B/B practice

A researcher is attempting to overexpress a gene in a positive inducible operon. This operon is currently not transcribed because the activator proteins are unable to bind the DNA. To increase the transcription of this operon, the researcher could add: A. an inducer. B. an inhibitor. C. a repressor. D. a nuclease.

A is correct. Addition of an inducer will increase the rate of transcription. In a positive inducible operon, the activator protein is incapable of binding the DNA during the basal state. Upon binding of the inducer, the activator can undergo a conformational change that allows it to bind DNA and initiate transcription. B: Inhibitors disable activators, so this would not upregulate transcription in a positive inducible operon. C: Repressors bind to operators to inhibit transcription. D: Nucleases are enzymes that cleave nucleic acids. The addition of such enzymes would break down the DNA, which certainly is not what the researcher wants.

In what way is apoptosis distinct from necrosis? A. Apoptosis is a naturally-triggered event, while necrosis involves uncontrolled cell death. B. Apoptosis produces inflammation, while necrosis is not associated with any noticeable symptoms. C. Apoptosis is typically caused by bacterial or fungal infections, while necrosis is a process stimulated by signals within the cell. D. Apoptosis usually requires medical treatment, while necrosis typically resolves itself without external attention.

A is correct. Apoptosis, sometimes known as "programmed cell death," represents a naturally-occurring physiological process. In contrast, necrosis is generally triggered by an environmental stressor, such as extreme cold, injury, or lack of blood flow to an area. This promotes the death of otherwise healthy cells. B: Apoptosis is typically thought to be non-inflammatory. C: Again, apoptotic pathways are directly triggered by internal factors, not infection. In fact, these descriptions are reversed. D: This choice also switches the characteristics of apoptosis and necrosis.

While DNA and RNA are strikingly similar macromolecules, they also differ in a variety of ways. Which of the following statements regarding DNA and RNA is accurate? I. DNA contains a more stable sugar than RNA. II. RNA can catalyze biochemical reactions, but DNA cannot. III. DNA and RNA cannot base pair with each other. IV. DNA and RNA are found throughout the cell. A. I and II only B. II and III only C. III and IV only D. II, III, and IV only

A is correct. DNA contains a deoxyribose sugar, while RNA is formed with ribose, a sugar with a free 2' hydroxyl group. This small difference makes RNA more reactive and unstable; for this reason, it rapidly degrades once exiting the nucleus. Ribozymes are RNA-based enzymes, and as such, are able to that can catalyze biochemical reactions. Virtually all other enzymes are composed of protein. III: A DNA strand can certainly hybridize with an RNA polymer. In fact, this process must be able to occur for transcription to take place. IV: DNA is localized in the cell and found in either the nucleus or the mitochondria. However, RNA is found throughout the cell, whether rRNA (in the form of ribosomes), tRNA (in the cytosol), or mRNA (which is transcribed in the nucleus, then transported into the cytoplasm).

Interphase, which comprises the bulk of the cell cycle, is itself split into multiple phases. Which segment of interphase generally spans the longest period of time? A. G1 B. G2 C. G0 D. S

A is correct. During the G1 phase, the cell conducts protein and organelle synthesis at a high rate while the cell grows in size. The transition from the G1 to the S phase is termed the "restriction point" and constitutes the rate-limiting step in the cell cycle.

Back side

A is correct. Molecule A is absorbed at a rate directly correlated with its concentration outside the cell, suggesting that it can easily diffuse through the membrane under any conditions. In contrast, molecule B seems to display Michaelis-Menten kinetics, with the concentration of the compound impacting rate at low, but not high, concentrations. This implies that a protein is required to transport the molecule inside the cell. Therefore, molecule A is likely small and nonpolar, while molecule B must be large, polar, or both. B: Since aldosterone and cortisone are both steroid hormones, they would be expected to diffuse similarly. This does not agree with the results shown in the graph. C: Since glucose and propanol are both polar molecules, they would also show similar diffusion patterns. D: This is the reverse of the correct answer.

Which organisms are dependent on the electron transport chain to create ATP? A. All obligate aerobes B. Eukaryotic, but not prokaryotic, obligate aerobes C. Obligate anaerobes D. Aerotolerant anaerobes

A is correct. Obligate aerobes, whether prokaryotic or eukaryotic, cannot survive without creating ATP through aerobic means. Aerobic respiration involves the electron transport chain. B: This is a common point of confusion. Although prokaryotes do not possess mitochondria, they still utilize the electron transport chain if they are aerobic. C: This is the reverse of the correct answer; obligate anaerobes exclusively use anaerobic respiration and often die in the presence of concentrated oxygen. D: Aerotolerant anaerobes are anaerobic species for which oxygen is not lethal. However, as anaerobes, they do not use the ETC.

In which of these stages do homologous chromosomes separate into distinct cells? I. Anaphase of meiosis I II. Anaphase of meiosis II III. Anaphase of mitosis A. I only B. II only C. I and III only D. II and III only

A is correct. Remember the difference between homologous chromosomes and sister chromatids! Homologous chromosomes are the pairs of non-identical chromosomes that are present in every somatic cell. (For example, you inherit one version of chromosome 2 from your father and one from your mother; these structures are homologous.) These pairings separate during meiosis I, causing the daughter cells generated by this process to be haploid before the start of meiosis II. II, III: Both of these processes involve the separation of sister chromatids. Be especially careful when considering mitosis; as this event yields identical daughter cells, we would not want to segregate the different members of a pair of homologous chromosomes into separate cells

758 of 1000 individuals have a dominant trait in which they cannot differentiate between the colors periwinkle and lavender. What fraction of the population consists of people homozygous for this allele? A. 25% B. 50% C. 67% D. 75%

A is correct. The 758 individuals described are those who express the dominant allele, meaning that they can be either homozygous or heterozygous for the trait. If we round 758 to 750, we can estimate that 75% of the population has the inability to differentiate these colors; in other words, p2 + 2pq = 0.75. While this may be difficult to solve for p, we can easily use it to find q2 using the equation p2 + 2pq + q2 = 1. If 0.75 + q2 = 1, then q2 = 0.25 and q = 0.5. Thus, p + 0.5 = 1, and p = 0.5. Finally, we can use this value to find that p2 (the proportion of homozygous dominant individuals) is 0.25, or 25% of the population. B: This is q, not p2. D: This results from the mistaken assumption that all of the individuals described in the question are homozygous dominant. In reality, most are heterozygotes.

A neuroscientist observes that treatment of temporal neurons with a certain toxin causes a depolarization of approximately 8 mV as compared to the resting value, but no action potential. He hypothesizes that this is caused by the opening of additional voltage-gated sodium channels. Is this explanation reasonable, and if not, why? A. Yes; this explanation makes sense. B. No; the opening of additional sodium channels would cause a hyperpolarization, not a depolarization. C. No; such a depolarization must trigger an action potential, albeit a small one. D. No; the opening of additional sodium channels would cause the neuron to become more positive, not more negative.

A is correct. The opening of voltage-gated Na+ channels would allow sodium ions to enter the neuron. Such an action certainly should depolarize the cell. Furthermore, if only a small number of channels were opened, this effect would likely not be large enough to reach the threshold value. (Note: Be very careful when reading this question stem! The question stem did NOT state that the membrane potential reached +8 mV; it only said that it depolarized by 8 mV compared to the resting value. For example, if the resting potential were -70 mV, this depolarization would bring the potential to -62 mV.) C: This statement is not true. Action potentials are all-or-nothing in nature, so a subthreshold depolarization will not provoke a response. D: This answer mistakenly interprets the term "depolarization," which is used when a cell has become less negatively charged. In other words, the opening of Na+ channels certainly would cause the cell to become more positive, but that is exactly what the question states.

A certain protein release factor functions to specifically recognize stop codons and terminate translation. How many tRNA molecules bind to the same codons as this factor? A. None B. 1 C. 2 D. 3

A is correct. To ensure the proper termination of translation, stop codons are only recognized by protein release factors. In other words, no tRNA molecules bind to these codons, and they do not correspond to any amino acid residues. If tRNAs were able to recognize such codons, the translational machinery would be able to bypass them and produce inappropriately long protein chains.

What is a missense mutation?

A missense mutation occurs whenever one amino acid-coding codon is replaced with another.

Silent mutation

A mutation that changes a single nucleotide, but does not change the amino acid created.

Ploidy at the end of meiosis I

At the end of meiosis I, cells are already haploid, though they are still paired with their identical sister chromatids. For this reason, the 46 chromosomes present in a typical somatic cell have already been reduced to 23 by the time that prophase II is reached.

Bats, birds, and butterflies all developed wings in order to utilize the sky and all of its associated advantages. What type of evolution is involved here? A. Parallel evolution B. Convergent evolution C. Coevolution D. Divergent evolution

B is correct. Convergent evolution occurs when entirely separate lineages gradually appear more similar over time. Here, bats, birds, and butterflies are very distantly related, but all evolve to possess wings through different mechanisms. In the end, these species resemble each other despite their genetic differences. A: Parallel evolution happens when closely related species evolve in a similar way over time. This differs from the question stem, where the three types of organism did not previously resemble each other as strongly. C: Coevolution requires that two species evolve in response to each other. The question gives no indication that bats evolved in part "because" birds did so, for example. D: Divergent evolution is the opposite of the situation described. In this phenomenon, two closely related lineages gradually become more dissimilar.

Colchicine is a medication that is often used to treat gout. It is also used to make karyograms, for which cell division must be arrested at metaphase. What is a potential mechanism of this medication? A. Colchicine inhibits DNA polymerase, blocking DNA replication. B. Colchicine binds to tubulin and prevents the mitotic spindle from forming. C. Colchicine disrupts histone-DNA interactions, which keeps DNA from condensing. D. Colchicine prevents the nuclear envelope from re-forming.

B is correct. During metaphase, microtubules must locate and attach to kinetochores and pull chromosomes to the center of the cell. This requires proper functioning of tubulin-based microtubules, which corresponds to choice B. A: DNA replication occurs during the S phase of the cell cycle. If this were colchicine's mechanism of action, the cell would be arrested in interphase, not in metaphase. C: DNA condensation occurs during prophase. D: The nuclear envelope re-forms during telophase, not metaphase.

Other side

B is correct. From the table, we see that type C has the largest diameter, which is characteristic of microtubules. Additionally, microtubules are known for their rigidity, and type C is listed as the most rigid of the polymers. As some of the longest cells in the body, neurons rely on microtubules to transport neurotransmitter-containing vesicles and macromolecules along their axons. Therefore, one can expect to find the longest type C polymers in neural cells like those in the cerebral cortex.

Photo question on back

B is correct. From the third row of the table, we can see that long (LL) is a dominant trait over curly (ll), since two long-winged parents produced some curly-winged offspring. Thus the parents in that cross must both have had the genotype Ll, making them carriers for the "hidden" wing type, curly. A cross of pure-breeding long (LL) and pure-breeding curly (ll) would produce all heterozygous individuals (Ll) who would all display long wings.

Which of these groupings accurately connects the Gram status of a bacterial species with its structural hallmarks? A. Gram-positive - thick cell wall - outer membrane present B. Gram-positive - thick cell wall - outer membrane absent C. Gram-negative - thick cell wall - outer membrane absent D. Gram-negative - thin cell wall - outer membrane absent

B is correct. Gram-positive bacteria have a thick peptidoglycan wall. This allows them to stain purple, since crystal violet stain, once fixed with iodine, is unable to exit through this thick layer. Additionally, such a bacterium lacks a second membrane on the outside of its cell wall. This quality makes Gram-positive cells especially sensitive to antibiotics that target peptidoglycan. A: Gram-negative (not positive) cells possess a second, external cell membrane. C: Gram-negative bacteria do not have thick peptidoglycan walls. If they did, they, too, would stain purple instead of light pink. D: This choice wrongly states that Gram-negative cells lack an outer plasma membrane.

Back side

B is correct. Layer 2 is the basement membrane, which is connected to an epithelial layer (1) and the endothelium of the capillary (3). Epithelial cells such as these do not directly connect to collagen and other basement membrane fibers; instead, they contain transmembrane proteins known as integrins that can connect to collagen, or, more typically, to a bridging protein like fibronectin, which itself connects to collagen. A: Connexin is a gap junction protein and is not involved in this type of connection. C: Cadherins form cell-cell junctions, not junctions between the cell and the extracellular matrix. D: Occludin comprises tight cell-cell junctions.

A patient with persistent difficulties fighting off bacterial infections is found to have a genetic mutation that adversely affects the speed of phagocytosis in macrophages. This mutation most likely impairs which type of cytoskeletal polymer? A. Lamin filaments B. Actin filaments C. Microtubules D. Collagen fibers

B is correct. Macrophages must undergo rapid actin reorganization during phagocytosis. If a macrophage cannot engulf bacteria in this manner, its overall function would be considered very impaired. A, C: Lamin filaments and microtubules are not involved in phagocytosis. D: Collagen is not a cytoskeletal polymer; it is an extracellular fiber that is secreted by certain cell types.

During most of the duration of an action potential, another stimulus, regardless of its strength, cannot cause the neuron to fire again. This is termed the absolute refractory period, and it can be attributed to: A. the fact that sodium channels are closed. B. the fact that sodium channels are inactivated. C. the fact that potassium channels are closed. D. the fact that potassium channels are inactivated.

B is correct. Many voltage-gated channels have two gates, an activation gate and an inactivation gate. When either is closed, transmission through the channel is blocked; however, inactivation gates are generally closed immediately after a channel has been open. This coincides with the duration of an absolute refractory period - from immediately after sodium channels have opened through the repolarization phase. Additionally, this choice makes sense. Sodium channels cannot simply open once they have become inactivated; they must be "de-inactivated" first. For this reason, the neuron is incapable of firing again until this process is complete. A: Sodium channels are closed at the very beginning of an action potential, and an impulse can still be created. We must choose an answer that explains why, during an absolute refractory period, the neuron is completely incapable of firing. C, D: The closing of potassium channels actually prevents K+ efflux out of the neuron, keeping the cell from becoming more negative. Thus, these statements do not explain why action potentials are inhibited during the period described.

Which of the following enzymes must be present within the viral capsid of a (-)RNA virus for successful infection and replication to take place? A. Reverse transcriptase B. RNA-dependent RNA polymerase C. Protease D. Integrase

B is correct. Negative-sense RNA cannot be directly translated by host ribosomes to produce functional viral proteins. The (-) RNA must be used as a template strand for the production of (+)RNA by a viral RNA polymerase carried within the capsid. An enzyme that produces RNA from an RNA template is termed an RNA-dependent RNA polymerase. A: Reverse transcriptase is not necessary; once the new (+)RNA is synthesized, it can be directly translated into protein without a DNA intermediate. C: Protease and other viral enzymes will be produced by host ribosomes from the (+)RNA and need not be carried in the capsid initially. D: Integrase is an enzyme responsible for integrating viral dsDNA into the host genome. It is not required for (-)RNA viruses.

All of the following statements regarding reverse transcriptase are false EXCEPT: A. it is produced by viral ribosomes. B. it is carried within the retroviral capsid and released into the cytosol following viral penetration. C. it is unable to use DNA as a template. D. it is unable to use RNA as a template.

B is correct. Reverse transcriptase must be carried intact within the capsid in order to process the RNA genome. Without reverse transcriptase, the RNA genome cannot be converted into DNA. A: Viruses do not have ribosomes. All viral proteins are produced at some point during the replication cycle by host ribosomes. C: This is incorrect. Reverse transcriptase first synthesizes a single strand of DNA from the viral RNA template. It then synthesizes a second complementary strand of DNA using the first DNA strand as a template. D: Reverse transcriptase uses the viral ssRNA genome as the initial template for DNA synthesis.

All of the following are similarities between DNA and RNA EXCEPT: A. the capacity to hydrogen bond. B. the ability to catalyze certain biochemical reactions. C. the role of encoding genetic information. D. the direction of their synthesis.

B is correct. Ribozymes are RNA molecules that are capable of catalyzing biochemical reactions, including RNA and DNA ligation and peptide bond formation. DNA molecules are not able to function as enzymes. A: This is a characteristic shared by DNA and RNA and required for the synthesis of mRNA. C: The encoding of genetic material is an essential function of both molecules. D: Both DNA and RNA are synthesized 5'-3'.

A drug that disrupts hydrogen bonding would most directly affect what level of protein structure? A. Primary B. Secondary C. Tertiary D. Quaternary

B is correct. Secondary structure includes the formation of alpha helices and beta-pleated sheets. Both of these structures are defined by patterns of hydrogen bonding. A: Primary structure relates to the linear sequence of amino acids. It is held together by peptide bonds. C: Tertiary structure is driven mainly by hydrophobic interactions. D: Quaternary structure is stabilized by disulfide bonds and other noncovalent interactions.

Other side

B is correct. Since Sample 4 displays a much faster recovery time than the wild-type sample, it must possess a membrane with very high fluidity. Unsaturated fatty acids tend to increase the fluidity of membranes. More notably, overexpression of an omega-3 - a classic group of polyunsaturated fatty acids - also decreases recovery time, but not as much as the lipid overexpressed in Sample 4. Therefore, the overexpressed lipid must be polyunsaturated, probably more so than the tested omega-3 biomolecule. Polyunsaturated oils are liquid at room temperature and in the refrigerator. Common sources of polyunsaturated fat are safflower, sesame and sunflower seeds, corn and soybeans, many nuts and seeds, and their oils. Monounsaturated oils are liquid at room temperature but start to solidify at refrigerator temperatures. Canola, olive, and peanut oils, and avocados are sources of monounsaturated fat.

Phenylketonuria is a rare autosomal recessive disorder with a carrier rate of 1/50 within Caucasian populations. A family enters a clinic with a sick child. Although neither parent is experiencing symptoms, you suspect PKU after a family history revels that the child's paternal great-grandmother had this condition. What are the chances, respectively, that the father is a carrier and that the child expresses this trait? A. 100%, 1% B. 50%, <1% C. 50%, 1% D. 25%, <1%

B is correct. Since the child's great-grandmother had PKU, she must have possessed two copies of the recessive allele. She passed one of these alleles to her offspring (the father's mother or father), making him or her an obligate carrier. This individual then had a 50% chance of passing it to the father of the sick child. (Note that we do not need to consider the father's other parent in this calculation, as the carrier rate is extremely small.) Now, we know that the father had a 50% chance of inheriting one allele and thus has a 50% chance of giving it to his child. As we lack any specific information about the child's mother, we can assume that she, like the population in general, runs a 1/50 risk of being a carrier. If so, she, too, has a 50% chance of giving her disease allele to the child. Multiplying these proportions yields (1/2)(1/2)(1/50)(1/2), or a 1/400 chance that the child is affected by PKU. A one-in-four-hundred risk is less than 1%.

Which of the following statements are true regarding spermatogenesis? I. Meiosis I marks the transition of a spermatogonium into a primary spermatocyte. II. In a secondary spermatocyte, sister chromatids are still paired in the same cell. III. Immediately before the second meiotic division, cells are diploid but chromosomes lack replicated copies. A. I only B. II only C. I and III only D. I, II, and III

B is correct. Spermatogonia give rise to primary spermatocytes before meiosis even begins. Meiosis I, then, marks the division of a primary spermatocyte into two secondary daughter cells. At the end of meiosis I, which is also known as reductional division, cells are already haploid; however, chromosomes retain their identical copies in the form of attached sister chromatids. This makes statement II accurate. I: Again, meiosis I produces secondary spermatocytes. III: This choice can be evaluated using general knowledge about meiosis. Remember, the products of meiosis I are haploid cells.

Of these combinations, which could result in a successful transfer of genetic material through conjugation? A. An F+ bacterium with another F+ bacterium B. An Hfr bacterium with an F- cell C. An F+ bacterium with an Hfr cell D. An F- bacterium with another F- bacterium

B is correct. The F, or fertility, factor is a piece of genetic material that codes for the construction of a sex pilus. This bridge allows a bacterium to transfer the factor to another cell, promoting genetic diversity. However, only bacteria with the factor can initiate this transfer, and only to cells that lack it. Typically, the F factor is held outside the main genome in the form of a small plasmid. However, Hfr bacteria represent an interesting phenomenon; in these cells, the F plasmid has become incorporated into the organism's circular chromosome. As these cells do contain the F factor, they can initiate conjugation to F- cells, but they often transfer genomic material along with part or all of the factor. A, C: All four of the cells mentioned in these choices already possess the F factor, whether integrated into or situated outside the main genome. As such, no recipient is present and transfer is impossible. D: Neither of these cells possess the fertility factor required to initiate conjugation.

If a mutation caused the above tRNA to have a shortened acceptor stem, what would be the consequence? A. This tRNA would no longer be able to participate in translation. B. Aminoacyl tRNA synthetases would have more difficulty recognizing this tRNA. C. This tRNA would automatically bind the incorrect amino acid. D. This tRNA would recognize a different mRNA codon.

B is correct. The acceptor stem often plays a role in the recognition of tRNA by aminoacyl tRNA synthetase. tRNA molecules have acceptor stems of different lengths, and the synthetases enzymes use these distinctions to differentiate between tRNA substrates.

During observation of a particular chemical synapse, a researcher notices that neurotransmitter is released, and that the membrane potential of the post-synaptic neuron quickly rises from -68 to -64 mV. This synaptic potential is: A. an EPSP, possibly provoked by the neurotransmitter GABA. B. an EPSP, possibly provoked by the neurotransmitter glutamate. C. an IPSP, possibly provoked by the neurotransmitter GABA. D. an IPSP, possibly provoked by the neurotransmitter glutamate.

B is correct. The acronym "EPSP" stands for "excitatory post-synaptic potential." Since the post-synaptic neuron became depolarized (and thus more likely to reach the threshold required to execute an action potential), this term aptly describes the situation. Glutamate is the classic excitatory neurotransmitter and often promotes such responses. A, C: GABA is an inhibitory neurotransmitter. As such, it would likely hyperpolarize, not depolarize, a post-synaptic cell. D: "IPSP" stands for "inhibitory post-synaptic potential." Depolarization constitutes excitation, not inhibition. Additionally, glutamate should not produce an inhibitory response.

Of these pairings, which accurately group(s) an organelle with one of its functions? I. Rough ER - synthesis of transmembrane proteins II. Nucleolus - rRNA production and ribosome assembly III. Peroxisome - breakdown of proteins A. I only B. I and II only C. II and III only D. I, II, and III

B is correct. The rough ER is dotted with bound ribosomes that serve to produce a wide variety of proteins. These products include transmembrane proteins, making option I is correct. Additionally, option II accurately describes the main function of the nucleolus. This region within the nucleus is vital for the synthesis of ribosomes from protein and rRNA. III: Peroxisomes are small membrane-bound organelles that function mainly to break down lipids. In contrast, lysosomes facilitate the enzymatic catabolism of protein.

The activity of a Class I transposon, also known as a retrotransposon, involves: I. a conversion of DNA to RNA. II. a conversion of RNA to DNA. III. the production of an additional copy of the transposon. IV. an end result of the same total number of transposons. A. I and III only B. II and IV only C. I, II, and III only D. I, II, and IV only

C is correct. Class I transposons are known as "copy-and-paste" transposons, as they involve the creation of a new copy of the transposable element. These elements first undergo transcription into RNA using RNA polymerase; as their name implies, they are then reverse transcribed back into DNA and placed in a distinct location elsewhere in the genome. IV: This would be the case with a Class II, or "cut-and-paste," transposon.

Back side

C is correct. For this question, we simply need to know that the frontal lobe is part of the brain. Multipolar neurons like #3 predominate in the CNS and the cerebrum specifically; their many dendrites allow them to integrate into a complex neural network capable of complicated responses. A: This image depicts a unipolar neuron. Such simple neuron structures are rare in humans, though common in some insects. B: #2 shows a bipolar neuron; these cells are common in special sensory organs, and are most commonly remembered as components of the retina. D: Pseudobipolar neurons like #4 typically transmit sensory information from the periphery to the central nervous system. While you do not need to be intimately familiar with these structures, you should be able to tell from the figure that these do not resemble neurons found in the brain.

Which of the following is true of a typical viral capsid? A. It is typically amorphous and lacks discernible symmetry. B. Lipids comprise a substantial amount of the capsid material. C. Most viruses have only one or two genes coding for capsid proteins. D. The capsid monomers are assembled into a complete structure by host ribosomes

C is correct. Most viral capsids are composed of repeated identical monomers, reducing the need for a large genome containing multiple different genes for structurally distinct capsid proteins. A: Viral capsids tend to be spherical or otherwise geometrically regular. Since the structure is formed from hundreds of identically-shaped protein monomers, physical constraints are placed on the ultimate shape of the virus. B: The capsid is made from noncovalently-associated protein subunits, not lipids. D: In general, the capsid assembles spontaneously around the genetic material. It sometimes requires viral enzymes for complete assembly, but never ribosomes. Ribosomes assemble polypeptides form individual amino acyl tRNA molecules and have no impact on intact proteins.

For an upcoming experiment, you need to obtain an ampicillin-resistant strain of Streptococcus pneumoniae. Assuming that you begin with a strain that is sensitive to this antibiotic, which of these protocols is most likely to accomplish your goal? A. Plating Streptococcus on media infused with 1 g/ml ampicillin B. Plating Streptococcus on media infused with 1 μg/L ampicillin C. Plating Streptococcus on a gradient spanning from 1 μg/L to 1 g/mL of ampicillin D. Plating Streptococcus on regular media, knowing that the rate of random mutation will be sufficient

C is correct. Mutations always occur spontaneously, but we still must be able to detect cells that have mutated in the desired way. This can be accomplished by growing the bacteria on a gradient. 1 μg/L of ampicillin is extremely dilute, so colonies should grow regardless of resistance. In contrast, only those that are completely resistant to ampicillin will grow in media containing 1 g/ml. A: The antibiotic on this plate is too concentrated. It will likely kill all of the bacteria before they are able to replicate. B: This plate is too dilute; note that the ampicillin here is measured per liter, not per milliliter. Some (or even all) of the non-resistant cells might be able to grow here. D: Random mutation could be enough to acquire a resistant strain, but this choice gives no mechanism by which you can determine which cells these are.

Carpenter ants grow fungal gardens in their nests that provide them with nutrients and vitamins. In exchange, the ants carry leaves into their nests to feed these gardens. This relationship is most closely related to: A. commensalism. B. parasitism. C. mutualism. D. ectosymbiosis.

C is correct. Mutualism is a symbiotic relationship in which both parties benefit. Here, the ants feed the fungus and the fungus feeds the ants. A: Commensalism involves a benefit to only one participating species; the other species is unaffected. B: Parasitism occurs when one organism benefits from the relationship at the expense of the other. According to the question stem, neither participant appears to be harmed at all. D: Ectosymbiosis is a specific phenomenon in which one species lives on the surface of another.

Which of these components is / are involved in prokaryotic transcription and its regulation? I. Repressor proteins II. Promoter DNA sequences III. Activator proteins IV. Enhancer DNA sequences A. I only B. III only C. I, II, and III only D. I, II, and IV only

C is correct. Prokaryotic and eukaryotic transcriptional regulation requires transcription factors, including repressor and activator proteins. Repressors inhibit gene expression by blocking specific regions of the DNA or by preventing RNA polymerase from attaching to the promoter. In contrast, activators upregulate transcription by binding to either operators (in prokaryotes) or enhancers (in eukaryotes). Finally, promoters are essential in both types of organisms, since they function as the DNA sequences to which RNA polymerase initially binds. IV: Enhancers are regulatory DNA sequences found only in eukaryotes. In prokaryotes, the operator serves as the regulatory DNA region instead.

Latrotoxin (LTX), a large globular protein produced by Latrodectus spiders, acts to perforate the presynaptic membrane at the axon terminal. The resulting channel is large enough to permit the free influx of calcium, as well as the passage of water and other small molecules. A Latrodectus bite would most likely result in: I. swelling of the axon terminal. II. fusion of docked vesicles with the plasma membrane. III. widespread release of glutamate within the CNS. IV. widespread release of acetylcholine within the PNS. A. I and II only B. III and IV only C. I, II, and IV only D. I, II, III, and IV

C is correct. Swelling will result from the movement of water down its concentration gradient and into the neuron, which contains many large proteins and other solutes that contribute to its hypertonicity. Fusion of presynaptic vesicles is mediated by an increase in intracellular calcium, which the question mentions as a consequence of LTX action. Finally, in the peripheral nervous system, LTX will cause the widespread release of the principal motor neurotransmitter, acetylcholine. III: We are told that the toxin is a large globular protein. Thus, it is unlikely to cross the blood-brain barrier, which is only permeable to small, hydrophobic molecules or those with a specific transporter. The toxic effects of LTX are most likely the result of peripheral activity.

In an isolated cave, two bat species are discovered and found to share a distant ancestor. The ears of the two species seem to have adapted to find different sizes of prey using echolocation. When crossed, individuals of the distinct species are unable to produce a virile litter. This situation exemplifies: A. parapatric speciation. B. peripatric speciation. C. sympatric speciation. D. allopatric speciation.

C is correct. Sympatric speciation is that which occurs without a physical barrier. A population that diverges into two separate species in a single cave certainly falls under this form of speciation. A: Parapatric speciation occurs when segments of two distinct populations overlap. Due to environmental differences, these segments may develop into two species, but individuals in the overlapping areas can typically still interbreed. B, D: Allopatric speciation occurs when populations, or parts of the same population, are separated by a physical barrier. Peripatric speciation is a subtype of this concept that occurs specifically when one of the two populations is much smaller than the other.

If trauma causes a cell to dramatically shorten the length of its S phase, what compensatory mechanism might occur? A. The cell would not produce viable daughter cells. B. The cell would enter the G0 phase. C. The cell would spent a longer period of time in the G2 phase. D. None; the cell would be unable to correct any errors that occurred.

C is correct. The S phase is responsible for DNA replication. A drastically impaired S phase, then, is likely to yield a cell with an incomplete or inaccurately replicated genome. As a result, this cell would be prevented from passing the G2-M checkpoint, which ensures correctness in DNA replication. Instead of quickly entering mitosis, it will spend additional time in the G2 phase to correct the errors generated by the traumatic event. A: While this is possible, we cannot be certain that the damage is irredeemable. In contrast, choice C is virtually guaranteed to be true. B: The G0 phase is a period of "rest" that is favored by specific cell types, especially neurons and certain kinds of muscle. It would not be entered after impaired DNA replication. D: Like choice A, this is too extreme to be the best answer. Additionally, the cell does not lack compensatory mechanisms entirely; this type of situation is what the checkpoints are for.

Photo question on other side

C is correct. The mother in the first generation expresses the trait, as do all of her sons. However, none of this woman's daughters have the condition, although they must be carriers. This explains how the male in the third generation acquired the trait even though neither of his parents display it. The fact that more male than female offspring express this trait, coupled with the idea that males never seem to inherit it from their fathers, is a clear hallmark of an X-linked condition.

Backside

C is correct. The question is asking for the tubes that contain obligate anaerobes. We can infer that aerobic species would cluster close to the open top of the tube, while anaerobes would prefer to grow lower in the tube, away from oxygen. Sample 2 clearly consists of obligate anaerobes, as no cells are located anywhere near the air. In contrast, the bacteria in Sample 4 are spread evenly throughout the broth, but still do not gravitate toward the open top. This species is likely aerotolerant, meaning that it is an obligate anaerobe that does not die in the presence of oxygen. A: Tube 1 clearly contains an obligate aerobe. D: Tube 3 appears to consist of a facultative species, as it survives without oxygen but tends to cluster near the top. The question asked for the tubes that contained exclusively anaerobic bacteria.

A clinical trial for an experimental form of birth control involves injecting patients with progestin, or synthetic progesterone. If a woman receives this injection during the early part of the follicular phase, and if its effects persist for approximately three weeks, what can be expected to occur? A. The woman's menstrual cycle will proceed as usual. B. The woman's blood will contain augmented levels of LH and FSH. C. The woman will not experience ovulation at the usual time. D. The woman will actually face an enhanced probability of becoming pregnant.

C is correct. Typically, estrogen and progesterone negatively feed back on gonadotropin-releasing hormone (GnRH). As a result, high levels of these steroids lead to reduced release of LH and FSH. As a surge in LH is required to stimulate ovulation, persistently high progesterone levels near the beginning of the menstrual cycle will prevent this woman from ovulating - at least until the effects wear off. B: LH and FSH levels should be low, not high. Since estrogen and progesterone secretion is promoted by both LH and FSH, a large progesterone concentration is a clear signal for the anterior pituitary to slow its release of these two hormones. D: Progestin is actually a very common and fairly effective form of birth control.

Back side

D is correct. An adenine nucleotide is added after the third position, which will result in a shift to the reading frame during translation. Frameshifts occur whenever nucleotides are inserted or deleted in numbers other than multiples of three. A: Silent mutations are those that do not affect the final protein product. Frameshift mutations, in contrast, often render the protein nonfunctional. B: Missense mutations involve the mutation of one amino acid into another without a frameshift. C: Nonsense mutations involve the mutation of one amino acid into a stop codon.

Which of the following accurately lists a second messenger and its role in one or more signaling cascades? A. FSH serves to attenuate the pathway(s) in which it is involved. B. FSH serves to amplify the signal(s) in which it is involved. C. cAMP serves to attenuate the pathway(s) in which it is involved. D. Ca2+ serves to amplify the signal(s) in which it is involved.

D is correct. Calcium ion is a classic example of a second messenger. In a typical signaling cascade, a ligand serves as the "first messenger" by binding to a membrane receptor outside the cell. This binding triggers the activation of a second messenger, which typically amplifies the relevant signal by activating kinase molecules. A, B: Follicle-stimulating hormone (FSH) is a first, not a second, messenger. C: "Attenuate" means "weaken." Second messengers typically strengthen, or amplify, their associated signals.

A congenital defect in the synthesis of the proteins cloudin and occludin is most likely to impact the function of the: A. lung vasculature. B. peripheral nerves. C. gastric musculature. D. intestinal lining.

D is correct. Cloudin and occludin are the proteins that form tight junctions between epithelial cells. Therefore, any organ that relies on tight junctions will be negatively affected if their synthesis is impaired. If the intestinal lining is to properly absorb nutrients while leaving behind undesired materials, free diffusion cannot be permitted between the digestive ECM and the lumen of the digestive tract. Tight junctions seal the gaps between epithelial cells in the intestine, allowing cells to selectively control what passes through the epithelium through the use of transmembrane transport proteins. Without tight junctions, nutrient concentrations would equalize due to diffusion through the epithelium, vastly decreasing the efficiency of the GI tract.

Latrunculin, a toxin produced by marine sponges of the genus Latrunculia, acts by binding to actin monomers and preventing their polymerization. Which cell function would be most directly sabotaged by this toxin? A. Centrosome polarization B. DNA replication C. Mitotic spindle formation D. Cytokinesis

D is correct. Cytokinesis involves a contracting ring of actin microfilaments that pinches the two daughter cells apart. Preventing actin monomers from polymerizing will inhibit the filament reorganization necessary for cytokinesis to occur. Even if we did not know this information, none of the other answers are sensible. A: Centrosome polarization involves microtubules, not actin. Specifically, microtubule extension pushes the centrioles to opposite sides of the cell during mitosis. B: DNA replication is independent of the activity of actin. C: The formation of the mitotic spindle also involves microtubules, which organize themselves around the centriole.

A type of plant uses different insect pollinators depending on its height. Unfortunately, due to an increase in the use of pesticides, the medium-height pollinator has recently become extinct. What type of selection will occur in the plant species over the next several generations? A. Directional selection; future generations will include short and tall plants only. B. Disruptive selection; future generations will include short or tall plants, but not both. C. Directional selection; future generations will include medium plants only. D. Disruptive selection; future generations will include short and tall plants only.

D is correct. Due to the extinction of the associated insect species, medium-height plants now cannot be pollinated. Over time, these plants will die out before they can reproduce, leaving only the short and tall alleles to pass to the next generation. This represents disruptive selection, as only the extreme phenotypes are being evolutionarily favored. A: Directional selection favors either of the extremes, not both. Such selection could not possibly promote the prevalence of short and tall plants simultaneously. B: The question stem gives no indication that short or tall plants are affected; thus, both phenotypes could certainly be favored. C: This is the opposite of the correct prediction.

All of the following are true of genetic drift EXCEPT: A. it can lead to loss of alleles from the population. B. it can lead to alleles becoming fixed in a population. C. it is the result of random allele segregation from parents. D. it can increase the genetic diversity of the population.

D is correct. Genetic drift is simply the change in allele frequencies due to random processes. Specifically, random chance plays a role in determining which alleles are inherited by offspring from their parents. This can cause some alleles not to be passed down at all, leaving others "fixed" as the only alleles present for that locus. However, genetic drift does not relate to the introduction of new alleles (as mutation does), and cannot increase a population's genetic diversity.

Gram staining is a common laboratory technique used to classify bacteria. When stained, Gram-positive bacteria appear dark purple under a microscope; in contrast, Gram-negative bacteria appear pink. How can this visible difference between the two bacterial categories be explained? A. Gram-negative bacteria lack the enzyme that cleaves the stain, causing them to be lighter in appearance. B. Gram-negative bacteria lack the cell wall antigen that reacts with Gram stain antibodies. C. Gram-negative bacteria do not have a cell wall and therefore do not retain crystal violet stain in solution. D. Gram-negative bacteria have a thinner cell wall, which allows the crystal violet stain to exit the cell.

D is correct. Gram staining first involves exposure to crystal violet stain, which is then fixed with iodine. Gram-positive organisms possess a thick peptidoglycan wall which attaches to crystal violet particles, preventing them from leaving the cell when it is subsequently washed. As a result, these cells stain purple. Gram-negative bacteria, in contrast, possess a thin cell wall that does not fix to the crystal violet stain. Instead, these cells are later stained pink with a counterstain. A, B: Gram staining does not involve antibodies or enzymatic cleavage. C: Gram-negative bacteria, unlike many eukaryotic cells, do possess a cell wall. This structure is simply thinner than the wall present in a Gram-positive organism.

In a negative inducible operon, a repressor protein binds to the operon and the genes are not actively transcribed. In such an operon, which of the following could be added to restore the operon to a transcriptionally active state? A. An inducer B. RNA polymerase C. An inhibitor D. An activator

D is correct. In a negative inducible operon, transcription is inhibited by a repressor; this results in a basal transcription rate near zero. However, transcription can be "switched on" by the addition of an activator protein, which blocks the repressor from binding to the operon. A: Inducers are proteins that bind to activators. While they are known to upregulate transcription in positive operons, the question describes a negative system. Here, addition of an inducer could not remove the repressor protein from the operon, so it would not activate transcription. B: RNA polymerase is the enzyme that binds to the DNA and carries out transcription. However, the addition of this enzyme would not activate the operon if the repressor protein was still bound. C: Inhibitors deactivate activator proteins and decrease the level of transcription.

Unlike humans, birds use a ZW sex-determination system in which males have two Z chromosomes and females have one Z and one W chromosome. What is the most likely regulatory mechanism for the potential imbalance in gene expression between male and female birds? A. Destruction of one Z chromosome in males B. Post-transcriptional modification of half of the gene products of the male sex chromosomes C. Inactivation of the W chromosome in females D. Inactivation of one Z chromosome in males

D is correct. In humans, females have two of the same sex chromosome (XX), while males have only one X and one Y. However, the X chromosome is much larger and carries significantly more genes than the Y version. To balance their genetic load with that of males, one of each female's X chromosomes is inactivated. As far as we know, birds likely use the same mechanism but for the opposite gender. The male, who carries two Z chromosomes, must then inactivate one to balance his female ZW counterpart. A: Complete destruction of an entire chromosome is too drastic to serve as a likely mechanism here. B: Post-transcriptional modification would not necessarily silence the "extra" genes present in male birds. C: The W chromosome is a non-issue when dealing with genetic load; it is the additional Z chromosome held by males that is of interest.

Which of the following cytoskeletal proteins contributes the most to the skin's resistance to stretching and tearing? A. Actin B. Tubulin C. Collagen D. Keratin

D is correct. Keratin fibers extend across epithelial cells (in the skin, for example) to link adjacent cells via structures called desmosomes. As a type of intermediate filament, keratin possesses high tensile strength; as such, it can form a network of fibers that distribute mechanical stress among the cells of an epithelial layer. Keratin is therefore the cytoskeletal protein that contributes most to the skin's stretching ability. C: Collagen may be a tempting choice, as it contributes heavily to the tensile strength of connective tissue in the skin. However, it is a fiber in the extracellular matrix, not a component of the cytoskeleton.

Although Creutzfeld-Jakob Disease is perhaps the best-known prion disease, recent hypotheses suggest that neurodegenerative diseases such as Alzheimer's disease and Parkinson's disease may have prion components. Which of the following, if true, would best refute the claim that prions contribute to the pathology of Parkinson's disease? A. Knockout mice lacking the PrPC, a prion precursor protein, do not develop Parkinson's disease when fed intact prions. B. Crude brain lysates from mice suffering from Parkinson's disease reliably induce Parkinson's when fed to healthy mice. C. Parkinson's disease appears to run in families. D. Neurons from Parkinson's patients show absolutely no increase in expression of heat shock proteins or cellular chaperones.

D is correct. Prions tend to induce misfolding and aggregation of endogenous cellular proteins, forming highly stable amyloid fibers. The natural cellular response to the presence of misfolded proteins is the production of heat shock proteins, which help to properly fold the defective protein molecules. The absence of heat shock activity suggests that the cell is not experiencing any problems with protein aggregation, which is one of the hallmarks of prion disease pathology. A: This does not refute the idea of a prion component. Prions induce conformational changes in endogenous precursor proteins that result in their conversion to the prion form. If the precursor protein is absent, the prion has no target and cannot cause pathology. B: This suggests a prion component. CDJ is spread through the consumption of prion proteins, specifically from the brains of infected cows. If Parkinson's disease were contracted by a similar mechanism, it would imply that prions could be responsible. C: This statement does not refute the prion hypothesis. Prions are formed from precursor proteins produced in healthy individuals. This could result from genetic abnormalities in the encoding of the prion precursors, which would tend to result in a familial trend.

A sample of non-coding RNA is isolated from a cell's nucleolus. The sample most likely contains: A. snRNA. B. hnRNA. C. tRNA. D. snoRNA.

D is correct. Small nucleolar RNAs (or snoRNAs) are involved in the modification of rRNA. As such, they would be located in the nucleolus, where ribosomes are assembled. A: snRNA, or small nuclear RNA, is found in the nucleus. These molecules aid in the splicing of pre-mRNA. B: hnRNA is simply an alternative name for pre-mRNA. As such, not only is this type of nucleic acid found in the nucleus, it also (at least partially, in its exons) codes for peptide products. The question stem referenced a type of non-coding RNA. C: tRNA molecules are most likely to be found in the cytoplasm.

Which of the following is NOT a function of the Golgi apparatus? A. Packaging of proteins in vesicles for transport to other parts of the cell B. Targeting of proteins for excretion C. Production of lysosomes D. Production of the enzymes found within peroxisomes

D is correct. The Golgi apparatus functions like a cellular "post office." It receives proteins from the endoplasmic reticulum and packages them into vesicles; these membrane-bound sacs can then travel to specific locations within the cell or to the plasma membrane for excretion. Additionally, lysosomes are formed from vesicles that bud off the Golgi's larger structure. However, note that choice D references enzyme production. Since nearly all enzymes are protein-based, it is the ribosomes - whether free-floating or attached to the rough ER - that synthesize them.

Back side

D is correct. The arrow points to ectodermal cells. This germ layer differentiates into "external" features including the epidermis, lens of the eye, fingernails, and hair. It also gives rise to the nervous system via neurulation. A: The tissue types listed here come from all three germ layers. B: This choice relates much more to the endoderm than the ectoderm. C: Muscle, the cardiovascular system, and the skeletal system are classic mesodermal tissues. The arrow above indicates the ectoderm.

Development of the mesoderm leads to the growth of a coelom, inside of which growing organs are protected. The mesoderm can develop into multiple sub-mesodermal tissues. Which of the following pairings correctly matches a sub-mesodermal tissue layer with its corresponding tissue? A. Intermediate mesoderm - respiratory lining B. Paraxial mesoderm - spinal cord C. Lateral plate mesoderm - urinary bladder D. Chorda-mesoderm - notochord

D is correct. The chorda-mesoderm, as its name implies, develops into the notochord. The intermediate mesoderm develops into gonads and kidneys, the lateral plate mesoderm develops into the gut wall and circulatory system, and the paraxial mesoderm develops into skeletal muscle and cartilage. However, you can answer this question without knowing these facts, as the other three tissues are not mesodermal at all. A, C: The lining of the respiratory tract and the bladder are derived from the endoderm. B: The spinal cord, as well as the rest of the central nervous system, arises from ectodermal tissue.

Of the following statements, which correctly describes the fluid mosaic model? A. Plasma membranes have a consistent composition and an even distribution of lipids and proteins. B. Proteins and lipids tend to separate into "rafts" based upon size. C. Plasma membranes are formed of lipids and proteins that remain virtually motionless in their respective positions. D. Plasma membranes act as two-dimensional fluids that allow for the free diffusion of proteins and lipids within the leaflet.

D is correct. The mentioned model involves two main aspects: "fluid" and "mosaic." The "fluid" aspect refers to the ability of component molecules to travel laterally within their leaflet of the bilayer, while the word "mosaic" denotes the presence of proteins and other molecules that are scattered in a mosaic within its structure. A: The idea of a perfectly consistent distribution of membrane components contradicts the "mosaic" idea proposed as part of this model. B: This statement does not directly relate to the fluid mosaic model. C: This contradicts the idea of a plasma membrane as a fluid. In reality, components are able to drift as opposed to remaining stationary.

Back side

D is correct. The notochord is located ventral to the neural tube (in other words, below the circular structure marked in the figure). The notochord provides the primitive axis of the developing embryo and, in vertebrates, develops into the vertebral column. A, B: The neural tube is the elongated cylindrical structure found above the notochord in this figure. The neural tube develops into the central nervous system, which includes the brain and spinal cord.

In a certain population of wild coyotes, brown fur (B) is dominant over spotted fur (b) and approximately 80% of a given breeding population has brown fur. Over the course of two decades, a genetic change occurs such that nearly the entire population expresses the spotted fur phenotype. In this population: A. the trait for spotted fur has become dominant. B. having brown fur must have been strongly selected against. C. overall fitness has been reduced. D. spotted fur is the wild type.

D is correct. The term "wild type" refers to the traits an animal typically possesses when found in nature. This usually refers to a dominant trait, but not always. If nearly all of the coyotes now have spotted fur, then spotted fur is the wild type. A: "Dominant" refers to the interaction between the genes, not which trait is simply the most common. We have no reason to suspect that Bb individuals would stop displaying brown fur. B: Selection pressures can decrease the frequency of a trait even if that trait is not negative and the pressure is not acting directly on that trait. For example, coyotes with brown fur may have had another trait (e.g. shorter legs) that was being selected against. C: The question tells us nothing about the fitness levels associated with the two fur types.

Tetrodoxin (TTX), a toxin found in pufferfish and other marine species, binds to and inhibits transmission through Na+ voltage-gated channels in neurons. An individual has spent the past six months consuming a small, daily dose of TTX to calm his restless legs. If this individual were to suddenly cease this medication, what neurological response is most likely? A. His neurons will be unusually insensitive, since sodium will now be able to exit the cell. B. His neurons will be unusually insensitive, since he had become accustomed to a lower amount of sodium influx, which is integral for the initial depolarization phase of the action potential. C. His neurons will be unusually sensitive, since sodium will now be able to exit the cell. D. His neurons will be unusually sensitive, since he had become accustomed to a lower amount of sodium influx, which is integral for the initial depolarization phase of the action potential.

D is correct. This man (wise or not) has built up a tolerance to TTX. In other words, his neurons have become accustomed to firing even while sodium movement through voltage-gated channels is somewhat inhibited. If he suddenly stops taking TTX, sodium will be able to enter cells normally, and extremely sensitive neurons are the most likely result. The man will now experience action potentials more easily than usual. A, C: When Na+ voltage-gated channels are opened, sodium will flow down its concentration gradient. This results in ion movement into, not out of, the cell.

Which of these statements accurately identify a function of transmembrane proteins? I. They act as receptors for hormones and initiate signal transduction pathways. II. They allow for transport of charged molecules across the cell membrane. III. They are responsible for the production of the majority of the ATP synthesized in eukaryotic cells. A. I only B. III only C. I and II D. I, II, and III

D is correct. Transmembrane proteins, such G protein-coupled receptors, bind hormones and other ligands to activate signal transduction pathways. This activation ultimately results in a change in gene expression. Additionally, statement II is accurate, as such proteins may form channels that allow charged molecules to pass. Finally, while not embedded in the overall lipid bilayer of the cell, ATP synthase spans the inner mitochondrial membrane and is therefore a transmembrane protein.

A researcher is attempting to create an artificial cell membrane that retains its fluidity at extremely low temperatures. Which features should he incorporate in this membrane? A. High levels of unsaturated fatty acids B. High levels of sterols C. High levels of saturated fatty acids D. Both A and B

D is correct. We are looking for characteristics of a membrane that increase fluidity, especially in cold environments. Unsaturated fatty acids increase fluidity in general, due to the presence of "kinks" in their structure. Cholesterol and similar molecules, which act as a type of buffer system for membrane fluidity, are slightly more complex. Large amounts of cholesterol decrease fluidity at high temperatures but increase it at low temperatures. Therefore, this researcher would need high levels of cholesterol to best accomplish his goal. C: Saturated fatty acids, which stack easily due to their unkinked structures, tend to promote rigid plasma membranes.

If all rRNA was suddenly targeted for enzymatic degradation, which statement most accurately reflects the effect on the cell? A. Protein translation would be inhibited due to lack of amino acids. B. Protein translation would be inhibited due to lack of transcripts. C. Protein translation would be inhibited due to the impairment of required enzymes. D. Protein translation would be inhibited due to lack of ribosomes.

D is correct. rRNA serves as a major component of ribosomes, which also contain protein. Ribosomes are the organelles at which protein synthesis occurs. A, B, C: Amino acids and enzymes would not be immediately affected by the lack of rRNA. The transcripts used as templates for translation would also be unaffected, as they are composed of mRNA.

Proper eukaryotic DNA replication requires a number of enzymes. The elongation step of replication involves: I. DNA polymerase. II. Ter protein. III. initiation proteins. IV. the Dicer enzyme.

I only A is correct. The elongation process of DNA replication involves the addition of nucleotides to form a daughter strand complementary to the parent strand. DNA polymerase directly catalyzes this synthesis. II, III: While these molecules do serve functions in DNA replication, they are not involved in elongation. Ter protein acts during the termination step, while initiation proteins function in the initiation step. IV: The Dicer enzyme is involved in producing siRNA, not replicating DNA.

What is special about silver chloride AgCl?

Its insoluble

Where to prokaryotes carry out aerobic respiration at?

Prokaryotes lack membrane-bound organelles and therefore cannot use either mitochondrial membrane. In addition, not all prokaryotes have cell walls. This leaves the plasma membrane as the most likely structure across which a proton gradient is established.

Promoter

The promoter is a DNA sequence upstream of its associated gene. RNA polymerase binds to this sequence to initiate transcription. If the promoter is deleted, transcription will not begin and the gene will not be transcribed.

Do prokaryotes have DNA

Ya

Is yeast a prokaryote or eukaryote?

eukaryote

Aminoacyl tRNA

is tRNA to which its cognated amino acid is chemically bonded (charged).