This question prompts us to identify the type of immune response miRNAs of AR kidney transplants likely regulate based on the passage. We will need to rely on the passage’s description of these miRNAs to answer this question.

Experiment 2 found that “there was a positive correlation between … mRNA levels [of T and B cells] and the levels of miR-142-5p, miR-155, and miR-223.” Innate immunity provides the body with a response to a wide range of pathogens. Adaptive immunity provides a specific immune response to pathogens. Adaptive immunity involves two types of immune responses: cell-mediated and humoral. Cell-mediated immune responses occur upon infection by an intracellular foreign pathogen, such as a virus, and other intracellular pathogens. Humoral immunity responses occur in the extracellular spaces and involve the production of antibodies by plasma cells. Humoral immunity also confers long-term protection via memory B cells that provide long-lasting immunity to reinfection. T cells play a role in activating cell-mediated immunity, whereas B cells play a role in activating humoral immunity.

As T and B cell mRNA is positively correlated with certain miRNAs of AR kidney transplants, these miRNAs likely regulate the expression of genes implicated in the adaptive immune response.

The passage indicates a positive correlation between CD3 (T cell) and CD20 (B cell) mRNA levels and that of miR-142-5p, miR-155, and miR-223. Although B cells contribute to humoral immunity, this answer choice does not mention T-cell contributions to immunity. T cells play a role in activating cell-mediated immunity. Option A is a stronger choice as it generally describes adaptive immunity, which includes B and T cell activity.

Though T cells play a role in cell-mediated immunity, neither B nor T cells play a role in innate immunity as they respond to the presence of specific pathogens.

B and T cells play a role in adaptive immunity, not innate immunity.

This question prompts us to identify a patient who is least likely to benefit from immunosuppressive therapy. We will need to rely on the information from Figure 1 to answer this question.

High levels of miR-155 are associated with acute rejection, indicating that these individuals would benefit from immunosuppressors as they could reduce the immune system’s attack on the transplanted organ.

Low levels of let-7c are associated with acute rejection, indicating that these individuals would benefit from immunosuppressors as they could reduce the immune system’s attack on the transplanted organ.

Acute rejection occurs when the body’s immune system attacks a transplanted organ as if it is a foreign pathogen. We will need to identify a highly expressed miRNA sequence with a high success rate, as this individual would be less likely to experience acute rejection and consequently not benefit from immunosuppression.

Figure 1 presents data regarding transplant success for individuals with different miRNA sequences. As the y-axis represents normalized miRNA levels, we are looking for miRNA sequences in which the black bar (representing normal transplants) is high, indicating high expression associated with successful transplants.

As high levels of miR-30a-3p are associated with normal transplants, we would not expect these individuals to benefit from immunosuppressors.

Low levels of miR-10b are associated with acute rejection, indicating that these individuals would benefit from immunosuppressors as they could reduce the immune system’s attack on the transplanted organ.

This question prompts us to identify the miRNA represented by the right-most curve on the graph provided in the question. As melting temperature is partially determined by nucleotide sequence, we must rely on the passage’s description of the miRNAs to answer this question.

The miR-142-5p sequence only has 7 G-C base pairs and will denature at a lower temperature than miR-30a-3p.

The miR-233 sequence only has 9 G-C base pairs and will denature at a lower temperature than miR-30a-3p.

The miR-10b sequence only has 10 G-C base pairs and will denature at a lower temperature than miR-30a-3p.

G-C base pairs are connected by three hydrogen bonds, whereas only two hydrogen bonds connect A-U and A-T base pairs. Consequently, genetic sequences with a higher ratio of G-C base pairs will be more stable and have a higher melting temperature. As the miRNA sequences provided are all of the same lengths, we will need to identify the sequence with the greatest number of G-C base pairs.

The miR-30a-3p sequence has the largest number (11) of G-C base pairs and is, therefore, the sequence with the highest melting temperature.

This question prompts us to identify the bond responsible for holding the backbone of regulatory RNAs together. We must rely on our content knowledge regarding nucleotides to answer this question.

Disulfide bridges are disulfide bonds that covalently link cysteine amino acids of proteins. Disulfide bridges are not found in nucleotides.

Though the question specifies regulatory RNAs discussed in the passage, we can rely on what we know about nucleotide bonds as we would not expect their nucleotides to function differently than standard ones. The backbone of DNA and RNA describes a single-stranded sequence of nucleotides. Phosphodiester bonds are covalent bonds that connect adjacent phosphate and sugar groups of adjacent nucleotides.

Therefore, we can conclude that phosphodiester bonds hold together the individual nucleotides within the backbone of the regulatory RNAs discussed.

Though hydrogen bonds are responsible for holding base pairs together in a duplex, they are not the bonds responsible for pairing nucleotides of a backbone.

Glycosidic linkages, also known as glycosidic bonds, are ether bonds that connect a molecule to a carbohydrate. Though a sugar, ribose, is indeed present in nucleic acids, the bond that links nucleic acids is called a phosphodiester bond.

This passage-based question asks us to identify which bacteria can use galactose as an energy source, prompting us to refer to Table 1 while applying content knowledge on carbohydrates.

According to Table 1, Ruminococci use polysaccharides (PS) and H+ as a source of energy.

Based on Table 1, Faecalibacteria use polysaccharides (PS) and glucosamine as a source of energy.

According to Table 1, Odoribacter can use glucose and its isomers as an energy source. An isomer is a molecule with an identical molecular formula but a different arrangement of atoms in space. Below, epimers of carbon are shown. Epimers are isomers that are also diastereomers.

Therefore, because galactose is an isomer of glucose, we may conclude that Odoribacter may use it as an energy source.

According to Table 1, Phascolactobacteria use succinate as a source of energy.

This passage-based question asks us to determine which metabolic process GI tract bacteria are directly involved in, prompting us to examine Table 1 while applying content knowledge on carbohydrate and lipid metabolism.

According to Table 1, GI tract bacteria convert monosaccharides and polysaccharides into acetate, propionate, and butyrate, which are short-chain fatty acids. Monosaccharides (also called simple sugars) are the monomers of carbohydrates, while polysaccharides (PS) are long-chain carbohydrate compounds consisting of numerous monosaccharide monomers.

Fatty acids are carboxylic acids with a hydrocarbon chain, serving as a major component of lipids. The hydrocarbon chains of fatty acids may be saturated (lacking double bonds) or unsaturated (containing at least one double bond).

Therefore, we may conclude that GI tract bacteria are directly involved in the conversion of polysaccharides into short chain fatty acids acetate, propionate, and butyrate.

The passage does not mention amino acids. Therefore, neither the passage nor Table 1 indicates GI tract bacteria have a role in amino acid absorption.

Fermentation is a metabolic process that occurs without oxygen. It involves the partial breakdown of glucose, producing ATP and byproducts such as lactic acid or ethanol. Fermentation is used by cells and microorganisms as an alternative way to generate energy when oxygen is not available. Based on the table, dietary fibers (PS) are converted into short chain fatty acids, not peptides.

As we can see in Table 1, GI tract bacteria use monosaccharides as an energy source. Monosaccharides include glucose, fructose, and galactose. In the table, we can see Phascolactobacteria and Clostridia use glucose as an energy source. However, neither the passage nor the table indicates the GI tract bacteria are involved in monosaccharide absorption.

This passage-based question asks us to determine which conditions are likely associated with a reduction in gram-positive bacteria, prompting us to refer to Table 1 while applying our scientific reasoning and problem-solving skills.

Table 1 demonstrates that Ruminococcaceae and Clostridia are both gram-positive bacteria and produce acetate. As a result, a decrease in these types of bacteria will cause a reduction, not an increase, in acetate production.

Table 1 demonstrates Roseburia is a gram-positive bacteria that use acetate as an energy source. Thus, a reduction of gram-positive bacteria would most likely cause a reduction, not an increase, in the use of acetate.

Gram-positive bacteria give a positive result in the gram stain test, indicating the presence of a thick peptidoglycan layer. They appear blue/purple after staining. In contrast, gram-negative bacteria appear pink/red upon staining and have a thin peptidoglycan wall.

Per Table 1, Ruminococcaceae (gram-positive) uses H+ as an energy source. Thus, it continually consumes H+, raising the surrounding pH. If the number of Ruminococcaceae is reduced, less H+ will be consumed, so the surrounding pH will decrease.

Therefore, we may conclude that a decrease in pH is most likely associated with a reduction in gram-positive bacteria.

None of the bacteria assessed, regardless of gram status, produce polysaccharides (PS). Consequently, reducing the number of gram-positive bacteria will not likely affect PS production.

This question asks us to identify which physiological change will result from the microbiome of CD-affected individuals, prompting us to think about the anatomy of the GI microbiome. Answering this passage-based question will require a reference to the second and third paragraphs.

Based on Table 1, none of the GI bacterial changes associated with CD impact polypeptide digestion.

The human intestine cannot absorb dietary fiber.

Based on Table 1, the two bacteria that produce propionate, Phascolactobacteria, and Odoribacter, are decreased in CD-affected individuals. Thus, the levels of propionate are most likely reduced, rather than increased, in CD-affected individuals.

The passage’s first paragraph states that Crohn’s disease is a chronic inflammatory disease and that research has shown the GI tract microbiome is different between CD-affected and healthy individuals. The passage follows up with the statement that researchers “hypothesized that differences in bacterial distribution and the host immune response to GI tract bacteria” might play a role in CD progression. The passage subsequently states that typically, “GI tract bacteria digest dietary fibers and converts them into butyrate, propionate, and acetate” and that “these three molecules modulate the innate immune system response by attenuating the inflammatory response to GI tract commensal bacteria.” Essentially, butyrate, acetate, and propionate inhibit the inflammatory response against commensal bacteria in the GI tract. In other words, these three molecules produced by certain bacteria have anti-inflammatory effects.

From Table 1, we can see the amount of Roseburia, Phascolactobacteria, Ruminococcaceae, Odoribacter, and Faecalibacteria are highly reduced in CD-affected individuals compared to healthy individuals. The products of the bacteria include butyrate, propionate, and acetate. As a result of these products being reduced, the GI tract will not have the same inhibition to host inflammatory responses. Therefore, the microbiome of CD-affected individuals may result in decreased immune tolerance.

This passage-based question asks us to determine which phylum Enterobacter likely belongs to, prompting us to refer to the third paragraph while examining Table 1.

The passage does not provide information regarding levels of Actinobacteria in CD-affected individuals. Thus, this answer is not sustainable.

Table 1 demonstrates that Firmiculate bacteria are highly reduced in individuals with CD, while Table 1 indicates that Enterobacter bacteria are highly increased in CD-affected individuals. Thus, Enterobacter likely does not belong to Firmiculate.

The passage does not provide information regarding levels of Bacteroidetes in CD-affected individuals. Thus, this answer is not sustainable.

The passage states that the microbiome of individuals with CD “have an increase in the number of the gram-negative Proteobacteria, whereas the number of Firmicutes bacteria is highly reduced.” According to Table 1, Enterobacter (gram-negative) levels are highly increased in CD-affected individuals.

Therefore, because it is a gram-negative bacteria that is highly increased in individuals with CD, we may conclude that Enterobacter most likely belongs to the phylum Proteobacteria.

This passage-based question asks us to determine which product is particularly reduced in CD-affected individuals, prompting us to analyze Table 1.

None of the bacteria analyzed in Table 1 produce glucose, so data from the passage does not support this option.

From Table 1, we can see only Enterobacter produces lactate. Since Enterobacter is increased in CD-affected individuals (denoted by two superscript stars: Enterobacter**), lactate production most likely increases, not decreases.

Enterobacter is the only LPS producer assessed and is increased during CD (includes the superscript stars **). LPS production is increased, not decreased, in CD-affected individuals.

Table 1 outlines bacteria of the large intestine that were analyzed in healthy individuals and CD-affected individuals. Several bacteria, including Roseburia, Odoribacter, and Faecalibacteria were “highly reduced” in CD-affected individuals compared to healthy individuals. Each of these bacteria produces butyrate from their respective energy source. Thus, if the bacteria are highly reduced, we would expect their products to also be highly reduced.

Therefore, we may conclude that butyrate is particularly reduced in CD-affected individuals.

This question asks us to identify the amino acid that exhibits a beta-branched side chain. We must use our content knowledge of amino acid structures to answer this question.

Alanine does not have any branches in its side chain, so it cannot exhibit a beta-branched side chain.

Leucine is branched but not on the correct carbon. The first carbon, the alpha carbon, is the carbon attached to the groups of interest, which is the amine and carboxylic acid groups in this case. The beta carbon would then be the first carbon in the side chain. In leucine, it is the third carbon that has a branch.

Beta-branched means that there is a branch on the beta carbon or second carbon from a designated group. Here, we are looking for a branch on the amino acid’s second carbon or the side chain’s first carbon. Neither alanine nor glycine has a branch in their side chain. Isoleucine had a branch on its second carbon, and leucine had a branch on its third carbon.

From this, we can determine that option C, Isoleucine, is correct.

Glycine does not have any branches in its side chain so it cannot exhibit a beta-branched side chain.

This question asks us to identify the statement that is unique to the chemical messengers in paracrine and autocrine signaling systems as compared to endocrine signaling systems. We must use our content knowledge of signaling systems to answer this question.

Paracrine and autocrine chemical messengers can be secreted by a wide range of cell types. Chemical messengers secreted from neurons are normally called neurotransmitters.

Endocrine, paracrine, and autocrine chemical messengers all bind to receptors on or in cells.

Exocrine signaling systems export their chemical messengers via ducts into an epithelium while endocrine, paracrine, and autocrine signaling systems do not.

Autocrine signaling systems are systems by which a cell releases chemical messengers that will interact with the cell’s own receptors. Paracrine signaling systems are systems in which cells release chemical messengers into the interstitial fluid which will then interact with cell receptors on cells that are nearby. Endocrine signaling systems are systems in which cells release chemical messengers that travel to distant cells to interact with cell receptors that are far away. Endocrine signals are released into the bloodstream to affect these distant, and often scattered cells.

Autocrine and paracrine signaling systems affect cells in the immediate environment and do not affect cells that are far away so we can conclude that they do not release chemical messengers into the bloodstream making option D correct.

This question asks us to identify the electrogenic membrane transporter that translocates a net charge across the membrane. We must use our content knowledge of membrane transporters to answer this question.

A Na+ − H+ exchanger trades a positive charge for a positive charge and does not change the electrical potential of a cell. Na+ − H+ exchanger is not electrogenic.

A Na+ − C- cotransporter will transport a positive and a negative charge in the same direction and thus will not change a cell’s electrical potential. Na+ − C- cotransporter is not electrogenic.

Electrogenic means producing a change in the electrical potential of a cell, as hinted at by the question stem. A Na+ − glucose cotransporter transports glucose and a positive charge from sodium into the cell, resulting in a change in the electrical potential of a cell.a

A Na+ − glucose cotransporter utilizes the energy stored in the sodium concentration gradient to transport both sodium and glucose from the extracellular fluid into the cell. The cell’s electric potential is changed due to a net positive charge flowing across the cell membrane from the extracellular space to the intracellular space.

We can conclude that a Na+ − glucose cotransporter is electrogenic; thus, option C is correct.

A GLUT2 facilitative glucose transporter only transports glucose, a neutral molecule, so there is no change in the electrical potential of a cell. GLUT2 facilitative glucose transporter is not electrogenic.

This question asks us to identify the type of cells that have a plasma membrane, lack a nucleus, lack most organelles, and are shaped like a flat disc with a concave center. We must use our content knowledge of cell types to answer this question.

Erythrocytes, or red blood cells, have a plasma membrane, lack a nucleus, lack most organelles, and are shaped like a flat disc with a concave center. Fibroblasts are connective tissue cells that produce collagen and other fibers and have a plasma membrane, a nucleus, abundant organelles (Golgi apparatus, endoplasmic reticulum, and ribosomes) involved in extracellular matrix synthesis, and an irregular shape. Monocytes are leukocytes with a plasma membrane and a nucleus, lack many organelles, and are bean-shaped. Neurons are nervous tissue cells with a plasma membrane, a nucleus, and all organelles that vary widely in shape.

From the types of cells listed, erythrocytes most closely resemble the description in the stem, so we can conclude that A is correct.

Fibroblasts are connective tissue cells that produce collagen and other fibers and have a plasma membrane, a nucleus, abundant organelles (Golgi apparatus, endoplasmic reticulum, and ribosomes) involved in extracellular matrix synthesis, and an irregular shape. This type of cell does not match the description in the question stem.

Monocytes are leukocytes with a plasma membrane and a nucleus, lack many organelles, and are bean-shaped. This type of cell does not match the description in the question stem.

Neurons are nervous tissue cells with a plasma membrane, a nucleus, and all organelles, and they vary widely in shape. This type of cell does not match the description in the question stem.

This question prompts us to determine a physiological variable whose increase would most likely cause an increase in the amount of EPO released by the kidneys in a healthy human adult. We must rely on the passage’s description of what causes EPO to be released.

The first paragraph of the passage states that “EPO is primarily produced and released by the kidneys in response to low tissue levels of oxygen” as it increases the “oxygen-carrying capacity of the blood.” During aerobic exercise, muscles rely on aerobic energy-producing pathways to function. As aerobic energy-producing pathways require oxygen to produce energy, we can conclude that aerobic exercise would result in low tissue levels of oxygen. These tissues would require an increased amount of oxygen, which is facilitated by erythrocytes.

Therefore, an increase in aerobic exercise would likely cause an increase in the oxygen demand by muscle tissue. The amount of EPO released would be increased during aerobic exercise in order to increase the amount of circulating erythrocytes to deliver oxygen to muscle tissue.

Hemoglobin is a protein found in red blood cells primarily responsible for carrying oxygen. If the total amount of circulating hemoglobin were to increase, the oxygen-carrying capacity of the blood would also increase. As EPO is released in response to low oxygen levels, we would not expect increased hemoglobin to result in EPO secretion.

The passage states that EPO causes erythrocyte precursor cells to “differentiate into mature erythrocytes,” consequently increasing the “oxygen-carrying capacity of the blood.” If the rate of erythrocyte maturation increased, we would expect the blood’s oxygen-carrying capacity to increase faster. This would not result in an oxygen deficit and would not cause EPO to be released.

An increase in cardiac output would result in more blood being pumped by the heart per minute. This increase would increase blood pressure and blood flow but would not alter the amount of oxygen in the tissues. Oxygen would only be expected to decrease if tissues began to use up their oxygen reserves at a faster rate.

This question prompts us to identify a benefit of a form of EPO that can act on CNS neurons without affecting erythrocyte production in the bone marrow. We will need to rely on the passage’s description of EPO, as well as our content knowledge surrounding erythrocyte production, to answer this question.

The passage states that EPO decreases apoptosis, contradicting the first half of this answer option.

The first paragraph of the passage states that “EPO is primarily produced and released by the kidneys in response to low tissue levels of oxygen” as it increases the “oxygen-carrying capacity of the blood.” Though neuronal cell death would likely be limited, the second half of this answer option is incorrect as EPO is not expected to decrease the oxygen-carrying capacity of the blood.

The passage states that “EPO has been shown to decrease the apoptosis of both erythrocyte precursor cells and CNS neurons,” indicating that one benefit of EPO in the CNS is a decrease in apoptosis. This supports the first half of this answer option, but we will also need to make sure the second half is supported.

Blood is composed of blood plasma, red blood cells (erythrocytes), white blood cells, and platelets. Consequently, blood viscosity will rely on the balance of these four components. The passage tells us that EPO “[causes premature erythrocytes] to differentiate into mature erythrocytes that are released into circulation,” effectively increasing the number of red blood cells in the blood and consequently influencing blood viscosity.

If this new form of EPO can act on the CNS without resulting in erythrocyte production, we can conclude that it would decrease apoptosis in the CNS without causing an increase in blood viscosity.

The passage’s last paragraph states that “a significant number of tumors express EPOR, even though the healthy tissue…does not,” indicating that once tumors are present, they can interact with circulating EPO. However, as EPO is not shown to promote tumor development, the second half of this answer option is not supported.

This question prompts us to identify a potential explanation for why the EPO protein produced by bacteria did not increase erythrocyte production when injected into humans. We will need to rely on the passage’s description of EPO, as well as our content knowledge regarding proteins to answer this question.

Translation involves interpreting mRNA to create proteins of a specific amino acid sequence. The ribosomes responsible for translation interpret mRNA as three-piece codons in both eukaryotes and prokaryotes.

The only structural description of EPO tells us that “erythropoietin (EPO) is a glycoprotein hormone.” Glycoproteins are created when sugars are added to the protein via glycosylation. Even though E. coli could have the same rHuEPO gene as humans, if they lack the enzymes necessary to glycosylate EPO identically to humans, the resulting glycoprotein will differ between the two organisms.

As the protein’s structure may differ, it is likely that bacterial EPO does not properly interact with human EPORs if E. coli cannot glycosylate EPO in the same way that it is glycosylated by eukaryotic cells.

Similarly to eukaryotes, bacteria are able to produce proteins by translating mRNA. Though the method of translation slightly differs, bacteria can still secrete proteins that are found in eukaryotic cells.

Recombinant proteins are proteins produced from recombinant (modified) DNA. As the enzymes necessary for modifying DNA are present in eukaryotes, prokaryotes, and viruses, we cannot state that only viruses contain the necessary cellular machinery to express recombinant proteins properly.

This question prompts us to identify a mechanism through which a mutation in the EPO gene could lead to a higher-than-normal number of circulating erythrocytes. We will need to rely on the passage’s description of EPO in order to answer this question.

The promoter region of a DNA sequence can be bound by proteins to initiate RNA transcription. If the promoter of the mutant EPO allele was defective, proteins would not be able to bind and the allele would not be transcribed. We would expect a lack of EPO to result in a less-than-normal number of circulating erythrocytes.

The passage states that “EPO binds the EPO receptor (EPOR) in erythrocyte precursor cells, causing them to differentiate into mature erythrocytes that are released into circulation.” If a mutant EPO allele results in a protein that has an increased affinity for EPOR, EPO would bind to EPOR more frequently and would consequently cause precursor cells to differentiate into mature erythrocytes at a faster rate.

Therefore, we can conclude that an increased affinity for EPOR could be a potential mechanism by which a dominant EPO allele could cause a higher-than-normal number of circulating erythrocytes.

If the mRNA produced by the mutant EPO allele were degraded before translation, EPO would not be produced, and there would be a lower-than-normal number of circulating erythrocytes.

As EPO must bind EPOR to cause precursor cells to differentiate into mature erythrocytes, a mutant allele that produces a protein that is unable to bind to EPOR would lead to a lower-than-normal number of circulating erythrocytes.

This question prompts us to identify the cellular location in which EPO likely binds EPOR in erythrocyte precursor cells. This requires us to rely on the passage’s description of EPO.

The cytosol is the fluid found inside the cell. As EPO is a glycoprotein, it could not independently cross the plasma membrane to reach the cytosol, making Option D a stronger choice.

The endoplasmic reticulum is an organelle that plays an important role in cellular transportation and protein folding. As EPO is a glycoprotein, it would not be able to independently cross the plasma membrane to reach the endoplasmic reticulum, making Option D a stronger choice.

The nucleus is the structure inside the cell that contains genetic information. As EPO is a glycoprotein, it would not be able to independently cross the plasma membrane to reach the endoplasmic reticulum, making Option D a stronger choice.

The passage states that “erythropoietin (EPO) is a glycoprotein hormone.” As the intermembrane space is nonpolar, glycoproteins cannot pass through the plasma membrane due to their size and polarity.

Therefore, we can conclude that EPO likely binds EPOR on the surface of the plasma membrane.

This question prompts us to identify a likely consequence of the continuous production of EPO. We will need to rely on the passage’s description of the role EPO plays in order to answer this question.

The passage states that though “EPO is primarily produced and released by the kidneys… several other tissues, including the liver and neurons of the central nervous system (CNS), can produce EPO.” This statement tells us that the liver is also able to produce EPO. However, as the tumors described in the question will still be secreting EPO, an excess of EPO will be present in the body regardless of which organ is responsible for erythrocyte production.

As the passage tells us that EPO results in an increase in erythrocytes, we would expect erythrocyte production within the bone marrow to increase.

The passage tells us that EPO “[causes premature erythrocytes] to differentiate into mature erythrocytes released into circulation.” If EPO were continuously being produced, we would expect erythrocytes to continuously be released into circulation. Red bone marrow is spongy tissue located within bones responsible for creating red blood cells (erythrocytes), white blood cells (leukocytes), and platelets. If erythrocytes are continuously released into circulation, we would expect red blood cells to be continuously created by the bone marrow.

Therefore, tumors that constantly secrete EPO likely result in constant stimulation of erythrocyte production that will occur within the bone marrow.

As the passage tells us that EPO results in an increase in erythrocytes, we would not expect tumors that consistently secrete EPO to have no impact on erythrocytes.

This passage-based question asks us to identify which class of proteins 5-HTR4 belongs to, requiring us to carefully examine the second paragraph and Figure 2.

Protein kinases are enzymes that catalyze the addition of a phosphate group to proteins. Although protein kinases are involved in the GPCR signaling pathway, the passage states that 5-HTR4 is a transmembrane receptor, which does not describe protein kinases.

Tyrosine kinase receptors are transmembrane receptors that cause an enzymatic reaction within the cell upon the binding of a ligand. Figure 2 shows that 5-HT activation of 5-HTR4 results in increased cAMP levels, which is a hallmark of GPCRs, not tyrosine kinase receptors. Tyrosine kinases selectively phosphorylate tyrosine residue in different substrates.

Ligand-gated ion channels, also known as ionotropic receptors, are transmembrane proteins that allow the passage of certain ions through the membrane in response to the binding of a chemical ligand. Figure 2 shows that 5-HT activation of 5-HTR4 results in increased cAMP levels, which is a hallmark of GPCRs, not ligand-gated ion channels.

The second paragraph states that 5-HTR4 is “a 5-HT transmembrane receptor,” or a transmembrane protein that binds 5-HT. To determine which type of transmembrane receptor 5-HTR4 is, we need to understand how the interaction between 5-HT and 5-HTR4 affects cAMP levels.

Figure 2 demonstrates that when 5-HTR4 is present, the addition of 5-HT increases cAMP levels. This is consistent with the expected response of a G protein-coupled receptor (GPCR). GPCRs are transmembrane protein receptors that activate an intracellular G protein in response to an extracellular ligand binding. Typically, activation of the G protein leads to the activation of adenylyl cyclase, which produces the second messenger, cyclic adenosine monophosphate (cAMP).

Therefore, because 5-HTR4 is a transmembrane protein that is associated with increased cAMP levels in the presence of 5-HT, we can conclude that it is a G protein-coupled receptor.

This question asks us to determine which amino acid 5-HT is synthesized from, given its structure and information about its synthesis pathway. This pseudo-discrete question does not require a reference to the passage.

The structure of 5-HT contains an indole group. Histidine contains an imidazole ring, not an indole group. Furthermore, histidine cannot be converted directly into 5-HT by hydroxylation and decarboxylation.

The structure of 5-HT contains an indole group. Proline is characterized by its pyrrolidine ring, not an indole group. Furthermore, proline cannot be converted directly into 5-HT by hydroxylation and decarboxylation.

The structure of 5-HT contains an indole group. Tyrosine contains a phenol group and an aromatic benzene ring, not an indole group. Furthermore, tyrosine cannot be converted directly into 5-HT by hydroxylation and decarboxylation.

The question states that 5-HT is synthesized “from a single amino acid by a short metabolic pathway consisting of a hydroxylation reaction followed by a decarboxylation reaction.” To visualize the structure of the amino acid, we can reverse this reaction to imagine the structure before the removal of the carboxyl and the addition of the hydroxyl group. From the structure, we can see that 5-HT contains an indole group, which tryptophan also contains. Moreover, the structure is identical to tryptophan’s, except for the addition of a hydroxyl group and the omission of a carboxylic acid group.

Therefore, we can conclude that 5-HT is synthesized from tryptophan.

This question asks us to determine the proper control group in an experiment in which IBU is administered to mice to determine whether it can cause symptoms of depression. This pseudo-discrete question does not require the passage to answer.

Selective serotonin reuptake inhibitors (SSRIs) are antidepressants that would likely inhibit symptoms of depression in mice. The experiment within the question stem is to test if long-term treatment of IBU alone can cause behavioral symptoms of depression. Consequently, this would not be an appropriate comparison for researchers to assess whether IBU causes depression.

The experiment aims to determine whether IBU causes depression in normal mice. To accurately examine this question, these mice should be compared to other normal mice, not depressed mice.

In an experiment, a control is a standard group to which comparisons are made. Specifically, a control allows researchers to ensure that changes observed in the mice are caused by IBU and are not an artifact of the experiment. Treatment of normal mice with a placebo will provide a baseline for them to observe normal behavioral characteristics in the absence of IBU. Using this control group, they can determine whether IBU causes behavioral symptoms of depression.

Therefore, we can conclude that normal mice treated with a placebo would be the appropriate control group.

The experiment aims to determine whether IBU causes depression in normal mice. To accurately examine this question, these mice should be compared to other normal mice, not depressed mice. Furthermore, the administration of cytokines should be avoided as it is not known whether it will affect depressive symptoms.

This question asks us to identify which comparison would allow us to determine whether IFNγ is necessary for the increased expression of p11 induced by SSRIs. This experimental reasoning question does not require reference to the passage.

Treating mice with p11 instead of with an SSRI would make it impossible to measure the antidepressant-induced change in p11 expression. Instead, both groups should be treated with an SSRI, not p11.

Although the passage states that “IFNγ directly induces the expression of the protein p11 in neighboring neurons,” we are asked to determine how this relationship could be concluded.

For this question to be examined, both test groups must be given an SSRI so that antidepressant-induced increases in p11 can be observed. However, one group should have IFNγ (wild-type), and the other should lack IFNγ (IFNγ knockout). Then, p11 expression levels can be compared between groups to assess the effect of IFNγ.

Therefore, we can conclude that this experimental setup would allow us to determine whether IFNγ is necessary for the antidepressant-induced increases in the expression of p11.

In order to examine changes in antidepressant-induced expression of p11, both groups must be treated with an SSRI.

In order to examine changes in antidepressant-induced expression of p11, both groups must be treated with an SSRI.

This passage-based question asks us to identify the accurate relationship between pH and charged functional groups based on the information in Table 1, prompting us to think about isoelectric points.

The isoelectric point (pI) is the pH at which a molecule has no net charge. Although the molecule can contain positively and negatively charged functional groups, the net charge of the molecule will be zero. The pI of each V. cholerae Na+-NQR subunit is given in Table 1. At a pH of 8.50, the NqrD subunit is at its isoelectric point and will thus have a net charge of zero. The number of cationic and anionic function groups must be equal because the net charge is zero.

Therefore, we can conclude that at pH 8.50, the ratio of the cationic to anionic function groups in the NqrD subunit is equal to 1.

Table 1 shows that the isoelectric point of the NqrE subunit is 5.35, so the majority of functional groups will be deprotonated.

Table 1 shows that the isoelectric point of the NqrA subunit is 6.30, so at pH 6.30, the net charge of cationic and anionic functional groups will be zero.

Table 1 shows that the isoelectric point of the NqrF subunit is 5.25, so at pH 6.00, the majority of functional groups will be deprotonated.

This question asks us to determine the likely effect of introducing a sodium ionophore to a culture of V. cholerae, prompting us to consider the role of ionophores. This question requires reference to the passage and application of content knowledge.

The sodium ionophore would degrade the sodium gradient established by Na+-NQR, but would not directly interfere with Na+-NQR.

Ionophores are compounds that bind to ions, allowing them to cross the plasma membrane. The passage states that “An electrochemical gradient of Na+ known as the ‘sodium motive force’ provides V. cholerae with the energy for key functions.” This implies that V. cholerae uses the sodium motive force to generate ATP, the conventional energy source of cells.

Introducing sodium ionophores to the culture would permit the movement of Na+ across the membrane, dissipating the sodium motive force that allows ATP to be generated.

Therefore, we may conclude that adding a sodium ionophore to a culture of V. cholerae would likely lead to decreased ATP production.

Sodium ionophores affect the permeability of sodium ions, not protons, across the membrane, so this would not alter pH. Moreover, the passage states that V. cholerae uses an electrochemical gradient of Na+, not protons, to generate energy.

By diminishing the electrochemical gradient created by Na+-NQR, the ionophore will cause decreased ATP production. To compensate, V. cholerae is likely to increase its respiratory rate, corresponding to increased, not decreased, oxygen consumption.

This question asks us to determine which subunits of Na+-NQR can be separated by gel filtration but not by ion exchange chromatography, prompting us to consider protein separation methods. To answer this passage-based question, we must carefully examine Table 1.

NqrA and NqrB have very similar molecular weights (47.6 vs 47.4 kDa), so they cannot be separated effectively by gel filtration chromatography.

NqrC and NqrE have a large charge difference, as shown by their isoelectric points (7.30 and 5.35). Therefore, they can be separated by ion exchange chromatography.

Although NqrD and NqrF have a large enough difference in molecular weight (22.8 vs 47.1 kDa) for separation by gel filtration chromatography, their isoelectric points (8.50 vs 5.25) are also sufficiently dissimilar for effective separation by ion exchange chromatography.

Chromatography is a laboratory technique to separate a mixture into its individual components. Gel filtration chromatography (also known as size-exclusion chromatography) is a method that separates compounds by size. Using gel containing specially formulated porous beads, larger molecules pass through the chromatography column unhindered while small molecules become stuck in beads, leading to separation based on molecular weight.

Ion exchange chromatography separates molecules according to their charge using a column containing oppositely-charged molecules. By altering the pH of the column, proteins can be separated on the basis of their isoelectric point.

Thus, our task is to identify the subunits that have a large difference in molecular weight (for gel filtration separation) but a very slight difference in isoelectric point (pI). NqrE and NqrF have drastically different molecular weights (20.5 vs 47.1 kDa) but similar isoelectric points (5.35 vs 5.25).

Therefore, we can conclude that NqrE and NqrF can be separated by gel filtration but not by ion exchange chromatography.

This question asks us to determine which enzyme of the citric acid cycle is not involved in generating the dinucleotide required for Na+-NQR activity. This passage-based question requires us to refer to the second paragraph.

Malate dehydrogenase catalyzes an oxidation-reduction reaction to convert malate and NAD+ to oxaloacetate and NADH. Therefore, it is involved in the generation of NAD+/NADH.

The passage states that Na+-NQR requires NAD+/NADH to function. Thus, we must identify the enzyme that is NOT involved in the generation of NAD+/NADH.

Succinate dehydrogenase catalyzes the conversion of succinate to fumarate while reducing FAD to FADH2. Neither NAD+ nor NADH are involved in this reaction.

Therefore, we can conclude that succinate dehydrogenase is not directly involved in the generation of the dinucleotide required for Na+-NQR activity.

Isocitrate dehydrogenase catalyzes oxidative decarboxylation to convert isocitrate and NAD+ to α-ketoglutarate, NADH, and CO2. Therefore, it is involved in the generation of NAD+/NADH.

Α-Ketoglutarate dehydrogenase catalyzes oxidative decarboxylation to convert α-ketoglutarate, NAD+, and CoA to succinyl CoA, CO2, and NADH. Therefore, it is involved in the generation of NAD+/NADH.

This question asks us to identify a biochemical technique that relies on a pH gradient. Here, we need to understand the basics of each technique mentioned in the answer choices, specifically how they work and whether they need a pH gradient to operate.

Limited proteolysis is a technique that involves using a protease enzyme to selectively and partially cleave a protein. Proteases cleave proteins at specific sites, so limited proteolysis can be used to study the structure of proteins by observing where the protease is able to cleave the protein. This technique does not rely on a pH gradient.

Southern blotting is a method used for DNA identification and involves the transfer of DNA fragments from an agarose gel to a membrane, followed by detection with a labeled probe. The separation of DNA fragments is based on size, not pH.

Isoelectric focusing is a technique that separates proteins based on their isoelectric point (pI), which is the pH at which a particular molecule or surface carries no net electrical charge. The technique involves applying a pH gradient along a gel, then applying an electric field across the gel. This causes proteins to move in the electric field until they reach a region of the gel where the pH is equal to their isoelectric point. At this point, the protein will have no net charge and will stop moving, allowing it to be isolated.

The key here is understanding that a pH gradient is required to separate molecules based on their isoelectric point. While all the options involve biochemical techniques, only isoelectric focusing uses a pH gradient for molecule separation, making option C the correct answer.

SDS-PAGE is a common laboratory technique used to separate proteins according to their electrophoretic mobility. While it does involve the application of an electric field, it does not require a pH gradient. In SDS-PAGE, proteins are denatured and covered with negative charges proportional to their length, ensuring that the separation of proteins is based primarily on size, not charge or pH.

This question prompts us to identify the expected phenotype of the children of a color-blind woman and a man who is not color-blind. We will need to rely on the question’s description of red-green color blindness to answer this question.

Though the couple’s daughters would be expected to have normal vision, all of their sons would have red-green color blindness. Refer to Option B for a more in-depth explanation.

The question states that “the gene for red-green color blindness is recessive and X-linked.” Therefore, we would expect a color-blind woman to hold two affected X chromosomes and a man who is not color-blind to hold zero affected X chromosomes. This couple’s sons will inherit their Y chromosome from their father and their X chromosome from their mother, resulting in a red-green color-blind phenotype. On the other hand, this couple’s daughters will inherit one affected X chromosome from their mother and one unaffected X chromosome from their father. As red-green color blindness is X-linked recessive, all of these daughters will have a normal phenotype.

Therefore, we can conclude that all of the couple’s daughters will have normal color vision, and all the sons will be color-blind.

We would expect the opposite of this answer option, with all daughters having normal vision and all sons having red-green color blindness.

As there is no possibility of this couple’s daughters inheriting two affected X chromosomes, we would expect all of their daughters to have normal vision.

This question prompts us to identify the type of reaction that would have Keq > 1 and be kinetically fast.

Endergonic reactions describe reactions in which the products contain more energy than the reactants. As these reactions are not thermodynamically favored, they would have a Keq < 1. Furthermore, reactions with high activation energies would be kinetically slow as more energy is necessary for them to occur.

Though a reaction with low activation energy would be kinetically fast, Keq > 1 are associated with endergonic reactions.

Though an exergonic reaction would have Keq > 1, a reaction with high activation energy would be kinetically slow.

Keq, also known as the equilibrium constant, is the ratio of products over the reactants of a reversible reaction. If Keq is greater than 1, we would expect the products to be favored over the reactants. An exergonic reaction describes a reaction that results in a release of free energy. As exergonic reactions generally increase entropy by releasing free energy, these reactions are favored and would be associated with an equilibrium constant that favors its products.

Activation energy describes the energy required to get molecules into a high-energy transition state and, consequently, for a reaction to occur. Reactions with lower activation energies will be kinetically faster as less energy is needed for them to occur.

Therefore, we can conclude that a reaction that has a Keq > 1 and is kinetically fast would be exergonic and have a low activation energy.

This question prompts us to identify what events occur at time points A and B according to the graph provided. We will need to rely on our content knowledge regarding glucose regulation to answer this question.

Glucagon is secreted by pancreatic alpha cells, not hepatocytes (liver cells). Furthermore, insulin is secreted by pancreatic beta cells, not pancreatic alpha cells.

Though this answer option correctly pairs beta cells with insulin and alpha cells with glucagon, we would not expect insulin to be secreted at point B as blood glucose levels are already low. Similarly, we would not expect glucagon to be secreted at point A as blood glucose levels are already high.

This answer option incorrectly attributes glucagon secretion to hepatocytes and insulin secretion to pancreatic alpha cells.

As the y-axis of the graph describes blood glucose levels, we can interpret point A as an instance of high blood glucose and point B as an instance of low blood glucose. Insulin is a hormone secreted by pancreatic beta cells in response to high blood glucose in order to promote the uptake of glucose into cells and out of the bloodstream. Glucagon is a hormone secreted by pancreatic alpha cells in response to low blood glucose to promote the release of glucose from cells and into the bloodstream.

Therefore, we can conclude that pancreatic beta cells secrete insulin at point A to lower blood glucose levels and that pancreatic alpha cells secrete glucagon at point B to raise blood glucose levels.

This question asks us to identify how the Unit Membrane Model differs from the Fluid Mosaic Model, requiring us to carefully analyze the information presented in the passage.

According to the Unit Membrane Model, the passage states, “An extended monomolecular layer of protein coats both the outside of the membrane and the inside of the bilayer, but the protein does not penetrate the bilayer.” According to the Fluid Mosaic Model, proteins are “inserted into the bilayer” with hydrophobic regions of the protein “found on the inside surface.” The Unit Membrane Model posits that proteins coat the membrane’s outer surface, whereas the Fluid Mosaic Model concludes that proteins can span the entire membrane.

Therefore, the Unit Membrane Model differs from the Fluid Mosaic Model by the location of proteins.

While it is true that the Unit Membrane Model has a monomolecular layer of protein on both sides of the membrane, the Fluid Mosaic Model does not have proteins forming a bimolecular layer of protein on each surface. It posits that proteins can span the phospholipid bilayer.

In both models, the membrane is made up of a phospholipid bilayer.

This answer choice reverses the characteristics of each model. The Unit Membrane Model has a coated, monomolecular layer of protein on each side of the membrane, and the Fluid Mosaic Model suggests that proteins are dissolved in the bilayer.

This question asks us to identify the observation that would invalidate the Unit Membrane Model, prompting us to employ our scientific reasoning and problem-solving skills. This question requires a thorough understanding of the passage to answer.

This would prove that the membrane is composed of a phospholipid bilayer, which would not contradict the Unit Membrane Model.

This would support the idea that phospholipid heads are more hydrophilic than tails, which aligns with the Unit Membrane Model.

The Unit Membrane Model suggests that membranes are composed of a phospholipid bilayer with a monomolecular layer of protein that coats but does not penetrate the bilayer. If a membrane were frozen and split from surface to surface, the middle of the membrane would be exposed, revealing the inner contents. If proteins were observed within the hydrocarbon chains, this would suggest that proteins penetrate the lipid bilayer, contradicting the Unit Membrane Model.

Therefore, we can conclude that observing proteins within hydrocarbon chains in the middle of the bilayer would invalidate the Unit Membrane Model.

This would prove that phospholipid tails are more hydrophobic than the heads, which aligns with the Unit Membrane Model.

This question asks us to determine how the discovery of cholesterol embedded within the phospholipid bilayer aligns with the Fluid Mosaic Model, prompting us to employ our scientific reasoning and problem-solving skills. This question requires a thorough understanding of the passage to answer.

The discovery of cholesterol in the middle of the lipid bilayer does not exclude the possibility of proteins being embedded more than halfway through the membrane. Transmembrane proteins could remain embedded within the membrane leaflet that was removed.

The discovery of cholesterol within the membrane would not suggest that proteins cannot also be found within the membrane.

The Fluid Mosaic Model states that hydrophilic regions of proteins and lipids are found on the outer surfaces of the membrane, while hydrophobic regions are found on the inside. Because the middle of the bilayer is hydrophobic, the lipids do not need to be embedded in the proteins.

According to the Fluid Mosaic Model, proteins are “inserted into the bilayer.” This theory is supported by an experiment that split the middle of the lipid layers, revealing “small bumps … assumed to be proteins.”

During the electron microscopy technique known as freeze-fracture, cell membranes are split apart at the center of the bilayer, allowing internal components of the membrane to be visualized. Here, the freeze-fracture assay revealed “small bumps … assumed to be proteins.”

If these small bumps were chemically shown to be cholesterol, not proteins, then the Fluid Mosaic Model would have less supporting evidence. However, this discovery would not contradict the model, which states that hydrophobic regions of both proteins and lipids are found on the inside surface of the membrane. Because cholesterol is a lipid, this aligns with the Fluid Mosaic Model.

Therefore, we can conclude that this discovery would not require the Fluid Mosaic Model to be modified, but it would invalidate previous supporting evidence.

This question asks us to identify a weakness of the Unit Membrane Model, prompting us to consider various aspects of the model. This question requires a thorough understanding of the passage as well as the application of content knowledge.

The membrane is extremely thin and thus does not primarily determine cell size. This is not a weakness of the Unit Membrane Model.

For nutrients and waste to enter or exit the cell, the cell membrane must be crossed. Whereas the Fluid Mosaic Model explains how membrane-spanning proteins could mediate these processes, the Unit Membrane Model does not offer a mechanism to explain the exchange of nutrients or waste. According to the passage, it states that no proteins penetrate the lipid bilayer, failing to explain how hydrophilic substances can pass across the membrane.

Therefore, we can conclude that a weakness of the Unit Membrane Model is the lack of explanation for how the exchange of nutrients and wastes occurs.

The Unit Membrane Model does explain how the membrane acts as a barrier between the contents of a cell and the external environment, suggesting that the hydrophobic interior of the membrane will not permit the transport of hydrophilic molecules.

The Unit Membrane Model describes a highly hydrophobic interior consisting of hydrocarbon tails that impede the transport of polar molecules. Thus, this is not a weakness of the Unit Membrane Model.

This question asks us to identify which enzyme UCP1 reduces the activity of, prompting us to think about the enzymes involved in oxidative phosphorylation. This passage-based question requires a reference to the first paragraph.

The passage states, “UCP1 directly reduces the proton gradient driving oxidative phosphorylation.” Oxidative phosphorylation is the method by which cells produce ATP during cellular respiration. Fueled by energy from the proton gradient, the enzyme ATP synthase converts ADP + Pi to ATP.

Uncoupling proteins like UCP1 dissipate the proton gradient to create heat, ‘uncoupling’ the energy-releasing event of proton transport from the energy-requiring event of ATP synthesis. Without the proton gradient, ATP synthase cannot function properly.

Therefore, we can conclude that UCP1 reduces the activity of ATP synthase.

Pyruvate kinase is responsible for converting phosphoenolpyruvate, ADP, and H+ to pyruvate and ATP during glycolysis. The activity of pyruvate kinase does not depend on the mitochondrial proton gradient.

Na+/K+ ATPase is involved in ion transport. It is not involved in oxidative phosphorylation and would thus not be directly affected by UCP1.

Succinyl-CoA synthetase combines succinyl CoA, Pi, and GDP to form succinate, CoA, and GTP during the citric acid cycle. The activity of succinyl-CoA synthetase does not depend on the mitochondrial proton gradient.

This question asks us to identify which hypothesis is best supported by information from the passage, prompting us to employ our scientific reasoning and problem-solving skills. This question requires a careful understanding of the passage to answer.

The passage indicates that exercise is correlated with increased FNDC5 expression and that mice overexpressing FNDC5 exhibit lower nonfasting glucose levels. This is likely due to higher, not lower, cellular glucose uptake.

The passage demonstrates that during exercise, expression of UCP1 is increased, which in turn “reduces the proton gradient driving oxidative phosphorylation.” Cellular respiration is the process by which cells oxidize fuels, like glucose, to produce energy in the form of ATP. The final step of cellular respiration depends on the mitochondrial proton gradient, which drives oxidative phosphorylation. Thus, during exercise, the proton gradient is partially dissipated by increased UCP1 levels, which causes more energy to be lost as heat and less to be used to synthesize ATP.

Therefore, the information in the passage best supports the hypothesis that exercise promotes less effective cellular respiration.

The passage indicates that irisin is secreted into the blood before acting on adipose tissues, making it an endocrine, not exocrine, secretion of skeletal muscle.

The passage indicates that irisin is formed by cleaving the extracellular domain of FNDC5 in skeletal muscle before acting on adipose tissue, not the other way around.

This question asks us to identify which phenotypical difference is likely exhibited in mice that overexpress PGC-1α in their skeletal muscle, requiring us to refer to the passage and apply content knowledge.

Table 1 demonstrates that mice with skeletal muscle PGC-1α overexpression have elevated UCP1 mRNA levels in subcutaneous fat cells during rest, which will lead to more energy being dissipated as heat and less energy being stored as fat. Due to decreased fat storage, these mice are likely to have lower body weight.

Therefore, we can conclude that mice overexpressing PGC-1α in their skeletal muscles are likely to have lower body weight than wild-type mice.

According to the passage, mice that overexpress PGC-1α in skeletal muscle have elevated UCP1 expression in fat, leading to the energy dissipating as heat. Thus, these mice are likely to exhibit higher, not lower, body temperatures relative to wild-type mice.

According to the passage, higher physical activity leads to increased UCP1 expression. Although mice that overexpress PGC-1α in skeletal muscle have elevated UCP1 expression in fat, this change is due to PGC-1α overexpression, not higher physical activity. Physical activity and exercise are upstream of skeletal muscle PGC-1α expression, not downstream.

PGC-1α overexpression in skeletal muscle leads to increased FNDC5 expression, which is associated with lower, not higher, nonfasting glucose levels.

This question asks us to identify the process by which the viruses that encode FNDC5 or control protein transfer genetic information prompting us to think about methods of genetic transfer. This question does not require information from the passage to answer.

Transformation refers to the acquisition of genetic material from the environment by bacteria, while this question asks about viruses.

Transduction is the process by which foreign DNA (e.g., from viruses) is introduced into a cell. Transduction doesn’t require physical contact between the cell donating the DNA and the cell receiving it. Transduction is a common method of transferring genetic information between bacteria.

Therefore, we can conclude that the viruses that encode either FNDC5 or control protein transfer genetic material via transduction.

Contamination refers to impurities in a substance or solution. This does not apply to the viruses that encode FNDC5 or control protein.

Conjugation is the exchange of nucleic acids between bacteria across conjugation bridges, while this question asks about viruses.

This passage-based question asks us to determine the likely outcome of exposing OGG1-/- mice to TNFα, prompting us to employ our scientific reasoning and problem-solving skills.

The passage states that exposure to TNFα leads to the production of “reactive oxygen species (ROS), which can react with guanine bases in DNA to generate 8-oxo-G.” 8-oxo-G can participate in Hoogsteen base pairs, leading to DNA mutations. OGG1 is involved in the base excision repair of 8-oxo-G, preventing these potentially cancer-causing mutations. OGG1-/- mice would not have OGG1 available to excise 8-oxo-G, so these DNA mutations would not be prevented.

Therefore, we can conclude that exposure of OGG1-/- mice to TNFα would result in an increased incidence of cancer.

The passage states that OGG1 initiates transcription for Cxcl family members, so OGG1-/- mice would have decreased Cxcl2 expression.

The passage states that pro-inflammatory agents such as TNFα cause cells to generate ROS. Thus, we would expect ROS levels to increase, not decrease.

The passage states that TH5487 is a competitive inhibitor of OGG1. TH5487 will have no effect on knockout mice, which lack OGG1.

This question asks us to identify which type of inhibitor TH5487 is classified as, prompting us to consider different types of inhibition. This question requires a reference to the third paragraph.

In competitive inhibition, an inhibitor molecule is structurally similar enough to the substrate that it can bind to the active site, preventing the substrate from binding. It ‘competes’ with the substrate to bind with the enzyme.

The passage states that “TH5487 is a compound that mimics 8-oxo-G structure and binds to the active site of OGG1.” Because TH6487 is similar in structure to the substrate (8-oxo-G) and binds to the active site of OGG1, it is a competitive inhibitor.

Therefore, we can conclude that TH5487 acts as a competitive inhibitor.

Mixed inhibitors may bind to the enzyme or the enzyme-substrate complex at an allosteric site. The passage states that TH5487 binds to the active site of the enzyme, not an allosteric site.

Noncompetitive inhibitors bind to an allosteric site on the enzyme to induce a conformational change that reduces the enzyme’s affinity for its substrate. The passage states that TH5487 binds to the active site of the enzyme, not an allosteric site.

Uncompetitive inhibitors bind only to the enzyme-substrate complex. The passage states that TH5487 binds to the enzyme’s active site, preventing the substrate from binding.

This passage-based question asks us to determine the effects of OGG1 and TH5487 on Cxl1 expression, requiring us to carefully analyze Figure 2 while employing our data-based statistical reasoning skills.

Compared to OGG1 knockout mice, wild-type mice have considerably increased Cxcl1 expression in the presence of TNFα, suggesting that OGG1 stimulates, not reduces, the TNFα induced Cxcl1 expression.

Compared to OGG1 knockout mice, wild-type mice have considerably increased Cxcl1 expression in the presence of TNFα, suggesting that OGG1 stimulates, not reduces, the TNFα induced Cxcl1 expression.

To determine the effect of OGG1 on the TNFα induced Cxl1 expression, we should direct our attention to the third column of Figure 2, where TNFα is present, and TH5487 is absent in the presence of wild-type and OGG1 knockout mice. In wild-type mice, Cxl1 expression is considerably elevated, suggesting that OGG1 stimulates the TNFα induced Cxl1 expression.

Furthermore, examination of the fourth column reveals that when TH5487 is included, expression of Cxcl1 is considerably reduced in wild-type mice, suggesting that TH5487 reduces the OGG1-mediated Cxcl1 expression.

Therefore, we may conclude that OGG1 stimulates the TNFα induced Cxcl1 expression, and TH5487 reduces the OGG1-mediated Cxcl1 expression.

Compared to untreated wild-type cells, TH5487-treated wild-type cells exhibit reduced OGG1-mediated Cxcl1 expression, contradicting the notion that TH5487 stimulates the OGG1-mediated Cxcl1 expression.

This question asks us to identify the correct depiction of the results when researchers measure the NF- κB occupancy of the Cxcl2 promoter following exposure to TNFα both in the presence and absence of TH5487.

NF-κB recruitment increases following OGG1’s binding to 8-oxo-G. Thus, because TH5487 competitively inhibits OGG1 to prevent the binding of 8-oxo-G, we expect to see diminished NF-κB occupancy when TH5487 is added.

NF-κB recruitment increases following OGG1’s binding to 8-oxo-G. Thus, because TH5487 competitively inhibits OGG1 to prevent the binding of 8-oxo-G, we expect to see diminished NF-κB occupancy when TH5487 is added.

According to the passage, Increased ROS production in the presence of TNFα leads to the generation of 8-oxo-G, and “NF-κB recruitment increases dramatically following OGG1 binding to [8-oxo-G].” TH5487 is an inhibitor of OGG1 that prevents the OGG1/8-oxo-G binding event, so we would expect it to negate this increase in NF-κB recruitment, leading to decreased NF-κB occupancy of the Ckcl2 promoter region.

Therefore, we can conclude that option C is correct because it depicts diminished NF-κB occupancy when TH5487 is added.

OGG1 binding of 8-oxo-G within the promoter drives NF-κB recruitment to the promoter. This means that NF-κB occupancy of the promoter should increase when the cell is exposed to TNFα. Furthermore, because TH5487 competitively inhibits OGG1 to prevent the binding of 8-oxo-G, we expect to see diminished NF-κB occupancy when TH5487 is added.

This question prompts us to identify the Roman numeral that represents the site that integrates communication in the central nervous system. We will need to rely on our understanding of the central nervous system and reflex arcs to answer this question.

Roman numeral I represents the dorsal root ganglion (collection of cell bodies of the afferent sensory fibers). This plays a role in bringing sensory information from the peripheral nervous system to the central nervous system but does not play a role in integrating this information into a response.

A reflex arc is the physiological basis of a reflex (an involuntary response to a stimulus). A reflex arc is composed of a receptor, afferent nerve fibers, interneurons, efferent motor neurons, and an effector organ. Interneurons are neurons exclusive to the CNS (in the brain and spinal cord) that are responsible for integrating signal reception to motor action in a reflex arc.

Therefore, as Roman numeral II is located in the spinal cord, we can conclude that this likely represents an interneuron that integrates signals in a pain reflex arc.

Roman numeral III represents the ventral root of the efferent motor neuron. This plays a role in sending information from the central nervous system to an effector organ to carry out a reflex action.

Roman numeral IV represents an effector organ, the organ responsible for actually carrying out the reflex action.

This question prompts us to identify the cause of the resting potential of a neuron. This requires us to rely on our content knowledge regarding membrane potentials.

A neurotransmitter is a signaling molecule secreted by neurons to communicate with other neurons. Neurotransmitters are released in order to transmit signals and do not contribute to a neuron’s resting membrane potential (membrane potential when the neuron is at rest and not excited).

The membrane potential describes the difference in electrical potential inside and outside of the cell. The resting membrane potential of a neuron describes the membrane potential when the neuron is not stimulated. The resting potential is around -70 mV, indicating that the inside of the cell is more negatively charged than the outside of the cell. This is primarily determined by the sodium and potassium gradients of neurons. As neuronal plasma membranes contain potassium leak channels, these positively charged ions exit the cell and contribute to a relatively negative internal environment.

Therefore, we can conclude that the resting potential of a neuron is primarily a result of the distribution of ions across the plasma membrane.

As neurotransmitters are signaling molecules, they cause ion channels of the postsynaptic cells to open and close. This change in ion channels moves the cell away from its resting potential.

The active transport of Ca2+ ions would result in the movement of positively charged ions outside of the cell. This is associated with signal transmission and does not contribute to the cell’s resting potential.

This question prompts us to identify a consequence of diaphragm contraction. We must rely on our content knowledge regarding respiration to answer this question.

The diaphragm is a dome-shaped muscle located between the thoracic and abdominal cavities that are necessary for respiration. As the diaphragm contracts, it flattens in shape, resulting in increased space in the thoracic cavity. As space increases, intrathoracic pressure (the pressure inside the thoracic cavity) decreases and creates a vacuum in the lungs and pulls air in (inhalation). The opposite occurs during exhalation when the diaphragm relaxes, reduces intrathoracic volume, increases intrathoracic pressure, and pushes air out of the lungs (exhalation).

Therefore, we can conclude that contraction of the diaphragm results in decreased intrathoracic pressure and inhalation.

Though diaphragm contraction results in decreased intrathoracic pressure, this will lead to inhalation, not exhalation, as air will be pulled into the lungs.

Diaphragm relaxation, not contraction, causes an increase in intrathoracic pressure. This answer option also incorrectly pairs increased intrathoracic pressure with inhalation rather than exhalation.

Diaphragm contraction results in decreased intrathoracic pressure as volume increases upon contraction.

This question prompts us to identify a kinetic parameter that will increase with the addition of more substrate, given that the initial substrate concentration is relatively low. We will need to rely on our content knowledge surrounding enzyme kinetics to answer this question.

KM, the rate constant of a reaction, describes the substrate concentration in which half of the active sites available are occupied. Km is dependent on the affinity of the enzyme for a substrate and is not dependent on substrate concentration.

kcat, also known as the reaction turnover number, dictates the amount of product produced per unit of time by one enzyme. kcat is a quality of the enzyme and is not dependent on substrate concentration.

Vmax describes the maximum velocity of an enzyme-catalyzed reaction when the solution is fully saturated with substrate. Though adding additional substrate will increase the initial velocity, the maximum velocity of the reaction is constant and will not change.

In enzyme kinetics, V0 (initial velocity) describes the initial rate of product formation. If substrate concentration is relatively low, the enzymes will have a lower probability of reaching substrates and, therefore, will have a lower rate of product formation. If additional substrate is added to the solution, substrate concentration will increase, and the probability of each enzyme reaching and interacting with a substrate will increase. This will lead to an increased rate of product formation and, therefore, an increased V0.

Therefore, we can conclude that the initial velocity will increase as additional substrate is added to the reaction.

This passage-based question asks us to identify which action is not characteristic of the mutant RB proteins, prompting us to employ our scientific reasoning and problem-solving skills while carefully analyzing Figure 1.

In the experiment, researchers sought to examine whether both E7 and E2F bind to RB’s pocket domain. As depicted in Figure 1, wild-type (WT) and pocket domain mutant RB were expressed in an RB-deficient cell line. Then, the change in the percentage of cells in G1 was measured after the cells were transfected with RB or with RB and E7. Transfection is the process of introducing genetic material into eukaryotic cells.

Figure 1 shows that in each pocket domain mutant RB cell line, the presence of E7 does not affect the change in the percentage of cells in G1. This indicates that the mutants are insensitive to E7 and do not likely interact with it. Furthermore, it suggests that E7 does bind to the pocket domain because RB in the mutant cells is identical to RB in wild-type cells, except for the mutant pocket domain.

Therefore, we may conclude that the experiments suggest that the mutant RB proteins do not bind to E7.

As stated in the first paragraph of the passage, E2F is a transcription factor “needed for further progression through the cell cycle.” In addition, the passage further explains that RB proteins decrease human cell division by arresting cells in the G1 phase of the cell cycle. The last sentence states, “… progression through the cell cycle is accomplished when cyclin-dependent kinases (CDKs) phosphorylate RB, preventing its binding to E2F.” Consequently, we can assume that RB decreases cell division by directly binding E2F.

Figure 2 demonstrates that cells containing RB mutants can be arrested in G1, so they must be able to bind E2F.

Figures 1 and 2 demonstrate mutant RB proteins are still able to arrest cells in G1 phase of the cell cycle.

As indicated in Figure 2, in the presence of CDK4, which phosphorylates RB, the number of cells in G1 is reduced. Thus, mutant RB proteins permit exit from G1 when phosphorylated.

This question asks us to determine what will happen if the RB mutants cannot bind to E2F, requiring us to refer to the first paragraph and apply content knowledge.

The passage states that E2F is a transcription factor required for progression through the cell cycle. Normally, RB arrests cells in the G1 phase by binding and inactivating E2F. When RB is prevented from binding E2F by CDK phosphorylation, cells progress through the cell cycle instead of being arrested in the G1 phase.

Therefore, we can conclude that if the RB mutants cannot bind to E2F, then cell division arrest will not occur.

There is no evidence to suggest that E7 and E2F bind RB by the same mechanism. In fact, Figure 2 demonstrates that cells containing the RB mutant can exit G1 if CDK4 is administered, suggesting that E2F binds at a site other than the pocket domain, where E7 binds.

Phosphorylation of RB by CDK4 controls whether RB can bind to E2F. There is no evidence to suggest that E2F binding influences whether the RB mutants can be phosphorylated by CDK4.

There is no information in the passage that suggests E2F directly binds E7. E2F and E7 work independently in regulating RB function.

This question asks us to identify which conclusion based on Figure 2 is correct, prompting us to employ our scientific reasoning and problem-solving skills. This question requires a careful understanding of the passage to answer.

The percentage of WT cells arrested in G1 was similar to the percentage of mutant RB cells arrested in G1. This was observed both in the presence and absence of CDK4.

This experiment did not attempt to examine how CDK4 transfection affects the phosphorylation of mutant RB. Instead, it measures the percentage of cells in G1.

Figure 2 indicates that fewer cells are G1-arrested when transfected with CDK4.

As depicted in Figure 2, wild-type (WT) and pocket domain mutant, RB was expressed in an RB-deficient cell line. Then, the change in the percentage of cells in G1 was measured after the cells were transfected with RB or with RB and CDK4. Transfection is the process of introducing genetic material into eukaryotic cells.

When cells were transfected with RB and CDK4, fewer cells were in the G1 phase in every experimental group compared to transfection with RB alone. This observation was consistent among all of the RB proteins assessed.

Therefore, we can conclude that Figure 2 supports the conclusion that CDK4 transfection results in fewer cells being G1 arrested.

This question asks us to identify which characteristic is true of highly conserved parts of the genome, prompting us to think about natural selection. This pseudo-discrete question does not require specific information from the passage to answer.

A highly conserved sequence of the genome is a section that remains identical or similar across species, indicating that it has been maintained by natural selection. Natural selection is the process by which beneficial phenotypes are passed on via reproduction while harmful phenotypes are not. For example, a camouflage-patterned butterfly is less likely to be hunted than a brightly colored one. The camouflage-patterned butterflies are thus more likely to live and reproduce, and the genes encoding for the pattern will be passed on.

If the LXCXE motif is “highly conserved,” as described in the passage, it must be important for survival. Mutations to the LXCXE motif would likely lead to a decreased fitness of the organism, preventing it from reproducing.

Therefore, we can conclude that certain parts of the genome are highly conserved because they are vital to an organism’s survival.

Mutations can affect any portion of DNA; there are no areas that are chemically incapable of mutation.

DNA is not stored in vesicles. DNA is stored in the nucleus.

DNA is stored in the nucleus and is unlikely to be secreted. Genomic conservation refers to maintaining sequence similarity across species, not conserving DNA within the cell.

This passage-based question asks us to identify the genomic location of the mutation that causes the LP phenotype, prompting us to refer to the final paragraph while applying content knowledge on gene expression.

Introns are regulatory segments of transcribed RNA that do not encode functional proteins. Introns are removed from the pre-mRNA transcript during a process called splicing. The LP phenotype is more likely caused by a mutation to the enhancer region, not an intron.

A mutation affecting the coding sequence would result in a mutated form of lactase, which is not observed in LP.

The passage states that adults exhibiting lactase persistence (LP) “continue to produce lactase into adulthood,” permitting them to consume dairy without gastrointestinal distress. Furthermore, “sets of alleles located closely together on the same chromosome (haplotypes) associated with the LP phenotype contain several single nucleotide polymorphisms (SNPs).”

Enhancers are short DNA regions that increase the rate of transcription of a particular gene by binding transcriptional activators. Because LP is caused by the continuous synthesis of lactase, it is likely that the mutation targets the enhancer region. Specifically, the mutation could make it easier for transcriptional activators to bind the enhancer sequence, leading to persistent lactase synthesis into adulthood.

A mutation affecting the stop codon would result in a mutated form of lactase, which is not observed in LP.

This question asks us to identify how the lactase enzyme is classified, prompting us to think about the classification of enzymes. Although the passage provides information that can help to answer this question, it can also be answered using discrete content knowledge.

An isomerase catalyzes the rearrangement of bonds within a molecule to produce an isotope. They do not break down chemical bonds as lactase does.

The passage states that lactase “converts lactose into glucose and galactose.” Lactose is a disaccharide consisting of one glucose monomer and one galactose monomer. Thus, lactase cleaves the glycosidic linkage to produce glucose and galactose. Glycosidic linkages in disaccharides are cleaved via a hydrolysis reaction. A hydrolase is an enzyme that breaks a covalent bond using water during a process called hydrolysis.

Therefore, we may conclude that lactase can be classified as a hydrolase.

A transferase catalyzes the transfer of a functional group, like phosphate, from one molecule to another. They do not break down chemical bonds as lactase does.

An oxidoreductase catalyzes the transfer of electrons during a redox reaction. They do not break down chemical bonds as lactase does.

This question asks us to identify which cells express lactase, prompting us to recall and apply content knowledge on carbohydrate digestion. This pseudo-discrete question does not require details from the passage to answer.

After food is ingested through the mouth and partially digested in the stomach, it passes through the small intestine. The small intestine is a long, tube-like organ with a highly folded surface containing finger-like projections called villi, which increase the surface area of the walls. The duodenum is the first section of the small intestine and the site of the majority of chemical digestion. Although most digestive enzymes are secreted from the pancreas, cells of the small intestine also secrete digestive enzymes.

Enterocytes are absorptive epithelial cells found in the villi of the small intestine. They are the cells responsible for the secretion of digestive enzymes, including hydrolases that break down disaccharides, like lactase.

Therefore, we may conclude that lactase is expressed by enterocytes of the duodenal villi.

Cells of the colon do not produce digestive enzymes, as chemical digestion occurs by the time ingested food reaches the large intestine. However, bacteria of the colon may break down unusual sugars, like oligosaccharides.

Parietal cells of the stomach lining produce hydrochloric acid (HCl), not digestive enzymes.

Bile is a substance that aids in the digestion of lipids, not carbohydrates. Furthermore, hepatocytes do not produce digestive enzymes.

This question asks us to determine which type of mutation produced the single nucleotide polymorphism referenced in the question stem, prompting us to consider the classification of nucleotides. Answering this question requires an understanding of the passage and question stem.

The genomic radiograph indicates an SNP containing both cytosine and thymine, both pyrimidines. Thus, the SNP resulted from a mutation from pyrimidine to pyrimidine.

The genomic radiograph indicates an SNP containing both cytosine and thymine, both pyrimidines. Thus, the SNP resulted from a mutation from pyrimidine to pyrimidine.

The genomic radiograph indicates an SNP containing both cytosine and thymine, both pyrimidines. Thus, the SNP resulted from a mutation from pyrimidine to pyrimidine.

Nucleotides are organic molecules composed of a nitrogenous base, a pentose sugar, and one or more phosphate groups. The nucleotides that comprise DNA are grouped by their nitrogenous base: purines (adenine and guanine), which consist of two fused rings, and pyrimidines (cytosine, thymine), which include a single 6-member ring.

A single nucleotide polymorphism (SNP) refers to a substitution of a single nucleotide at a specific location in the genome of a population. To examine this nucleotide substitution, genomic DNA and cDNA containing the SNP were sequenced and presented on a radiograph.

The radiograph in the question stem indicates that in the genomic DNA, one position is occupied by both cytosine and thymine, indicating the presence of an SNP. Thus, the SNP resulted from a mutation between pyrimidines.

Therefore, we may conclude that the SNP resulted from a mutation from pyrimidine to pyrimidine.

This question prompts us to identify potential explanations for why adding a ligand does not result in a current between two ionic solutions. We must rely on our content knowledge surrounding ligand-gated channels and concentration gradients to answer this question.

Though explanation I could explain why no current was generated, option III is also a valid explanation.

If potassium ions were only present in one solution, the addition of the ligand would cause the ligand-gated potassium channel to open and consequently generate a current as potassium ions would flow from one solution to the other.

A ligand-gated channel is a type of protein that allows specific molecules to move across a membrane in response to binding with a specific ligand. In the question, the ligand-gated potassium channel would be expected to open and allow the free movement of potassium ions upon the addition of the ligand. Upon opening, potassium ions would move down their concentration gradient (from areas of high potassium concentration to areas of low potassium concentration) until the concentration of potassium ions is the same on both sides of the membrane. This movement of potassium ions would result in a current as charged ions would be moving in one direction across the membrane.

If neither solution contained potassium, there would be no potassium ions crossing the membrane despite a ligand-gated potassium channel opening. Similarly, if there were equal concentrations of potassium in the solutions, potassium ions could move freely across the membrane, but there would be no net movement of ions as the concentration of potassium ions was already equal.

Therefore, we can conclude that explanations I and III could explain why no current was generated upon the ligand’s addition.

Similar to answer option B, this answer option mistakenly includes explanation II. Refer to option B for a more in-depth explanation.

This question prompts us to identify a molecule not formed during glycolysis. We will need to rely on our content knowledge surrounding this metabolic pathway to answer this question.

Phosphohexo isomerase, also known as glucose-6-phosphate isomerase, converts glucose 6-phosphate to fructose 6-phosphate in the second step of glycolysis.

Glycolysis is a metabolic pathway in which glucose molecules are broken down to create pyruvate and release free energy. The steps of glycolysis are detailed in the image below:

Oxaloacetate is a metabolite that plays an important role in the Krebs cycle and gluconeogenesis. Though pyruvate, the final product of glycolysis, can be converted to oxaloacetate in order to enter the gluconeogenesis pathway, oxaloacetate itself does not play a role in glycolysis.

Therefore, we can conclude that oxaloacetate is a molecule that is not formed in the steps of glycolysis.

Phosphoenolpyruvate is converted to pyruvate by pyruvate kinase in the final step of glycolysis. We could have also eliminated this answer option by remembering that the purpose of glycolysis is to produce pyruvate and release energy.

3-Phosphoglycerate is converted to 2-phosphoglycerate by phosphoglycerate mutase in the eighth step of glycolysis.

This question prompts us to identify the number of different types of gametes the organism described in the question can produce. We will need to rely on our understanding of gamete formation to answer this question.

The organism in question is described as carrying 3 genes. However, as a gamete is a haploid cell, and the organism is heterozygous for each gene, two alleles per gene could potentially be found in the gamete. This would result in 8 gamete options, not 3.

This answer option likely results from a miscalculation. Refer to option D for an explanation of how to correctly calculate the number of gametes.

This answer option likely results from adding the two potential alleles of each gene instead of multiplying them. The numbers must be multiplied in order to find the total number of combinations.

A gamete is a reproductive haploid (n) cell that carries one copy of each gene. We are told that the diploid organism in question is heterozygous in the three unlinked genes. This means that the organism carries both a dominant and a recessive allele for each gene and that these genes are assorted independently without influencing one another. From this information, we can conclude that each allele has an equal likelihood of being represented by the gamete. As there exist two options for each gene found in the organism, we can calculate the total number of potential gametes by multiplying 2 x 2 x 2, yielding 8 potential gametes in total.

Therefore, we can conclude that this organism can produce 8 different types of gametes.