genes. In some cases, this disruption can result in a genetic defect affecting lipoprotein receptors. This, in turn, can lead to extremely high cholesterol levels and heart failure.

HIV is believed to have originated from which virus? A. Simian immunodeficiency virus (SIV) B. Maedi-Visna virus (MV) C. Walleye dermal sarcoma virus (WDSV) D. Equine infectious anemia virus (EIAV)

A. Simian immunodeficiency virus (SIV) Explanation: HIV is a lentivirus that evolved from simian immunodeficiency virus (SIV) infecting African monkeys.

When intracellular levels of tryptophan are low, the _________ stem loop forms, which results in ___________ of transcription. A. 3:4 / progression B. 3:4 / termination C. 2:3 / progression D. 2:3 / termination

C. 2:3 / progression Explanation: When intracellular concentrations of tryptophan are low, the ribosome stalls when attempting to decode the two adjoining tryptophan codons in complementary region 1. With the ribosome stuck in this position, regions 2 and 3 base-pair to form the anti-termination loop. Region 3 is now prevented from base-pairing with region 4, so the termination loop is not formed, and the operon is expressed.

Within a single HIV-infected patient, the high mutation rates generate mutiple virus strains with differing properties of replication, tissue tropism, and resistance to antibiotics. What is this dynamic population of diverse mutant strains called? A. HIV-1 B. HIV-2B C. quasispecies D. HIV genome

C. quasispecies Explanation: Within a single infected patient, the high mutation rates generate multiple virus strains with differing properties of replication, tissue tropism, and resistance to antibiotics. These diverse mutant strains are called quasispecies. The quasispecies form when an infective virion commences replication with rapid mutation and trait diversification.

Which of the following drugs is NOT used to treat HIV? A. protease inhibitors B. raltegravir C. azidothymidine D. oseltamivir

D. oseltamivir Explanation: This is the drug used to treat influenza, as it blocks neuraminidase. Azidothymidine is incorporated into the growing DNA chain in place of thymidine, but because its 3' OH is replaced by an azido group (-N3), no further nucleotides can be added. Raltegravir inhibits integrase. Upon entering the nucleus, the DNA copy of the HIV genome integrates its sequence as a provirus at a random position in a host chromosome. Integration is catalyzed by integrase. Proteases that cleave Gag-Pol can be important drug targets for protease inhibitors.

Rank the following images in the correct order (first to last) to correctly show the steps in the formation of recombinant DNA molecules.

Explanation: The formation of recombinant DNA molecules involves the insertion of foreign DNA into an appropriate plasmid, or cloning vector. The plasmid and foreign DNA are cut with the same restriction endonuclease. Next, the cut vector and foreign DNA fragments are mixed and cohesive ends annealed. Finally, DNA ligase seals the foreign DNA fragments into the cut vector to produce a plasmid chimera.

Transcriptional regulation is normally accomplished by the reversible binding of regulatory proteins. The proteins that activate genes typically bind to sequences _________ the promoter, whereas those involved in repression usually bind _________ the promoter.

- upstream of - downstream of Explanation: Activator proteins usually bind upstream of the promoter and enhance the ability of RNA polymerase to interact with the promoter (either by DNA bending or direct interaction with the enzyme complex). In contrast, repressors usually bind to regulatory sequences (e.g., "operators") that are positioned downstream of the promoter, but before the transcriptional start site. In this configuration, repression is accomplished primarily by physically blocking access to, or preventing progression of RNA polymerase from, the promoter.

What is the role of the first structural gene (lacZ) in the lac operon? Select all that apply. A. It breaks down lactose into glucose and galactose B. It isomerizes lactose into allolactose C.It binds to and inactivates the lac repressor D. It transports lactose into the cell E. It causes cAMP levels to rise in the presence of increasing concentrations of lactose

A. It breaks down lactose into glucose and galactose B. It isomerizes lactose into allolactose Explanation: The lacZ gene product, β-galactosidase, has two main functions in lactose metabolism. Its primary role is to break down the disaccharide lactose into glucose and galactose, which are then metabolized (via glycolysis) for energy. It also converts lactose into the isomer allolactose. This form of the sugar induces expression of the lac operon by binding to and inactivating the lac repressor.

What is another term used for the electron-dense core particle of HIV? A. capsid B. phospholipid envelope C. spike proteins D. host cell envelope

A. capsid Explanation: The structure of HIV as visualized by TEM consists of an electron-dense core particle (also known as a capsid), surrounded by a phospholipid envelope.

In a CRISPR-Cas9 system, a synthetic csRNA (now called a guide RNA, or gRNA) guides Cas9 endonuclease to bind to a particular gene in a eukaryotic cell. What provides the vehicle that allows the Cas9 and gRNA genes to enter the cell? A. plasmid B. promoter-less reporter gene C. chromosome D. bacteriophage

A. plasmid Explanation: The Cas9 and gRNA genes are added to plasmids, which then enter the eukaryotic cell through the process of transformation. After entering the cell, the added genes are expressed in the nucleus. gRNA can then assist Cas9 in finding the desired gene sequence in the eukaryotic DNA.

How many proteins are made from the 8 RNA nucleotide segments? A. 5 B. 10 C. 6 D. 8

B. 10 Explanation: There are 10 proteins synthesized for the replication of influenza virus.

If the adjacent tryptophan codons in the trp operon leader sequence are mutated to alanine codons, what is the predicted effect on trp operon regulation? A. Attenuation may occur in response to low alanine levels B. Attenuation may occur in response to high alanine levels C. Attenuation will no longer occur in response to high tryptophan levels D. Repression can no longer occur in response to low tryptophan levels E. Repression can no longer occur in response to high tryptophan levels

B. Attenuation may occur in response to high alanine levels C. Attenuation will no longer occur in response to high tryptophan levels Explanation: The main level of regulation of the trp operon (which codes for proteins that synthesize tryptophan) is through a repressor protein. As is the case with the lac operon, repression is never 100% and some transcription will occur even under conditions where tryptophan is present. While this leakiness is useful in the lac operon, as the cell needs some lactose permease and B-galactosidase to allow lactose into the cell and convert it to the allolactose inducer, this leakiness is not beneficial for the trp operon. Thus the trp operon has another level of control, termed attenuation, which prevents transcription into the structural genes when tryptophan levels are high. Attenuation relies on a leader sequence that has two adjacent tryptophan codons. When tryptophan levels are low, the ribosome pauses at this leader sequence and prevents the formation of the attenuator transcription terminator loop. In contrast, when trp is abundant, the ribosome does not pause at the leader sequence and transcription is aborted.

What is the primary receptor for HIV? A. sialic acid B. CD4 C. X4 D. chemokine

B. CD4 Explanation: Like other viruses, HIV needs to recognize specific receptor molecules on the surface of its target cells. The primary receptor for HIV is the CD4 surface protein on CD4 T lymphocytes (T-cells).

Genes for which of the following functions would be constitutively expressed in a bacterial cell such as E. coli? A. heat shock B. virulence C. osmotic stress D. DNA replication E. nitrogen limitation

D. DNA replication Explanation: Certain genes (aka "housekeeping" genes) are always "on" (i.e., constitutively expressed) because they encode functions essential to the cell during normal growth. In addition to DNA replication, such activities include nutrient transport, glucose metabolism, cell wall synthesis, ATP synthesis, lipid catabolism, and so on. The RNA polymerase-RpoD holoenzyme has a strong affinity for these promoters, and no additional transcription factors (e.g., activators or inducers) are needed. Other gene sets, particularly those that enhance survival under various types of nutritional or environmental stress, have promoter sequences that are recognized by alternative sigma factors.

When ribosomes translate full length RNA molecules, what is the predominant protein made? A. spike proteins B. Gag-Pol C. envelope proteins D. Gag

D. Gag Explanation: Gag proteins are used for the synthesis of core structural proteins that are needed for the new virion.

The boundaries of an open reading frame (ORF) are defined by which of the following? A. ribosomal binding and release sites B. transcription start and terminator sequences C. enhancer and operator sequences D. distances between adjacent structural genes E. start and stop codons

E. start and stop codons Explanation: An ORF defines a specific DNA sequence corresponding to a putative protein product from beginning to end. Thus, the first three nucleotides, if transcribed to mRNA, would produce a translation start codon (e.g., AUG), and the last three would produce a translation stop codon (e.g., UAG, UGA, or UAA). Recall that the start codon signals the ribosome to initiate polypeptide chain synthesis, and then every subsequent set of three nucleotides codes for incorporation of a specific amino acid until the stop codon is reached, releasing the mature protein.

Match the human disease or symptoms with the correct DNA virus. 1. HPVs 2. Epstein- 3. Variola 4. Herpes 5. Varicella - smallpox - chickenpox, shingles - infectious mononucleosis - genital and skin warts, cervical and penile cancer - skin and genital lesions, neuron latency

Explanation: DNA viruses have genomes of double-stranded DNA. Such viruses are responsible for many diseases like infectious mononucleosis (mono), smallpox, chickenpox, and many others. Table 11.3 in Section 11.5 of the text lists some common DNA viruses and the diseases and symptoms they cause in humans and animals. Herpes simplex virus 1 and 2, varicella-zoster virus, and Epstein-Barr virus are all herpes viruses that cause skin and genital lesions, as well as neuron latency, chickenpox and shingles, and infectious mononucleosis, respectively. Human papillomaviruses can cause genital and skin warts and cervical and penile cancer, whereas variola majora virus (a pox virus) leads to smallpox.

Outline the basic steps for annotating a genome from start to finish. - Search both strands in all six reading frames. - Map locations of individual ORFs. - Obtain and compile DNA sequence data. - Exclude regions based on minimum-length criteria.

Explanation: The first step in genome annotation is obtaining and compiling DNA sequence data from an organism or sample. Computer programs then scan the three potential reading frames on each DNA strand for the presence of translation start and stop codons. Putative open reading frames (ORFs) that are not long enough to encode a functional protein are excluded from the resulting map. Also, although not required, identification of a ribosomal binding site upstream of the ORF increases the possibility that the region represents an actual gene.

Match the HIV gene with its function. 1. vpr, vpu, nef, vif, tat 2. gag 3. pol 4. env 5. rev - provides capsid monomer - provides the envelope glycoprotein - necessary for exporting viral mRNA from the nucleus into the cytoplasm during the infection - provides reverse transcriptase - code for virulence factors for disease

Explanation: The genes vpr, vpu, nef, vif, and tat all code for virulence factors for disease, so they are removed from the lentivector to ensure its safety for human use. The gag (provides capsid monomer), pol (provides reverse transcriptase), env (provides the envelope glycoprotein), and rev (necessary for exporting viral mRNA from the nucleus into cytoplasm during infection) are all necessary for virion production. They are placed in DNA helper plasmids and are provided in tissue culture only for vector production.

Identify the components of a herpes simplex 1 virion. A. Tegument B. Lipid envelope from host C. Double-stranded DNA chromosome D. Nucleocapsid E. Envelope spike proteins

Explanation: The herpes simplex virion consists of an icosahedral capsid that houses the double-stranded DNA. The tegument, a collection of proteins encoded by the virus and the previous host, resides outside of the capsid. The virion derives its lipid membrane envelope from the host cell. This membrane is punctuated with spike proteins.

A "competent" bacterium is ready to undergo which of the following? - sporulation - conjugation - binary fission - transformation - transduction

Transformation Explanation: Competence refers to the ability of a cell to be transformed (i.e., acquire foreign DNA sequences that may be incorporated into the cell's genome). In Streptococcus, competence is controlled by quorum sensing. It is thought that by restricting expression of the transformation machinery to conditions of high cell density, it is more likely that free extracellular DNA (e.g., from dying cells) will be available for uptake.

You and your team are developing a gene therapy for cystic fibrosis. Which of the following changes would you make to ensure that the lentivector is safe for human use? A. replace viral envelope gene with envelope gene from another virus B. derive the vector from HIV C. incorporate a genetic enhancer sequence from WHV D. add human embryonic kidney 293T cells

A. replace viral envelope gene with envelope gene from another virus Explanation: The HIV envelope gene is usually replaced with the envelope gene from vesicular stomatitis virus (VSV). VSV's envelope has a broad tropism, meaning that it is capable of infecting a wide variety of cell types. If scientists wanted the gene therapy to be focused on a select or even an individual cell type, they would replace the vector envelope gene with an envelope gene from other viruses. This process is called pseudotyping, and it allows scientists to focus gene therapy on specific cells, preventing lentivectors from targeting host cells that are not of interest.

AZT targets which part of HIV? A. reverse transcriptase B. chemokine receptor C. spike protein D. transmembrane protein

A. reverse transcriptase Explanation: Reverse transcriptase is the target of the first clinically useful drug to treat HIV infection, the nucleotide analog azidothymidine (AZT). AZT is incorporated into the growing DNA chain in place of a thymidine, but because its 3' OH is replaced by an azido (-N3), no further nucleotides can be added.

HPV causes what disease? A. chickenpox B. cervical cancer C. smallpox D. cowpox

B. cervical cancer Explanation: Human papillomaviruses cause genital warts, cervical and penile cancer, and skin warts.

Arrange the following items in the order in which they receive electrons from glycolysis and TCA oxidations and harness the energy to establish a transmembrane proton gradient. - Oxidoreductase - NAD+ - Quinone - Oxygen - Terminal oxidase

1. NAD+ 2. Oxidoreductase 3. Quinone 4. Terminal oxidase 5. Oxygen Explanation: During aerobic respiration, NAD+ becomes reduced to NADH when "food" is oxidized by glycolytic and TCA reactions. NADH is recycled to NAD+ via transfer of the electrons to a membrane-bound substrate oxidoreductase. As electrons are passed through the redox centers in this complex, energy is released, which is used to pump protons to the exterior side of the membrane. The oxidoreductase eventually transfers the electrons to a diffusible quinone, reducing it to quinol. In the simplest scheme, the quinol is then oxidized by a terminal oxidase, which pumps additional protons. Oxygen serves as the final (terminal) electron acceptor, becoming reduced to water in the process.

Using the table above, calculate (in mV) the value of E°' when hydrogen gas (H2) is oxidized using molecular oxygen (O2) as a terminal electron acceptor. E°' = _______ mV

1240 mV Explanation: A complete redox reaction occurs when one substrate is oxidized and electrons are transferred to another molecule, which thereby becomes reduced. The amount of energy released can be stored as potential energy, much like a battery. To oxidize hydrogen gas using molecular oxygen, the following half reactions would have to be combined: 2H+ + 2e- → H2 (E°' = -420 mV) ½O2 + 2H+ + 2e- → H2O (E°' = +820 mV) They are both written as reductions, so we need to reverse the first half reaction for H2 to express it as an oxidation. This also reverses the sign of E°', so now we have: H2 → 2H+ + 2e- (E°' = +420 mV) ½O2 + 2H+ + 2e- → H2O (E°' = +820 mV) The complete reaction is simply the sum of these two E°' values, which equals +1240 mV. This is roughly equivalent to the charge in a 1-V battery.

Mitochondrial electron transport systems are able to translocate up to 12 H+ per NADH. What is the corresponding maximum theoretical yield of ATP? _____ ATP

4 Explanation: The theoretical maximum yield is 1 ATP for every 3 H+ that pass through the ATP synthase. Thus, if 12 protons are translocated for every 2 e- that NADH donates, 4 ATP could be produced (12 H+ / 3 H+ per ATP = 4 ATP).

You are studying a large virus with genes that encode for a wide range of viral proteins. What kind of virus is it? A. DNA virus B. Phage virus C. Retrovirus D. RNA virus

A. DNA virus Explanation: In most cases, DNA viruses are greater in size than RNA viruses. They usually encode a wider range of viral enzymes. For example, the vaccinia genome codes for 200 proteins. Some DNA viruses, like vaccinia and herpes, have levels of complexity comparable to small cells.

The yeast two-hybrid technique maps protein-protein interactions. It takes advantage of the modular nature of transcription factors that contain a DNA-binding domain and an activation domain. Two different hybrid proteins are made: one with a bait protein (labeled X) fused to the DNA-binding domain of the transcription factor, and another with the prey protein (labeled Y) fused to the activation domain of the transcription factor. Libraries of plasmids containing bait and prey fusion proteins are transformed into yeast cells. When the bait and the prey interact, the DNA-binding domain and the transactivation domain are brought together, leading to transcription of a reporter gene. As with many techniques, both false positives (a signal when the proteins don't actually interact) and false negatives (no signal when in fact the two proteins do normally interact) are possible. 2. What conditions could lead to a false negative result? A. Fusion of the activation domain to the prey protein prevents proper prey protein folding. B. The prey protein itself contains an activation domain. C. The prey protein is targeted to the plasma membrane. D. The bait protein needs a posttranslational modific

A. Fusion of the activation domain to the prey protein prevents proper prey protein folding. C. The prey protein is targeted to the plasma membrane. D. The bait protein needs a posttranslational modification for interaction with the prey protein. Explanation: The traditional yeast two-hybrid system depends on both the prey and bait proteins entering the nucleus. (If either has a membrane localization signal, then it won't be found in close proximity in the nucleus, near the reporter gene.) Different methods are used to screen for interactions among membrane proteins. Furthermore, if either the bait or prey protein does not fold properly or requires a posttranslational modification (such as phosphorylation) for interaction, and that PTM does not occur in yeast cells, then the protein interaction will not be observed.

What is the location of the attenuator region that controls the expression of the trp operon? A. It is between the transcription start site and first structural gene. B. It is located upstream of the promoter. C. It is part of the holorepressor. D. It overlaps the promoter. E. It overlaps the CRP-cAMP binding site.

A. It is between the transcription start site and first structural gene Explanation: The trp operon in E. coli is repressed under conditions of excess tryptophan. This regulation occurs primarily at the level of transcription initiation and is mediated by the Trp repressor. However, because binding of the holorepressor to the operator is reversible, some "unintentional" and wasteful transcription initiation may occur. The attenuator region is, therefore, strategically located within the leader just downstream of the operator and prior to the first gene. Termination here prevents the unnecessary production of a full-length polycistronic mRNA.

As an up-and-coming scientist, you decide to create mutant bacteriophages to better examine structural function. Your first mutant lacks a sheath motor protein, so you decide to call this MutS. To determine the loss of function, you add MutS at a multiplicity of infection (MOI) of 1 to Escherichia coli cells and examine the rate of infection. You're very surprised to find that none of the Escherichia coli cells are infected. What happened? A. MutS prevented the sheath protein from contracting. B. MutS damaged the bacteriophage tail fibers. C. MutS created ghost bacteriophages. D. You failed to add enough bacteriophages to infect the E. coli.

A. MutS prevented the sheath protein from contracting Explanation: Sheath proteins are responsible for generating the contractile force necessary to inject the bacteriophage DNA into E. coli. In particular, the T4 bacteriophage generates a pressure as high as 50 atmospheres to release DNA. MutS appears to have altered the ability of the sheath protein to contract, hence preventing infection of E. coli.

A bacterial cell is infected by a phage with an RNA genome. The bacteria has not encountered this phage before. Which of the following correctly describes host defense mechanisms that can help protect the bacterial cell? A. Neither CRISPR nor restriction endonucleases can provide defense. B. Only restriction endonucleases can provide defense in this case. C. Only CRISPR can provide defense in this case. D. Both CRISPR and restriction endonucleases can provide defense.

A. Neither CRISPR nor restriction endonucleases can provide defense. Explanation: Bacteria contain restriction endonucleases, enzymes that cleave DNA (such as invading viral DNA) at particular sequences. The bacterial DNA is spared by enzymes that methylate those same sequences within the bacterial genome. Only the unmethylated sequences are recognized and cleaved by the restriction endonuclease. Restriction endonucleases work on any viral DNA containing the specified sequence, even if it is the first time the bacteria has encountered that virus. CRISPR also works by cleaving viral DNA, but it only works for phages that the cell has previously encountered. A piece of phage DNA gets copied as a "spacer" into the host genome. If the bacterium survives infection, later reinfection by the same kind of phage causes transcription of the spacers into CRISPR RNA. A processed spacer (crRNA) joins the Cas complex to recognize and cleave the phage DNA. Neither restriction enzymes nor CRISPR will help against a phage with an RNA genome.

In what ways do virions resemble living cells? Select all that apply. A. Their genomes can be larger than some cellular genomes. B. They possess genes that can direct their own replication. C. They are capable of protein synthesis. D. Some virions exhibit flagellar motility. E. Some virions possess genes for tRNA.

A. Their genomes can be larger than some cellular genomes. B. They possess genes that can direct their own replication. E. Some virions possess genes for tRNA. Explanation: Viruses have long been considered "nonliving" because they are inert as virions—incapable of metabolism, gene expression, motility, and replication. However, during infection, the viral genome commandeers the life processes of its host cell for the production of new virions. This ability to direct one's own replication is very "lifelike." Moreover, viral genomes are being discovered that exceed the size of certain prokaryotes, and some viruses even possess genes involved in protein synthesis (e.g., tRNAs, some ribosomal components).

What makes HIV a suitable vector for gene therapy? Select all that apply. A. ability to integrate into a host cell's chromosome B. efficient infection of dividing cells only C. infection of dividing or nondividing cells D. ability to replicate outside of the host's DNA

A. ability to integrate into a host cell's chromosome C. infection of dividing or nondividing cells Explanation: As the video explains, the fact that HIV is so efficient at infecting cells makes it a strong candidate for acting as a vector in gene therapy. HIV delivers its genes in the form of RNA into the cell's cytoplasm where HIV reverse transcriptase molecules turn the RNA into double-stranded DNA. This DNA goes into the host cell nucleus and integrates into the host cell genome. HIV, unlike other viruses, can infect both dividing and nondividing cells, which makes it a great candidate for gene therapy.

What factors are taken into account when looking for bacterial open reading frames (ORFs)? Select all that apply. A. minimum length of codons bracketed by start and stop codons B. presence of a transcription termination signal C. evidence of a ribosome-binding site D. comparison to other genes in the databases E. evidence of N-terminal signal sequences F. intron splice sites

A. minimum length of codons bracketed by start and stop codons C. evidence of a ribosome-binding site Explanation: When searching for open reading frames (ORFs), both a minimum length of codons bracketed by start and stop codons and evidence of a ribosome-binding site are used. If the ORF is only 12 amino acids, it is too small to be considered an ORF. Usually, a minimum number of amino acids, such as 50, is used. All ORFs have to contain a start and a stop codon. Intron splice sites are usually only involved in eukaryotic genes. Gene comparisons would occur after ORFs have been identified. N-terminal signal sequences are not used to identify ORFs since they are only associated with a subset of proteins. And transcription termination signals are used in transcription, not translation, so would not be useful in identifying ORFs, although they may be used to help verify potential genes, as would promoter sequences.

You want to test the effectiveness of a drug cocktail in the treatment of HIV. What two drugs would you include in your experiment if you wanted to target both the attachment and replication phases of HIV infection? A. azidothymidine (AZT) B. protease inhibitors C. antiviral agents targeting accessory proteins D. maraviroc

A. azidothymidine (AZT) D. maraviroc Explanation: When an HIV virion attaches to a host cell, it binds to CD4 receptor proteins. The HIV envelope spike protein SU also binds to secondary receptors called chemokine receptors (CCRs). CCR5 is one such receptor. People who lack the CCR5 protein due to a genetic defect were found to be resistant to HIV infection. This led to the development of maraviroc, which blocks CCR5, preventing virion attachment. HIV is reliant on reverese transcriptase to create DNA that can be incorporated into the host genome from viral RNA. Azidothymidine (AZT) takes the place of a thymidine as the reverse transcriptase creates a DNA chain from the viral RNA. AZT has an N3 azido group in place of a OH group at its 3' end, which prevents the addition of further nucleotides to the chain, stopping replication. These two would be most useful to study how a drug cocktail can block both attachment and replication. Protease inhibitors target the enzymes responsible for cleaving Gag-Pol, which is necessary for a virion to reach maturation and leave the host cell. Accessory proteins have a wider array of functions that assist in the viral replication, so drugs targeting these proteins must be specific to that particular protein and its function.

The main function of the dideoxynucleotides (ddNTPs) used in the Sanger method of DNA sequencing is to A. be incorporated into newly replicated DNA strands and stop elongation. B. denature the DNA to separate the strands. C. serve as primers. D. serve as a fluorescent tag.

A. be incorporated into newly replicated DNA strands and stop elongation. Explanation: While the dideoxynucleotides used in the Sanger sequencing method are often fluorescently tagged in order to identify the nucleobase, dideoxynucleotides are modified nucleotides that do not have a 3′ OH group. As a result, once a dideoxynucleotide is added to a newly synthesized DNA molecule, no additional nucleotides can be added and elongation stops. The result is a population of DNA strands of varying length, each one truncated after a dideoxynucleotide is added.

What function does the targeted endonuclease Cas9 from Streptococcus pyogenes perform in the CRISPR-Cas9 system? A. breaking the double-stranded target DNA B. transcribing RNA into DNA for incorporation into the target genome C. expressing a replacement gene to edit the target sequence D. recognizing the target eukaryotic sequence

A. breaking the double-stranded target DNA Explanation: In the CRISPR-Cas9 system, a programmed, synthetic csRNA (guide RNA) directs Cas9 endonuclease to a specific gene. When it reaches that gene, the Cas9 endonuclease creates a double-stranded break in the target DNA. The eukaryotic cell attempts to repair the break through nonhomologous end joining (NHEJ). Mutation often happens during this process, as bases can be deleted. Scientists can then study how a mutation affects the cell. This process can also be used to replace the target gene. CRISPR-Cas9 can achieve any type of insertion, deletion, or change in genetic sequence.

Which of the following elements must be present for attenuation of the trp operon to occur? Select all that apply. A. charged tRNA^Trp B. holorepressor C. ribosomes D. sigma factors E. DNA polymerase F. RNA polymerase

A. charged tRNA^Trp C. ribosomes F. RNA polymerase Explanation: Attenuation involves the activities of both RNA polymerase and ribosomes. As RNA polymerase transcribes the leader region, a ribosome can very quickly associate with the transcript to begin translation. When tryptophan is present in excess, there is no shortage of the corresponding charged tRNAs. This allows the ribosome to progress quickly through two adjacent trp codons before reaching a stop codon between the first two regions capable of forming a step loop. The temporary pause of the ribosome here (before it releases) prevents regions 1 and 2 from participating in complementary binding, and the alternate terminator stem loop (between regions 3 and 4) is thereby favored. Sigma factors and the Trp holorepressor control initiation of transcription, whereas DNA polymerase catalyzes DNA replication.

Picornaviruses can avoid detection by synthesizing virally induced vesicles, or replication complexes, formed from the A. endoplasmic reticulum B. lysosome C. nuclear membrane D. Golgi apparatus

A. endoplasmic reticulum Explanation: Picornaviruses replicate their viral genome and mRNA templates inside of virally induced vesicles, also known as replication complexes, formed from the endoplasmic reticulum. The unique genetics and replication intermediates possessed by picornaviruses make this virus less likely to be detected in the replication complex, thus allowing this virus to avoid the host immune system.

Which of the following mechanisms do prokaryotes use to generate phase variation? A. gene inversions B. transcriptional activators C. repressor-corepressor complexes D. slipped-strand mispairing E. post-translational modification

A. gene inversions D. slipped-strand mispairing Explanation: Both gene inversions and slipped-strand mispairing can produce alternate states of gene expression (i.e., phase variation). Invertible regions typically contain a promoter sequence that must be in the correct orientation to promote transcription of a given gene. In contrast, genes may contain regions of short multiple repeats that cause the DNA polymerase to "slip" during replication and produce single base-pair insertions or deletions. When this occurs, the altered mRNA produced during transcription will contain a frameshift that either restores or abolishes protein function.

The three tubes in the photograph contain a fermentable carbohydrate (e.g., glucose) and the pH indicator phenol red. The middle and right tubes were inoculated with bacteria, and the left tube was left uninoculated as a control. Which of the tubes shows evidence of fermentation? A. left tube B. middle tube C. both D. none E. impossible to tell without further analysis

A. left tube (yellow tube) Explanation: Fermentation often results in the accumulation of acidic end products in the growth medium. In the current test, the resultant drop in pH causes the pH indicator (phenol red) to turn yellow. In addition, the inverted tube (called a Durham tube) has collected gases (likely CO2 and/or H2), which are indicative of fermentation.

You have utilized transposons to select for acid-sensitive E. coli mutants. You want to find the glutamate decarboxylase enzymes most effective at removing a carboxyl group from glutamate within this sample of mutants. This will allow you to use this enzyme in a variety of biotechnological applications. What technique would allow you to do this? A. phage display B. two-hybrid analysis C. ChIP-seq D. synthetic biology

A. phage display Explanation: While phage display is used primarily to isolate peptide variants with alternative properties, this technique has been applied to find peptides with a high affinity for binding to attachment proteins of viruses that infect eukaryotic cells with applications for preventing infection. Additionally, phage display can find variant enzymes with desired traits, isolating them from natural enzymes in order to find the enzymes most suitable for biotechnological applications.

Phase variation contributes to the virulence of pathogens such as Salmonella and Neisseria by which of the following? A. prolonging the development of antibody-based immunity against the pathogen B. allowing the pathogens to survive fever and inflammation responses C. eliminating their need for iron D. conferring resistance to antibiotics such as penicillin E. rendering them insensitive to lysozyme

A. prolonging the development of antibody-based immunity against the pathogen Explanation: It is the job of certain immune cells to detect foreign antigens and produce antibodies to inactivate them. By periodically switching the type of surface proteins they produce, pathogens change how they appear to the immune system, and the old set of antibodies is rendered ineffective. The result is a prolonged infection because the host must now detect and produce new antibodies against the alternate antigen.

For what purpose does the bobtail squid (Euprymna scolopes) use the light produced by its symbiotic A. fischeri? A. to avoid predation B. as a warning that it tastes bad C. to attract a mate D. to attract food E. to see better at night

A. to avoid predation Explanation: When foraging at night, the squid adjusts the ventrally projected light to match the moonlight that diffuses down into the water. This behavior (called "counterillumination") prevents the slow-moving squid from casting a shadow that would be detected by hungry fish below

Under what condition would a temperate bacteriophage infecting Staphylococcus aureus (an aerobic bacteria commonly found on the human body) be shifted by environmental cues from lysogeny to a lytic cycle? A. when phage-infected cells are incubated at 50°C B. when phage-infected cells are incubated at 37°C C. when phage-infected cells are incubated at atmospheric levels of oxygen D. when phage-infected cells are placed on nutrient-rich media

A. when phage-infected cells are incubated at 50°C Explanation: S. aureus is a mesophile that grows best in the range of 20°C-40°C and especially well at human body temperature (37°C). When the temperature increases to 50°C, S. aureus undergoes heat stress, which causes the phage to shift from lysogeny to a lytic cycle.

What is the minimum number of NADH molecules that must be oxidized for oxygen to be reduced to water? A. 1 B. 2 C. 3 D. None; the electrons to reduce water are picked up free from the cytoplasm.

B. 2 Explanation: The difference between O2 and H2O is 2 hydrogen atoms. But because we need to reduce each oxygen atom in O2 to water, a total of 4 hydrogen atoms are needed. Recall that a single hydrogen atom consists of one proton (H+) and one electron (e-). Thus, it takes 4 e- and 4 H+ for the terminal oxidase to reduce molecular oxygen to water. Each donor NADH carries 2e-, so 2 molecules of NADH are required to produce one molecule of water. (The required protons at this step are picked up from solution.)

What percentage of Americans are infected with herpes simplex? A. 20% B. 60% C. 15% D. 45%

B. 60% Explanation: Herpes simplex virus strains HSV-1 and HSV-2 cause one of the most common infections in the United States. Approximately 60% of Americans acquire herpes simplex, usually HSV-1, in epithelial lesions commonly known as cold sores.

Which step of CRISPR-Cas9 initially affects the genetic material of the target cell? A. transferring the plasmid to the target cell via transformation B. Cas9 endonuclease creating a break in the double-stranded target DNA C. guide RNA (gRNA) binding to the Cas9 endonuclease D. nonhomologous end joining (NHEJ) repair deleting target DNA bases

B. Cas9 endonuclease creating a break in the double-stranded target DNA Explanation: The text refers to the Cas9 endonuclease creating a break in the double-stranded target DNA as "a critical feature of this gene-editing tool" (page 450). When the endonuclease creates this gap in the DNA, it allows for researchers to edit the genetic material. They can replace or delete part of the target gene, or even replace that part with another genetic sequence. Scientists may also allow nonhomologous end joining (NHEJ) repair to take place without altering the DNA in order to see how potential gene mutations that can be a result of NHEJ affect the cell.

What are the components common to all virions? A. envelope B. DNA or RNA genome C. DNA and RNA genomes D. capsid

B. DNA or RNA genome D. capsid Explanation: All viruses are comprised of a protein capsid and the genetic information necessary to replicate progeny virions. Viruses only carry DNA or RNA and use host ribosomes to create the viral enzymes needed to continue the replication process. Some viruses simply hijack host replication enzymes, now that's a sneaky little virus.

A protein's intracellular location or movement can be visualized by which of the following? A. ChIP-seq B. GFP fusion C. two-hybrid analysis D. northern blot

B. GFP fusion Explanation: GFP fusions allow for the study of individual organisms within a colony. In a study of the potato virus X, scientists fused GFP to TBGp3 protein, the viral protein that allows the virus to move between plant cells. Scientists could then observe the fluorescence of the GFP-TBGp3 fusion to trace its path through the potato plant.

Which of the following changes is NOT a change scientists make to convert viruses into gene therapy vectors? A. virulence genes removed B. HIV vpr, vpu, nef, and vif genes replaced with same genes from VSV C. insertion of a CMV promoter D. protein-encoding genes put into DNA plasmids

B. HIV vpr, vpu, nef, and vif genes replaced with same genes from VSV Explanation: Virulence genes are removed from lentivectors to prevent them from causing disease in the host. Any protein-encoding genes necessary for virion production are put into DNA helper plasmids so they may be used only in tissue culture, not in the host. A CMV promoter helps assist the expression of the gene of interest once the lentivector has entered the host. VSV provides an env gene to HIV lentivectors, allowing the vector to infect a broad range of host cell types. The vpr, vpu, nef, and vif genes are entirely removed from the lentivector, as they are some of the genes responsible for disease in HIV.

The final step of genome processing for HIV requires which of the following? A. DNA-dependent DNA synthesis B. Integration into host DNA C. RNA primer D. reverse transcription

B. Integration into host DNA Explanation: The final step of genome processing requires integration into the host DNA. The double-stranded DNA copy of the HIV genome circularizes. The circular molecule then integrates into the host chromosome by site-specific recombination at a host target sequence, mediated by the HIV integrase protein.

What happens during fluorescence resonance energy transfer (FRET)? Select all that apply A. Reporter prevents quencher from fluorescing. B. Quencher prevents reporter from fluorescing. C. Taq DNA polymerase begins synthesizing cDNA. D. Target DNA strand is denatured by heat. E. Reporter transfers energy to the quencher. F. Quencher transfers energy to the reporter.

B. Quencher prevents reporter from fluorescing E. Reporter transfers energy to the quencher Explanation: During the FRET process, the reporter dye of the fluorescent oligonucleotide probe is stimulated by light and transfers the energy to the quencher dye. The quencher dye absorbs that energy and prevents the reporter dye from fluorescing. The reporter dye will not fluoresce until it is separated from the quencher dye later in the RT PCR process.

What unique cloning strategy can join multiple adjacent PCR fragments in a single step? A. Fragments are randomly joined, but only those in the correct order can be cloned into the plasmid vector. B. Specific PCR primers generate distinct regions of overlap between desired adjacent fragments. C. Single-stranded linkers are added to PCR products and used to join desired adjacent fragments. D. Fragment mixtures are generated by restriction enzyme digestion and then rejoined using sticky end overlaps. E. PCR-amplified products are mixed, sheared into blunt-ended fragments, and then rejoined with DNA ligase.

B. Specific PCR primers generate distinct regions of overlap between desired adjacent fragments Explanation: During PCR, incorporation of specific primers generates regions of homology between the fragments to be joined. Each fragment end, therefore, shares homology to only one other fragment end in the mixture. In the example shown here, four distinct primer sequences are used to join five unique fragments in a specific order. A fifth primer sequence is included that shares homology with the insertion site on a plasmid vector.

Viruses express tissue tropism and host specificity. In particular, the avian influenza strain H5N1 has rare incidences in humans due to the reduced distribution of alpha-2,3 sialic acids in the lower respiratory tract. As a consequence, when the H5N1 strain does infect a human, the disease is very severe due to the risk of pneumonia. In contrast, human influenza strains (H1N1, among others) attach to alpha-2,6 sialic acids, which are readily expressed in the upper and lower respiratory tract. Which of the following would not be true if a mutation were to allow the H5N1 strain to attach to alpha-2,6 sialic acids? A. An epidemic would likely ensue due to the lack of immunity to the new avian influenza strain. B. The avian influenza strain would only infect the lower respiratory tract of humans. C. The avian influenza strain could be transmitted directly from birds to humans. D. The avian influenza strain could be transmitted easily from human to human.

B. The avian influenza strain would only infect the lower respiratory tract of humans. Explanation: If the avian influenza strain were adapted to bind to alpha-2,6 sialic acids, the virus would be capable of infecting the upper and lower respiratory tract. As a consequence, this virus could be transmitted easily from birds to humans and humans to humans. The ease of transmission would likely lead to an epidemic and possibly a pandemic outbreak depending on the virulence of the avian influenza strain.

Prions are aberrant proteins that cause Creutzfeldt-Jakob disease in humans,conformation of the PrPSc explains why the protein can easily form aggregates. mad cow disease from contaminated beef, and scrapie in sheep. Unfortunately, prions have demonstrated the highest resistance to methods of chemical and physical control, including the ability to survive nucleases and UV irradiation. Prions propagate aggregates of prion proteins by altering the normal quaternary conformation of PrPC in the brain. Based on the quaternary structure shown below, which of the following hypotheses could be supported? A. The disordered quaternary B. The increased presence of beta pleated sheets confers resistance to PrPSc. C. The increased presence of alpha helices confers resistance to PrPSc.

B. The increased presence of beta pleated sheets confers resistance to PrPSc Explanation: PrPSc prion proteins show increases in the areas of beta pleated sheets and an anti-parallel arrangement of alpha helices. This altered conformation allows for protein aggregates to form in the brain and enables propagation of prion proteins by altering the conformatation of normal PrPC found in the infected organism. Therefore, it would be plausible that the increased presence of beta pleated sheets plays a role in resistance to methods of microbial control.

What is the correct interpretation of the results of the real-time PCR experiment shown in the figure? A. If more dNTP molecules are present, the fluorescence increases above background sooner. B. The number of cycles required to detect fluorescence varies inversely with the amount of template DNA. C. The final fluorescence (at cycle 35) is used to calculate the amount of starting template DNA. D. For every 10-fold increase in the number of starting molecules, there is 10 times the amount of fluorescence at any given time. E. Reaction mixtures containing fewer than 103 molecules do not produce detectable fluorescence.

B. The number of cycles required to detect fluorescence varies inversely with the amount of template DNA Explanation: Real-time PCR is also known as quantitative PCR (qPCR) because increased amounts of starting template DNA result in earlier detection of fluorescence. While there are several formats for qPCR, this particular experiment was run using a probe whose signal is quenched until it is incorporated into an amplification product (see Figure 12.3A below). You will note that a reaction containing 107 initial copies of the target sequence results in detectable fluorescence after just 15 cycles of PCR, while it takes at least 28 cycles to reach the threshold level when only 1000 starting DNA molecules are present.

The video explains how O2 becomes reduced to H2O during aerobic respiration. In the example of the bacterial ETS that was given, how many protons were actually lost from the cytoplasm for every O2 molecule that was consumed? A. 4 B. 28 C. 20 D. 8 E. 16

C. 20 Explanation: For every NADH that is oxidized, 2 e- enter the ETS, and they are used to power the extrusion of 8 protons. You can review the details of this process near the end of the animation. Because 2 NADH are required to reduce 1 molecule of O2 to H2O, a total 16 protons are lost via translocation across the membrane. An additional 4 protons are picked up from the cytoplasm and used by the terminal oxidase during reduction of O2 to H2O, for a total of 20. These last 4 protons balance the ones associated with the corresponding reduction of 2 NAD+ during catabolism (i.e., to 2 NADH + 2 H+), and therefore do not contribute to the Δp.

How would removing the operator region from the E. coli tryptophan operon affect transcription of the trp genes? A. Transcription of the aporepressor gene would decrease. B. Transcription of the operon would occur even if tryptophan were abundant. C. Transcription of the operon could not occur, and the cells would require tryptophan in the growth media. D. Transcription of the operon would begin normally but terminate prematurely. E. Transcription of the operon would be greater than normal in absence of tryptophan.

B. Transcription of the operon would occur even if tryptophan were abundant Explanation: The trp operon in E. coli encodes several structural genes required for the synthesis of tryptophan. Similar to other biosynthetic operons, it is normally transcribed but subject to repression by the resulting end-product when it is present in excess. In the current example, tryptophan binds to a DNA-binding protein (the aporepressor) to increase its affinity for the trp operator. When the operator is occupied by this holorepressor complex, it blocks progression of RNA and the operon is not transcribed. Removal of the operator, therefore, would remove any hindrance to transcriptional control, and the operon would be expressed under all conditions.

In a virus expressing immediate-early RNA genes, the virus is preparing for which of the following? A. a latent infection B. a lytic infection C. latency in the trigeminal ganglion D. the expression of LAT proteins

B. a lytic infection Explanation: VP16 (a viral protein) works with host factors to activate a set of viral genes called immediate-early genes. The immediate-early RNA enters the cytoplasm of the host cell and ribosomes translate it into proteins necessary for lytic infection and the production of new virions.

Investigations leading to a better understanding of viruses have provided much insight into the benefits of viral infection to human survival. All of the following are proteins or regulators that provide beneficial functions to humans except A. syncytin-1 (for placental fusion). B. botulin toxin (for wrinkle removal). C. apolipoprotein C1 (for liver function). D. beta 1,3-galactosyltransferase (for colon and mammary gland function).

B. botulin toxin (for wrinkle removal) Explanation: Humans have certainly benefited from the colonization with certain viruses. In particular, human endogenous retroviruses (HERV) have integrated several genes that assist in placental fusion and development, as well as liver, colon, and mammary gland function. Although humans have hijacked the botulin toxin for cosmetic treatments to reduce wrinkles, the host benefit conferred by the phage C1 is for Clostridium botulinum bacteria, not humans.

Identify the function(s) of the viral capsid. A. is necessary for RNA virus attachment B. determines the shape of the virion C. protects the viral genome D. contains virulence factors that increase the severity of viral disease

B. determines the shape of the virion C. protects the viral genome Explanation: All viruses are comprised of repeating units of capsomeres (protein units) making up the viral capsid and either a viral RNA or DNA genome. Keep in mind that viral genomes are often varied in structure and can include: (+) sense single-stranded RNA (resembling eukaryotic RNA), (-) sense single-stranded RNA, double-stranded RNA, double-stranded DNA, and single-stranded DNA genomes.

Which component is unique to RT PCR? A. target DNA B. fluorescent oligonucleotide probe C. Taq DNA polymerase D. oligonucleotide primers

B. fluorescent oligonucleotide probe Explanation: The fluorescent oligonucleotide probe is the component of a real-time PCR that allows scientists to monitor the results of PCR as they occur. Taq DNA polymerase, oligonucleotide primers, and target DNA are all components found in standard, as well as real-time, PCR. These components, in addition to free nucleoside triphosphates, are all necessary for RT PCR to occur.

Which of the following research topics would require studying organisms at the single-cell level, rather than the population level? Select all that apply. A. observing the effects of an anaerobic environment on the translation of proteins in a colony of Mycobacterium tuberculosis (aerobic bacteria) B. locating where in a liver cell the protein albumin is produced C. following the trail of Campylobacter jejuni once it enters the human body D. selecting acid-resistant mutant E. coli cells to study acid resistance in genes

B. locating where in a liver cell the protein albumin is produced C. following the trail of Campylobacter jejuni once it enters the human body Explanation: GFP and YFP fusions have a broader range of functions than LacZ fusions, allowing scientists to study individual cells in a population. These fusions can track gene expression, identify the location of proteins inside cells, and follow the movement of pathogens within host animals. Scientists could use these fusions to follow the route of infection of Campylobacter jejuni in the human body, as well as discover the exact location in the liver cell that produces the protein albumin. They can also track gene expression, but a LacZ fusion would be more suited to studying a colony-wide response to stressful environments. Selecting acid-resistant mutant E. coli cells would also need to happen at the colony level in order to collect enough acid-resistant mutants for further study.

What does the (-)RNA need to make mRNA? A. envelope proteins B. poly-A tail C. 5' cap D. DNA polymerase

C. 5' cap Explanation: Within the nucleus, each genomic (-)RNA segment with its prepackaged polymerase synthesizes a (+) strand RNA for mRNA. Each mRNA synthesis initiates with a 7-methylguanine-"capped" RNA fragment.

Which features are desirable in a plasmid-cloning vector? A. restriction endonuclease gene(s) B. multicloning site C. antibiotic resistance gene(s) D. ligase gene E. ability to self-replicate

B. multicloning site C. antibiotic resistance gene(s) E. ability to self-replicate Explanation: The specialized plasmids used for gene cloning generally share some common features. First, plasmid vectors should be able to replicate independently of the host chromosome to high numbers (e.g., 100 or more per cell). Second, it is useful to have a locus that contains one or more restriction endonuclease recognition sequences (the multicloning site). In the presence of a corresponding restriction enzyme, the plasmid is cut (i.e., becomes linearized), and a foreign piece of DNA, cut with the same enzyme, can be inserted (see figure). The researcher then adds DNA ligase to seal the nicks flanking the insert. At this point, the recombinant vector is introduced into a host cell such as E. coli. If the plasmid carries a selectable marker (e.g., an antibiotic resistance gene), only cells containing the plasmid will be able to grow on the corresponding selective media.

Which of the following are features of operons? Select all that apply. A. riboswitch B. promoter C. protein-coding genes D. operator E. poly (U) tail F. start codon (AUG

B. promoter C. protein-coding genes D. operator Explanation: Prokaryotes tend to cluster genes required for particular functions or pathways into functional units of DNA known as operons. Transcription proceeds from a single promoter and ends at a single terminator sequence. This produces a "polycistronic" mRNA that is then translated to produce the encoded proteins. The level of gene expression may be controlled by repressors that bind to an operator site located between the promoter and start of transcription. Codons, riboswitches, and poly (U) tails are all features of RNA (not DNA).

During ATP synthesis, the energy in the Δp is used to drive the physical rotation of which portion(s) of the enzyme? A. the alpha and beta subunits of the F1 portion B. the gamma subunit connecting Fo and F1 C. the a and b subunits connecting F1 to the membrane D. the c subunits of the Fo portion

B. the gamma subunit connecting Fo and F1 D. the c subunits of the Fo portion Explanation: During ATP synthesis, energy in the Δp is converted to the rotational movement of the c subunits that comprise the Fo portion of the complex. This action also causes the gamma subunit (analagous to a drive axle) to turn within the F1 knob, producing the conformational changes in the catalytic subunits necessary for binding and phosphorylation of ADP.

One of the main drugs available to treat influenza is Tamiflu. How does it work? A. It blocks the replication of (-) strand RNA B. It blocks nucleocapsid proteins C. It blocks neuraminidase D. It blocks matrix proteins

C. It blocks neuraminidase Explanation: When a newly formed virion exits its host cell, neuraminidase acts as an enzyme to cleave a cell membrane protein, thus releasing the virion outside the cell. Neuraminidase can be blocked by the antiviral agent Tamiflu.

What is the hallmark characteristic of a phenotype whose expression is controlled by a quorum-sensing circuit? A. The phenotype is only expressed in mixed cultures (of one or more species). B. The phenotype is only expressed at the transition to stationary phase. C. Levels of expression initially do not reflect cell numbers but then rapidly rise past a critical concentration. D. Levels of expression increase as a direct function of cell concentration. E. Levels of expression decrease once a particular cell concentration is reached.

C. Levels of expression initially do not reflect cell numbers but then rapidly rise past a critical concentration Explanation: Quorum sensing involves small molecules (originally referred to as autoinducers) that are continuously produced and released into the surrounding medium during growth of a bacterial culture. As cell density increases, these molecules accumulate until they reach the threshold concentration required to induce expression of the target genes. This mechanism allows the cell to control cooperative phenotypes such as bioluminescence, virulence, and biofilm formation.

In Sample 1, you observe an increase in fluorescence over background levels after 18 cycles of RT PCR. In Sample 2, you observe the same change after 27 cycles. What does this tell you about Samples 1 and 2? A. Sample 2 had more starting DNA than Sample 1. B. Sample 1 is a reference sample and Sample 2 is an experimental sample. C. Sample 1 had more starting DNA than Sample 2. D. Sample 1 is an experimental sample and Sample 2 is a reference sample.

C. Sample 1 had more starting DNA than Sample 2 Explanation: As the video explains, "The time at which fluorescence increases over background levels reflects the amount of DNA in the original sample, with the earliest detectable reaction occurring in the sample with the most starting material." A sample that has 107 molecules of DNA will take fewer cycles to overcome background fluorescence than a sample with 103 molecules of DNA because more starting DNA molecules means more opportunity for RT PCR reactions to occur.

T-VEC is a promising potential treatment for tumors. What makes it especially effective at treating inoperable melanoma tumors? A. T-VEC can infect neurons. B. T-VEC can target epithelial cells. C. T-VEC releases tumor cell fragments after infecting a cell. D. T-VEC is an engineered form of HSV-1.

C. T-VEC releases tumor cell fragments after infecting a cell Explanation: T-VEC is an engineered form of herpes simplex virus 1 that infects only cancer cells, not healthy cells. It is thus considered an oncolytic virus, as it has an affinity for tumor cells. T-VEC was approved for treatment of melanoma in the United States in 2015. When T-VEC infects tumor cells, it releases tumor cell fragments. The host immune system is triggered by these fragments and learns to recognize these tumors. The immune system can then attack tumors found throughout the body. This technology is promising for the treatment of metastatic tumors that cannot be dealt with using traditional methods.

Why do target cells in gene therapy not produce viruses when they are infected by virions created by lentivectors in gene therapy? A. The viral RNA replicates outside the host genome. B. The transgene prevents the virions from passing on viral RNA to the target cell. C. Viral RNA transferred from the virion lacks necessary components from the HIV genome. D. Kill switches stop the virion from continuing infection in target cells.

C. Viral RNA transferred from the virion lacks necessary components from the HIV genome Explanation: HIV virulence genes are removed from the HIV genome. The genes that are necessary for virion production (gag, pol, env, and rev) are inserted into helper plasmids. These helper plasmids are available only when the lentivector is produced in cell culture. They do not enter the host organism. As a result, the genetic material transferred by the virions contains only the transgene, LTRs, and a few other genes, but not the genes necessary for producing virus particles, so the target cell merely uptakes the altered DNA and does not become a virus factory.

In E. coli, eight enzymatic steps are involved in the biosynthesis of arginine from glutamate. The synthesis of the required enzymes is subject to control by the ArgD protein, which, on its own, binds weakly to a regulatory sequence just downstream of the promoter regions for these genes. Upon the addition of excess arginine to the growth medium, the respective mRNA molecules are no longer produced. What is the likely role of arginine in the control of these genes? A. operator B. repressor C. corepressor D. activator E. inducer

C. corepressor Explanation: The ArgD protein exhibits significant affinity for DNA only when it is bound to its corepressor, arginine. Together, these bind to the operator region upstream of the arginine biosynthesis genes and decrease the level of gene expression. The activity of the first enzyme required for synthesis of arginine is also subject to feedback inhibition by this amino acid. Thus, the pathway for arginine biosynthesis is under both transcriptional and post-translational controls.

The above illustration is an example of phage display technology. This tool is used for which of the following? A. DNA shuffling B. biopanning a library of M13 display phage C. directed evolution of peptides D. linker peptides

C. directed evolution of peptides Explanation: Phage display is a tool for the directed evolution of peptides. Discovering peptides that bind to specific target proteins is simplified by phage display techniques.

Influenza is capable of genetic reassortment in which genetic components from two influenza strains reassort between themselves. Which animal strains of influenza prove especially virulent when reassorted with human strains? A. dogs and cats B. cows and sheep C. ducks and swine D. goats and horses

C. ducks and swine Explanation: Influenza strains that infect animals such as ducks or swine can reassort with human influenza strains if they are both infecting the same host. This can lead to a drastic increase in virulence and mortality. Ducks and swine live in close proximity to humans, which increases the likelihood and chances of coinfection and reassortment. The 1968 Hong Kong flu strain killed 33,000 people in the United States and contained three segments derived from avian strains. The H1N1 (swine flu) had segments derived from strains of influenza that infected pigs.

Viruses are most commonly known for infectious diseases such as influenza or measles; however, not all viruses pose harm to the human host, and, in some cases, they can contribute to human survival. Which of the following is an example of the benefits conferred by endogenous viruses? A. producing enzymes necessary for human digestion B. encoding the enzyme glycoside hydrolase used to digest milk C. encoding placental proteins that are essential for early development of human embryos D. acting as a component of the first line of immune defense educating immune cells

C. encoding placental proteins that are essential for early development of human embryos Explanation: Endogenous viruses are permanently embedded in the human genome, becoming germ-line encoded as they pass from the human host to their progeny. Many of these endogenous viruses provide beneficial functions that ensure the survival of the host (humans) and hence the virus. Examples of this are shown in Table 6.2. In particular, note the number of endogenous viruses that help facilitate the success of placental fusion, development, and function. Clearly, these viruses have a vested interest in the survival of human progeny!

What process does the host use to take up the virus? A. penetration B. expulsion C. endocytosis D. fusion

C. endocytosis Explanation: endocytosis is the process used for the influenza virus to enter the host cell. The influenza virion binds to its sialic acid receptor and a host protease cleaves hemagglutinin to form a fusion peptide. The fusion peptide involves a portion of an envelope protein that changes conformation to allow fusion with the host cell membrane.

Animals and most microorganisms are classified as chemoorganoheterotrophs. The prefixes "chemo," "organo," and "hetero" refer to how the organism meets its needs, respectively, for: A. carbon, electrons, and energy B. carbon, energy, and electrons C. energy, electrons, and carbon D. electrons, carbon, and energy E. electrons, energy, and carbon

C. energy, electrons, and carbon Explanation: Similar to animals, many microorganisms obtain energy from chemical (as opposed to light) reactions. The prefix "organo" means that the source of electrons (i.e., reducing power) is organic, rather than inorganic. Finally, the prefix "hetero" indicates that the organism requires carbon for biomass production, which has been prefixed by other organisms, such as plants. It is not really that complicated when you consider that chemoorganoheterotrophs typically meet all of these nutritional needs using just one food source - namely, fixed carbon supplied by other organisms. For this reason, they are usually just referred to as heterotrophs.

The time during which the inserted phage genome directs the production of progeny virions is called the A. eclipse period B. rise period C. latent period D. burst size

C. latent period Explanation: To observe one cycle of phage reproduction, phages are added to host cells at a multiplicity of infection (MOI, ratio of phage to cells) such that every host cell is infected. The phage particles immediately adsorb to surface receptors of host cells and inject their DNA. As a result, intact virions are virtually undetectable in the growth medium. This short period after infection is called the "eclipse period."

Consider this image illustrating the various protein-DNA interactions that occur in the promoter region of the lac operon. The configuration shown in panel A would occur in cells with _________ levels of lactose and _________ levels of glucose. A. high / high B. high / low C. low / high D. low / low

C. low / high Explanation: In panel A, the LacI repressor is shown binding as a tetramer to the operator regions, which partially overlap the lac promoter (lacP). Because LacI binds only in the absence of its inducer, lactose levels must be low. Additionally, binding of the CRP-cAMP complex is not observed in panel A. Because levels of cAMP rise only when glucose levels drop, sufficient amounts of this sugar must still be available.

The temperate phage lambda can undergo two life cycles depending on the environmental cues. During which cycle does phage lambda form a prophage? A. slow-release cycle B. burst cycle C. lysogeny D. lytic cycle

C. lysogeny Explanation: The life cycle selected by phage lambda is highly dependent on environmental cues. The longer the phage remains in the host cell, the greater the number of progeny phages produced. If environmental conditions are favorable for bacterial growth, the phage benefits by remaining in the host cell. Consequently, the phage DNA genome will be inserted into the bacterial genome, generating a prophage.

What are the primary mechanisms by which E. coli regulates the amount and activity of its heat-shock sigma H? Select all that apply. A. RNA degradation B. gene inversion C. proteolysis by chaperones D. translational control E. transcriptional controls F. anti-sigma factors

C. proteolysis by chaperones D. translational control Explanation: Sigma H activity in E. coli is regulated by two mechanisms. First, the actual sequence of the mRNA causes the molecule to assume a secondary structure at moderate temperature. The level of translation is therefore minimal because internal base-pairing masks the ribosomal binding site (RBS). Any σH that is produced becomes rapidly degraded by cellular chaperones (i.e., DnaK, DnaJ, and GrpE). If the cells experience subsequent heat shock (e.g., upshift to 42°C), the hydrogen bonding within the mRNA secondary structure is disrupted to expose the RBS, and σH is produced. The half-life of σH also increases because the chaperones are diverted to proteins that denature because of the thermal shock.

The final step in glycolysis catalyzed by pyruvate kinase couples the hydrolysis of phosphoenolpyruvate (PEP) to the formation of ATP via substrate-level phosphorylation of ADP. If the ΔG°' of ATP + H2O → ADP + Pi + H+ is -30.5 kJ/mol and the ΔG°' of the coupled reaction is -31.4 kJ/mol, what is the ΔG°' of PEP + H2O → pyruvate + Pi? A. -0.9 kJ/mol B. More information is needed C. +0.9 kJ/mol D. -61.9 kJ/mol E. +61.9 kJ/mol

D. -61.9 kJ/mol Explanation: Because ATP being formed rather than hydrolyzed in the reaction, the first reaction becomes: ADP + Pi + H+ → ATP + H2O. We then reverse the sign to indicate that it requires +30.5 kJ/mol to proceed. For this reaction to be driven by PEP hydrolysis, the ΔG°' of PEP + H2O → pyruvate + Pi must be greater than -30.5 kJ/mol. To calculate the actual value of PEP + H2O → pyruvate + Pi , we subtract the energy required for ATP formation (+30.5 kJ/mol) from the actual ΔG°' of the coupled reaction (-31.4 kJ/mol). Thus, -31.4 kJ/mol - 30.5 kJ/mol = -61.9 kJ/mol. You'll note that release of the phosphoryl group from PEP yields a little more than twice the energy compared to ATP. In fact, PEP has the highest-energy phosphate bond of any biological molecule.

Ultimately, respiration and fermentation are simply different ways to conclude sugar catabolism. Following the energy-yielding reactions, respiring cells pass electrons to oxygen (or other external inorganic oxidant), while fermenting cells dump electrons back onto a partially oxidized organic molecule (such as pyruvate), which then exits the cell as a waste product. Either way, the goal is to do which of the following? A. fully oxidize NADH into CO2 and H2O B. generate as much NADH as possible (then synthesize more NAD+ if necessary) C. recycle NADH back to NAD+ D. break down NADH into biosynthetic precursors E. stimulate de novo production of NAD+

C. recycle NADH back to NAD+ Explanation: During glycolysis and related pathways, energy-yielding oxidations reduce the electron carrier NAD+ to NADH. So, in addition to the 2 molecules of pyruvate and 2 net ATP that result from glycolysis, some energy is "stored" in NADH. That energy can be harvested via the electron transport chain, which, in aerobic respiration, ultimately transfers the electrons from NADH onto O2, making H2O. (Respiring cells generate additional NADH via the TCA cycle.) By donating 2e- + H+, NADH is reoxidized back to NAD+, and is again available for glycolytic oxidations. Cells lacking access to O2, or another suitable external electron acceptor, are left with no other alternative than to forgo the remaining energy in pyruvate. In the simplest type of fermentation, NADH is recycled by transferring electrons to pyruvate to make lactic acid. This waste product is then released by the cell into the surrounding medium. The cells loses out on some potential energy, but because NAD+ is regenerated in the process, glycolysis can continue.

The requirements of biosynthesis are carbon and other essential elements, as well as which of the following? A. oxidizing agents and energy for anabolic enzymes B. reducing agents and energy for catabolic enzymes C. reducing agents and energy for anabolic enzymes D. oxidizing agents and energy for catabolic enzymes

C. reducing agents and energy for anabolic enzymes Explanation: All organisms need carbon and other essential elements to build organic molecules. Because organic macromolecules are more reduced than the substrates used to make them, reducing agents such as NADPH are required for biosynthesis. Biosynthetic, anabolic enzymes require energy to build large organic molecules.

The energy cost of producing unneeded substrates in anabolic pathways has resulted in microbial species that A. have duplications of biosynthetic enzyme genes. B. constitutively express their biosynthetic enzymes. C. tightly regulate the expression and activity of their biosynthetic enzymes. D. constantly produce antimicrobial agents in the presence and absence of competitors.

C. tightly regulate the expression and activity of their biosynthetic enzymes. Explanation: Microbes face selective pressures to control the energy costs of anabolism. As a result, microbial species tightly regulate the expression and activity of biosynthetic enzymes to prevent the energy costs of producing unneeded products.

Which of the following is NOT a method by which latent viral infections, such as a latent herpes infection, persist inside their hosts? A. producing heterochromatin to inactivate viral proteins B. infecting long-lived cells, such as neurons C. using epithelial cells for viral replication D. mimicking immunosuppressive signals

C. using epithelial cells for viral replication Explanation: A lytic infection of the epithelial cells is usually the first step of an overall herpes infection. After infecting skin cells, the virus often moves to neurons as part of the latent infection process. These long-lived cells allow for a latent infection, as they are not destroyed and replaced as frequently as other cells. LAT proteins, an integral part of the latent infection, also use host cell proteins associated with the chromosome (heterochromatin) to inactivate all other viral proteins, thus preventing lytic replication. The viral DNA must also prevent apoptosis using a variety of methods, ensuring that the latent infection continues. Finally, herpes and retroviruses either express proteins or mimic signals that suppress the immune response, ensuring that the latent infection is maintained.

The Gibbs free energy change of a reaction can be calculated using the equation ΔG = ΔH - TΔS. A biochemical reaction that can provide energy for cellular metabolism would be one in which: A. ΔG was positive. B. all terms on the right are positive. C. ΔG was negative. D. all terms on the right are negative.

C. ΔG was negative Explanation: Reactions with a negative ΔG yield energy and can be used by the cell for growth, whereas reactions with a positive ΔG cannot proceed without the input of energy.

Where are capsid and envelope proteins translated? Select all that apply. A. Both are translated by ribosomes in the endoplasmic reticulum. B. Both are translated by free ribosomes in the cytoplasm. C. Capsid protein RNA is translated by ribosomes in the endoplasmic reticulum. D. Capsid protein RNA is translated by free ribosomes in the cytoplasm. E. Envelope protein RNA is translated by ribosomes in the endoplasmic reticulum. F. Envelope protein RNA is translated by free ribosomes in the cytoplasm.

D. Capsid protein RNA is translated by free ribosomes in the cytoplasm. E. Envelope protein RNA is translated by ribosomes in the endoplasmic reticulum. Explanation: DNA polymerase and other proteins replicate circular viral DNA using the rolling circle method. This produces new genomes for progeny viruses by generating a concatemer of many copies of the viral DNA. This newly synthesized DNA expresses late-stage mRNA that exits the nucleus for translation. Capsid proteins RNA is translated by free ribosomes, whereas envelope protein RNA is translated by ribosomes in the endoplasmic reticulum.

Sugars are used by many chemoheterotrophs as an important source of carbon and energy. Indeed, complete oxidation of glucose using oxygen as a terminal electron acceptor has a free energy change of -2,883.0 kJ/mol. What strategy do cells use to minimize the amount of that energy that is lost as heat during the catabolism of sugars? A. Oxidation is directly coupled to biomass synthesis. B. The oxidation reaction occurs within insulated organelles. C. The cells employ high-capacity energy carriers (e.g., ΔG°' of hydrolysis ~ -1000 kJ/mol). D. Energy is harvested using a series of smaller oxidations. E. Oxidations of sugars directly power membrane-bound ATP synthases.

D. Energy is harvested using a series of smaller oxidations Explanation: If glucose were oxidized in a single step, it would be much like igniting a stick of dynamite in the cell. It is simply too much energy for the cell to handle all at once. Even the most common energy currency in the cell (ATP) releases only 30.5 kJ/mol when it transfers its phosphate group to another molecule. To maximize efficiency and minimize heat loss, sugars are instead catabolized by cellular glycolytic reactions (such as those in the Embden-Meyerhof-Parnas pathway) that accomplish a stepwise harvesting and conservation of energy in carriers such as ATP and NADH.

The ATP synthase can be thought of as a type of molecule motor that is powered by the flow of ____________ through the subunits and __________ their concentration gradient. A. e- / up B. H+ / up C. e- / down D. H+ / down

D. H+ / down Explanation: The ETS of a cell uses the reducing power from central catabolism to pump protons from the cytoplasm to the exterior side of the membrane. The resulting proton motive force (i.e., Δp) represents potential energy that can be used by the cell for a variety of functions. In the case of ATP synthesis, protons move back into the cell and back down their concentration gradient through the subunits of the enzyme. The release of energy is coupled to phosphorylation of ADP.

All of the following characteristics are reasons why naturally occurring bacterial plasmids make good cloning vectors, except which one? A. Plasmids are generally small. B. Plasmids can carry genetic markers, such as antibiotic resistance genes. C. Plasmids can be transferred into appropriate host cells. D. Plasmids can carry multiple cloning sites (MCS), which contain numerous restriction endonuclease sites.

D. Plasmids can carry multiple cloning sites (MCS), which contain numerous restriction endonuclease sites. Explanation: Plasmids are naturally small, they carry genetic markers and survival genes, and they can be transferred between cells easily, all of which make them good cloning vectors. However, while plasmids often contain restriction endonuclease sites, cloning vectors are artificially constructed to include multiple cloning sites for convenience.

Some bacteria, such as Salmonella, use a ________ to regulate the half-lives of sigma factors during different stages of growth.

protease Explanation: Sigma factors are proteins, and would therefore be degraded by proteases. In Salmonella, the stress/stationary phase sigma factor (RpoS σ38) is degraded during exponential phase by the ClpXP protease.

The yeast two-hybrid technique maps protein-protein interactions. It takes advantage of the modular nature of transcription factors that contain a DNA-binding domain and an activation domain. Two different hybrid proteins are made: one with a bait protein (labeled X) fused to the DNA-binding domain of the transcription factor, and another with the prey protein (labeled Y) fused to the activation domain of the transcription factor. Libraries of plasmids containing bait and prey fusion proteins are transformed into yeast cells. When the bait and the prey interact, the DNA-binding domain and the transactivation domain are brought together, leading to transcription of a reporter gene. As with many techniques, both false positives (a signal when the proteins don't actually interact) and false negatives (no signal when in fact the two proteins do normally interact) are possible. 1. What condition could lead to a false positive result? A. Fusion of the activation domain to the prey protein prevents proper prey protein folding. B. Fusion of the DNA-binding domain to the bait protein prevents proper bait protein folding. C. The prey protein itself (protein Y) contains an activation domain. D. T

D. The bait protein itself (protein X) contains an activation domain. Explanation: False positives occur when the reporter gene is expressed in the absence of bait and prey interaction. This can happen when the bait protein contains an activation domain or if the prey protein contains a DNA-binding domain that binds a sequence near the reporter construct. When a signal is obtained in a yeast two-hybrid assay, control experiments examine the bait alone and prey alone. No signal should be observed. If either the bait alone or the prey alone increases reporter gene expression, then the original signal was a false positive.

Plaque assays are notoriously finicky assays, and whether for a bacteriophage or an animal virus, they will require a soft agar overlay. Which of the following best explains the rationale for this approach? A. The soft agar overlay allows for detection of the burst size B. The soft agar overlay allows the clumping of infected cells to pull away from the culture flask C. The soft agar layer provides a tissue-like matrix mimicking an organ D. The soft agar overlay prevents the free diffusion of progeny virions to cells

D. The soft agar overlay prevents the free diffusion of progeny virions to cells Explanation: Viruses can only infect neighboring cells on the tissue culture flask or agar plate (bacteriophages). The free diffusion of virions would allow progeny virions to infect distant cells across the culture plate. If this occurred, it would be impossible to determine the original viral titer. These assays are performed with a low enough multiplicity of infection to allow a single virion to infect only one cell, and the progeny virions can only travel to neighboring cells. This approach allows for areas of plaques of lysed cells to be counted and correlated back to the original viral load.

Which of the following does NOT contribute to the exceptional virulence of influenza? A. spread of strains between species B. random mutations in viral genome C. reassortment of segmented viral genome D. ability to form latent infections in neurons

D. ability to form latent infections in neurons Explanation: Influenza virus has a segmented genome that consists of multiple, separate nucleic acids, similar to the multiple chromosomes of a eukaryotic cell. The influenza virus has eight segments. If two different strains infect a host at the same time, these eight segments can reassort to form a new strain. This new strain can be created by two strains from two separate host species. This is how the H1N1 flu developed. Additionally, influenza viruses experience random mutations, just like other viruses. These random mutations, on top of the process of reassortment, make influenza an especially virulent virus.

Bioinformatics is a powerful tool that has enabled microbiologists to do which of the following? Select all that apply. A. quantify the infectious dose of a particular pathogen B. genetically engineer bacteria C. determine the biochemical composition of a cell D. characterize the species composition of a soil microbial community E. predict the function of a gene based on its nucleotide sequence

D. characterize the species composition of a soil microbial community E. predict the function of a gene based on its nucleotide sequence Explanation: Bioinformatics utilizes sophisticated mathematical algorithms for computer-based DNA sequence analysis. These approaches allow researchers to analyze the taxonomic composition of complex microbial communities and predict functions of genes with unknown functions. While results are generally predictive, they are important for generation of hypotheses that can then be experimentally tested. Genetic engineering (e.g., cloning) and investigations to determine the infective dose of a pathogen or the biochemical compositions of cells do not require large scale computer-assisted analyses.

Sialic acid attached to galactose provides a recognition site for which of the following? A. phospholipid bilayer B. neuraminidase C. matrix protein D. hemagglutinin

D. hemagglutinin Explanation: For the virus to enter the cell, the following has to happen: 1. An influenza virion attaches to a cell by its hemagglutinin (HA) envelope protein binding a receptor protein. 2. Host protease cleaves HA, forms fusion peptide. 3. The endocytic vesicle acidifies, and the lower pH induces a conformational change in HA. 4. The fusion peptides extend into the vesicle membrane, where they mediate fusion between viral and host membranes. 5. The membrane fusion releases the content of the virion into the host cytoplasm.

All strains of human papillomavirus (HPV) are capable of integrating their double-stranded DNA viral genome into host basal cells while waiting for keratinocyte differentiation before initiating viral replication for virion shedding. As a consequence of the integrated genome, HPV strains that more readily transform host cells are more likely to cause an increased expression of viral A. tumor suppressor genes B. lysogeny genes C. lytic genes D. oncogenes

D. oncogenes Explanation: Cells transformed by viral infection often express higher levels of viral oncogenes. Oncogenes are a result of a "gain-of-function" mutation that causes excess proliferation in host cells. In the case of human papilloma virus (HPV), genes E6 and E7 are viral oncogenes involved in cervical carcinogenesis that destabilize two cellular tumor suppressors, p53 and pRb, respectively.

Identify an infectious agent that is comprised of only a nucleic acid particle. A. virion B. bacteriophage C. prion D. viroid

D. viroid Explanation: A viroid is a plant virus that enters through the damaged plant cell wall and is responsible for significant economic losses in citrus crops. The potato spindle tuber viroid (Figure 6.13) consists of a circular, single-stranded molecule of RNA that doubles back on itself to form base pairs interrupted by short, unpaired loops. This unusual circularized form avoids breakdown by host RNase enzymes. In some cases, viroids have catalytic ability and thus can cleave themselves (similar to ribozymes) or other specific RNA molecules.

What process is illustrated by the diagram below? A. fermentation B. reverse electron flow C. glycolysis D. anaerobic respiration E. aerobic respiration

E. aerobic respiration Explanation: In this example, electrons originating from NADH are passed through a series of membrane-bound carriers (cytochromes) of increasing reduction potential (orange arrow) to oxygen. This process is known as aerobic respiration and is responsible for creating the proton motive force that most cells use to perform work. A similar electron transport chain also operates during anaerobic respiration, although in those systems, some terminal electron acceptor other than oxygen is reduced.

Because the reduction potential of the CO2/glucose redox pair is more negative than the Fe3+/Fe2+ redox pair, energy is released as electrons flow A. from Fe2+ (the donor) to CO2 (the acceptor). B. from CO2 (the donor) to Fe3+ (the acceptor). C. from Fe2+ (the donor) to glucose (the acceptor). D. from glucose (the donor) to Fe2+ (the acceptor). E. from glucose (the donor) to Fe3+ (the acceptor).

E. from glucose (the donor) to Fe3+ (the acceptor) Explanation: Glucose is a highly reduced form of carbon, and therefore it makes an excellent electron donor to the oxidized form of a redox pair with a more positive reduction potential (i.e., the molecules in the pink column of Table 14.1).

Which of the following modifications would likely cause constitutive expression of a gene that is normally regulated by an activator protein? A. modifying the activator protein so that it cannot bind the cognate inducer B. deleting the operator sequence downstream of the promoter C. deleting the transcription terminator D. deleting the regulatory region of DNA that binds the activator-inducer E. replacing the promoter with the consensus sequence for RpoD σ70

E. replacing the promoter with the consensus sequence for RpoD σ70 Explanation: Activator proteins with bound ligand (inducer) normally promote transcription by strengthening an otherwise weak association of RNA polymerase holoenzyme with the promoter. In contrast, expression from promoters that closely match the consensus sequence for RpoD σ70 is constitutive (although these may still be subject to repression by regulatory proteins bound to an operator).

Correctly order the steps necessary to perform a plaque assay with phages. A. Combine phage-infected bacteria with the top agar. B. Combine phages and bacteria. C. Pour immediately onto the agar plate. D. Incubate overnight at 37°C. E. Rotate to spread evenly and allow agar to solidify. F. (photo of plaque) Courtesy of Kema Malki

Explanation: A phage plate culture relies on the lysis of bacterial cells to indicate the viral load. At a low multiplicity of infection, each plaque will arise from a single phage infecting a bacterial cell and transferring to neighboring cells. First, phages are added to a suspension of bacterial cells and allowed to infect. Next, the phage-infected bacterial cells are added to a soft agar. The phage-infected bacterial cells and agar are then added evenly to an agar plate. The soft agar keeps the bacteria and phages from freely diffusing across the plate and ensures that phages can only transfer between connected bacterial cells. The bacteria then grow in a lawn covering the plate, and the phage-infected cells lyse, creating "plaques" or gaps between the cells. The plaques (areas where cells have lysed) can be used to determine how many plaque-forming units were present in the initial phage stock.

Please label the following image, which illustrates the effect of enzymes on reactions. A. Free energy B. Activation energy with enzyme C. ΔG D. Reactants E. Activation energy without enzyme F. Reaction progress G. Products

Explanation: As catalysts, enzymes accelerate reactions by lowering the activation energy required for a given reaction to proceed. Reactions that would otherwise take millions of years to occur may take only milliseconds when mediated by the appropriate enzyme.

Label the image illustrating how glucose is metabolized during aerobic respiration. A. 34 ATP B. 2 CO2 C. 2 ATP D. ETS E. 4 CO2 F. H+

Explanation: Catabolism of glucose to pyruvate yields 2 NADH in addition to 2 net ATP (2 ATP are expended in the preparatory reactions of glycolysis, and 4 are produced via substrate-level phosphorylations in the oxidative phase). Pyruvate dehydrogenase catalyzes the oxidative decarboxylation of pyruvate to acetyl-CoA, yielding 2 more NADH and 2 CO2 per glucose metabolized. The two molecules of acetyl-CoA are then oxidized by the TCA reactions to produce an additional 6 NADH and 2 FADH2 in addition to 2 molecules of ATP via substrate-level phosphorylation. Also, when operating as a complete cycle, a total of 4 atoms of carbon enter as 2 molecules of acetyl-CoA, while an equivalent amount is released as 4 CO2. The accumulated NADH and FADH2 then donate electrons to the electron transport system (ETS), which couples the stepwise release of energy to the extrusion of protons (H+). This electrochemical gradient powers the membrane-bound ATP synthase that produces most of the ATP resulting from aerobic respiration. The theoretical yield from oxidative phosphorylation alone is 34 ATP per glucose, although practically, it may be less.

Match each genome with the first polymerase it uses to initiate viral synthesis. 1. DNA 2. Host RNA 3. vRNA 4. Reverse - (+) sense single-stranded RNA - Double-stranded DNA - Double-stranded DNA pararetroviruses -Double-stranded RNA - Single-stranded DNA - (-) sense single-stranded RNA - (+) strand RNA retroviruses

Explanation: Double-stranded viruses will either use the host cell DNA polymerase or encode a viral DNA polymerase. RNA viruses are more complex. Double-stranded RNA viruses are unlike any part of the central dogma in host cells and therefore require an RNA-dependent RNA polymerase, whereas (+) sense single-stranded RNA viruses can use the host cell RNA polymerase. In contrast, (-) sense single-stranded RNA viruses require a viral RNA polymerase to create the (+) strand RNA. Although, retroviruses enter with (+) sense single-stranded RNA, these viruses encode a viral reverse transcriptase to convert the viral RNA to DNA, and then use host cell machinery. Pararetroviruses use a reverse transcriptase to convert their viral genome to an intermediate stage before replicating the genome.

Please label the following figure illustrating generation of the proton motive force and some processes that it drives. A. Symport B. Flagellar rotation C. Electron transport system D. ATP synthesis E. Antiport

Explanation: Electron transport systems use energy harvested from central metabolism to pump protons across the membrane. The resulting Δp is used to power ATP synthesis, flagellar rotation, and many types of transport. Note that export of drugs through an antiporter can be coupled to influx of protons, whereas other antiporters can use H+ influx to create secondary ion gradients (e.g., of Na+). Symporters couple the energy in ion gradients to the uptake of other molecules against their concentration gradients.

Phage display has recently been used to discover single-chain antibodies capable of binding to a unique surface protein found on cancer cells. The antibody chain interrupts the functions of this protein and weakens the cancer cell. In the phage portion of the figure below, identify the single-chain antibody. Within the magnified DNA strand (lower section of the figure), identify the sequence encoding the single-chain antibody. A. Single-chain antibody B. Sequence-encoding single-chain antibody

Explanation: Figure 12.20A (question above) shows the components of phage display. Single-chain antibodies take the same place as proteins of interest, connected to the linker peptides and the product of gene 3 (g3p) at the top of the phage. The genetic sequence for the single-chain antibody is incorporated into the single-stranded DNA in between the g3p signal peptide and the sequence encoding for the linker peptide. By genetically engineering the phage in this way, scientists have developed a potential tool in the fight against cancer.

Match the virus with the corresponding viral structure. 1. Filamentous viruses 2. Asymmetrical viruses 3. Icosahedral viruses - HIV - Vaccinia virus - Ebola - Influenza virus - HPV

Explanation: Filamentous viruses such as Ebola have a helical shape comprised of repeating capsomer subunits. Another example of a filamentous virus is that of the tobacco mosaic virus shown in Figure 6.10. Asymmetrical viruses often are complexed with additional protein subunits, such as that of influenza virus coated with nucleocapsid proteins or the very large vaccinia poxvirus depicted in Figure 6.12B. Symmetrical icosahedral viruses such as human immunodeficiency virus (HIV) and human papillomavirus (HPV) are comprised of 20 repeating capsid proteins and have 12 corners.

Drag the molecule labels to their correct location within the biosynthetic pathway. A. Lipids B. Peptides C. Glucose D. Acetyl-CoA

Explanation: Follow the "Gluconeogenesis" arrow in the figure. Glucose is the final product of gluconeogenesis. Follow the arrow leading away from "Fatty acids" in the figure. Lipids are formed from fatty acids and glycerol. Then follow the arrow leading away from "Amino acids" in the figure. Peptides are short chains of amino acids and proteins are longer chains of amino acids. Finally, follow the arrow leading away from the box labeled "Heterotrophy: Catabolism of Sugars, Amino acids, and Fatty acids." Acetyl-CoA is produced by catabolism of organic molecules.

Identify the steps of the retrovirus life cycle by dragging the labels to the appropriate targets. A. Uncoating of envelope B. (+) RNA reverse transcribed to dsDNA C. RNA polymerase transcribing viral RNA D. Translation of viral proteins E. Virion binding to host cell receptor F. dsDNA integration into host genome G. Uncoating of capsid and genome release

Explanation: Human immunodeficiency virus (HIV) is a retrovirus that upon binding to the host cell enters and undergoes envelope uncoating. Next, the capsid is uncoated from the viral genome. The encoded viral reverse transcriptase converts the (+) RNA genome to dsDNA, which is then integrated into the host genome. The host RNA polymerase II transcribes the DNA, creating mRNA. The mRNA is then translated in the cytoplasm by ribosomes. Since the viral DNA is integrated into the host genome, the viral mRNA and proteins are made simultaneously with the host mRNA and proteins.

HIV, which only infects humans, is considered to have a _________ host range, whereas rabies virus, which can infect a number of animals and humans, is considered to have a __________ host range.

narrow broad Explanation: HIV has a narrow host range, only infecting humans; even close relatives such as chimpanzees are not susceptible to the virus. On the other hand, rabies can infect humans, canines, bats, and a number of other species, creating a broad host range for infection.

Sort each item as a characteristic of 1. Respiration 2. Fermentation 3. Both - Low ATP yield; primarily from glycolysis - Pyruvate is major intermediate - Most carbon remains in organic form. - High ATP yield; most from oxidative phosphorylation - Important for catabolism of sugars - Electrons reduce an external electron acceptor (e.g., O2) - NADH major electron carrier - Electrons from catabolism end up back on partially oxidized substrate - Most carbon released as CO2

Explanation: In both respiration and fermentation, 6-carbon sugars are metabolized to pyruvate via glycolysis. This initial shared process involves oxidations using NAD+ to accept electrons, and yields a small amount of ATP by substrate-level phosphorylation. In respiring organisms, the pyruvate can then be fully metabolized to CO2. The accumulated NADH is also recycled by ridding itself of electrons (and protons) via the electron transport chain. The resulting proton motive force generates a lot of ATP (up to 38 ATP per glucose) via oxidative phosphorylation. During fermentation, no external electron acceptor is used, but NADH still must be recycled (back to NAD+) for glycolytic oxidations to continue. Thus, the partially oxidized organic substrate itself (or a derivative) serves as the terminal electron acceptor and a fermentation end product - with most of the carbon remaining - is dumped as waste.

Energy from cellular metabolism is converted to ATP by respiring organisms. Place the following steps in the correct order. - NADH and FADH are oxidized by electron transport proteins. - Influx of H+ through ATP synthase drives ATP production. - Glycolysis and TCA cycle generate NADH & FADH. - An electrochemical gradient of protons is established (Δp). - Electron transport releases energy that is used to translocate H+.

Explanation: In the simplest example (as shown in the hint), electrons from cellular oxidations of organics are supplied to the electron transport system. Mitochondrial respiration uses oxygen as the terminal electron acceptor, but many prokaryotes are also capable of using alternative electron acceptors in anaerobic respiration. In either case, the sequential redox reactions within the chain couple transfer of electrons to proton pumping, which establishes a proton motive force (Δp). Protons reentering the cell through ATP synthases provide energy for phosphorylation of ADP to ATP.

Label the illustration of the TCA cycle with the correct substrates and key intermediates. A. Succinyl-CoA B. Acetate C. Oxaloacetate D. Pyruvate E. Citrate F. 2-Oxoglutarate G. Acetyl-CoA

Explanation: Pyruvate from glycolytic pathways is first decarboxylated and converted to acetyl-CoA. Acetate is similarly funneled into the TCA cycle via acetyl-CoA. Incoming acetyl-CoA (2C) condenses with oxaloacetate (4C) to form citrate (6C). Citrate then isomerizes to isocitrate (not shown), which is oxidized to produce 1 NADH along with the release of 1 molecule of CO2. The resulting 2-oxoglutarate (5C) similarly undergoes oxidative decarboxylation and incorporation of CoA to produce succinyl-CoA (4C). A substrate-level phosphorylation then produces 1 ATP (or GTP) to produce succinate (not shown). Additional oxidations occur that regenerate oxaloacetate and provide an additional FADH2 and NADH.

Match the category of retroelement with its definition. 1. Endogenous 2. Retrotransposons 3. LINEs & SINEs - Retain vestigial retroviral genomic elements - Retain all retroviral genomic elements - Retain partial retroviral genomic elements and may have reverse transcriptase

Explanation: Retroviral genomes in varying states of decay are termed retroelements. Endogenous retroviruses (human endogenous retroviruses are called HERVs) maintain all the genomic elements of the original retrovirus, including the gag, env, and pol genes. Retrotransposons maintain only a partial set of retroviral genomic elements, but they may also have reverse transcriptase, which allows them to copy themselves into other locations in a genome. Long interspersed nuclear elements (LINEs) and short interspersed nuclear elements (SINEs) both have more vestigial remnants of the retroviral genome. These retroelements are responsible for almost half of the human genome.

Label the figure with the correct description of steps in a polymerase chain reaction (PCR) as well as the required temperature for each step by dragging and dropping the item into the appropriate circle. Place the temperature label in the top circle of each step. A. 55ºC B. 95ºC C. 72ºC D. Denaturation: Heat denatures and separates strands E. Repeats cycles 25-30 times F. Annealing: Primers anneal

Explanation: The cyclic polymerase chain reaction (PCR) makes a large number of copies of a small piece of DNA. The PCR reaction mixture is first heated to 95ºC to denature and separate the DNA strands. This mixture is then cooled to 55ºC to allow specialized primers to anneal to the separated strands, and subsequently heated to 72ºC to allow the polymerase to replicate and extend the DNA strands. The heat-stable DNA polymerase Taq, from the thermophile Thermus aquaticus, is used because the reaction mixture must undergo repeated cycles of heating and cooling. The optimal reaction temperature for Taq is 72ºC.

Match the virus with the mechanism of uncoating the viral genome. 1. Plasma Membrane 2. Cytosol 3. Endosome - Influenza virus - Papillomavirus - Picornavirus - Retrovirus

Explanation: The double-stranded DNA of papillomavirus is uncoated in the cytoplasm and then transported into the nucleus through the nuclear pore. The (+) sense single-stranded genome of a picornavirus is inserted directly into the cytoplasm from the plasma membrane, but is later sequestered in replication complexes comprised from the endoplasmic reticulum. Influenza virus, however, is uncoated in the late endosome due to a pH change allowing the dissociation of the capsid proteins. Finally, the (+) sense single-stranded retroviral genome is uncoated in the cytoplasm.

Label the diagram illustrating the general mechanism of transcriptional control by a regulatory protein. A. Regulatory sequence B. Promoter C. Target gene D. mRNA E. Regulatory protein F. Ligand binding

Explanation: The most common methods of transcriptional control involve regulatory proteins (activators or repressors) whose binding efficiency is often influenced by specific ligands. Transcriptional activators typically bind to regulatory regions upstream of the target promoter (as shown in this illustration). In contrast, repressors usually bind to regulatory regions (i.e., operator sequences) located between the promoter and transcriptional start site.

Order the terms to list the steps from HIV retroviral genome replication to protein synthesis. - Reverse transcriptase - Translation of viral proteins by host ribosomes - (+) ssRNA genome - dsDNA genome - RNA polymerase II transcription - Integration into host genome

Explanation: The retrovirus HIV begins with a (+) sense single-stranded RNA genome, which normally could be directly translated by ribosomes. However, as a retrovirus, HIV reverse transcribes the (+) sense single-stranded genome to dsDNA. The DNA is then integrated into the host genome and transcribed by RNA polymerase II alongside host mRNA. Finally, host ribosomes are used to translate viral mRNA into viral proteins, which are used to assemble the complete virion before budding.

Label the diagram illustrating how the cAMP-CRP complex activates transcription. A. α-NTD B. DNA C. cAMP-CRP D. α-CTD E. RNA polymerase

Explanation: This diagram illustrates how cAMP-CRP complex behaves as a positive transcriptional activator of otherwise weakly expressed genes. Typically, such genes (which include the lac operon) are involved in the catabolism of alternative carbon sources and thus are not expressed when the preferred carbon source (i.e., glucose) is available. This mode of control is also known as catabolite repression. Once glucose is depleted, however, cAMP accumulates and combines with CRP. As shown here, the corresponding regulatory sequence for this complex is located about 60 base pairs upstream from the transcriptional start site. When bound to this region, cAMP-CRP physically interacts with the α-NTD of the RNA to promote open complex formation and upregulate transcription.

Pseudomonas aeruginosa carrying a plasmid-borne N-acyl-homoserine lactonase gene would likely become _______

avirulent Explanation: Many pathogens use quorum sensing (QS) to carefully time expression of genes that are required for virulence. This strategy is analogous to an army increasing its chances for victory by building up a sufficient force before mounting an attack. In Pseudomonas aeruginosa, QS-regulated phenotypes include biofilm formation and production of pyocyanin, exoenzymes (e.g., proteases), and cytotoxins. Similar to Allivibrio fischeri, P. aeruginosa produces and responds to a specific acyl-homoserine lactone autoinducer. Inactivation of this molecule by acyl-homoserine-lactonase disrupts the circuit and prevents expression of the genes required for colonization and destruction of host tissue.

Increased synthesis of _______ is a direct result of decreasing glucose levels.

cAMP Explanation: As fewer molecules of glucose are transported into the cell, adenylyl cyclase activity increases, causing levels of cAMP to rise. This molecule can then combine with the cAMP receptor protein (CRP) to form a complex that binds to certain DNA regulatory sequences. Usually, as in the case of the lac operon, bound CRP-cAMP functions as an activator to upregulate gene expression.

The amino acid tryptophan functions as a(n) __________ in the control circuit of the trp operon.

corepressor Explanation: In contrast to the LacI repressor, the Trp repressor does not bind to DNA in the absence of a ligand. It is, therefore, referred to as an aporepressor. The corepressor is tryptophan, which, when present in excess, combines with the aporepressor to form the active holorepressor. Binding of the holorepressor to the corresponding operator sequence (adjoining the trp promoter) blocks RNA polymerase, resulting in an approximate 100-fold decrease in the level of transcription.

Temperate bacteriophages can undergo two routes of infection. During ________ the phage DNA is inserted into the bacterial genome and replicated each time the bacterial cell divides.

lysogeny Explanation: The lytic cycle takes place when a bacteriophage enters a bacterial cell, replicates progeny phages, and an enzyme lyses the bacterial cell wall for exit. In contrast, when conditions are favorable for the host cell, the bacteriophage may undergo the lysogenic cycle by integrating the phage genome and having the genome replicated each time the bacterial cell divides. In the case of lysogeny, if environmental conditions become unfavorable, the phage will excise the DNA genome and undergo the lytic cycle.

Certain archaea can also generate a small amount of energy by using the hydrogen released by E. coli to produce _________.

methane Explanation: Approximately 30% of people harbor methanogens as part of their gut microbiome. These archaea are able to generate a small amount of energy via the generation of methane from hydrogen and carbon dioxide.

The explosive gas, H2, may be produced by commensal E. coli metabolizing sugars via _______ fermentation.

mixed-acid Explanation: Mixed-acid fermentation by E. coli results in the production of several carboxylic acids (e.g., formate, acetate, succinate, and lactate) in addition to ethanol, hydrogen, and carbon dioxide.