JW Exam 1 Flashcards

C. 3 × 10⁻²⁰ J ✅

IR Radiation Energy Calculation (Step-by-Step)

Step 1: Understand the setup

• The passage says IR spectroscopy was used to monitor acetamide as substrate and acetic acid as product.
• It also states:
  “The characteristic absorbance band of the carboxylate group in the reaction product occurred at 1560 cm⁻¹.”

We interpret this to mean:
Wavenumber = 1560 cm⁻¹

Step 2: Convert wavenumber to wavelength

Use the formula:
λ = (1 / wavenumber) × 10⁻²

λ = (1 / 1560) × 10⁻² = 6.41 × 10⁻⁶ m

Step 3: Use mental math to estimate

Approximate 1560 cm⁻¹ as 1500 cm⁻¹:

• 100 / 150000 = 6.6 × 10⁻⁶ m
  So, λ ≈ 6.6 × 10⁻⁶ m

Step 4: Plug into the energy equation

E = (h × c) / λ

Substitute the known values:

• h = 6.6 × 10⁻³⁴ J·s
• c = 3 × 10⁸ m/s
• λ = 6.6 × 10⁻⁶ m

E = (6.6 × 10⁻³⁴) × (3 × 10⁸) / (6.6 × 10⁻⁶)
E ≈ 3.0 × 10⁻²⁰ J

D. 50 mM ✅

This question asks about the amount of acetamide hydrolyzed by 45 µM amidase after 30 minutes. We must refer to the passage and Table 1 to answer this question.

In the third paragraph, the passage states,
“Monitoring the change in the absorbance band intensity of amide I band at 1635 cm⁻¹ against time (min) yielded a reaction progress curve with the equation y = -0.004x + 0.22 for amidase at 45 µM enzyme concentration.”
Therefore, we can infer that x is the time in minutes while y is the band intensity. We can replace x with 30 minutes in the equation and solve it, where the equation will be:

y = -0.004(30) + 0.22 = 0.1

Then, we can use the band intensity of 0.1 and return to Table 1 to find the corresponding acetamide concentration, which would be 50 mM.

C. ✅

This question asks which compounds could be the monosaccharide based on the amount of CO₂ and H₂O produced during their metabolism. To answer this question, we must recall our background knowledge of balanced chemical equations.

The complete metabolism of a monosaccharide involves its breakdown into carbon dioxide (CO₂) and water (H₂O) with oxygen (O₂). For each carbon atom in the monosaccharide, the reaction produces an equal mole of carbon dioxide in aerobic respiration.

We are given that 0.1 mol of a monosaccharide is metabolized, producing 0.4 mol CO₂ and 0.4 mol H₂O.

Since 0.1 mol of the monosaccharide produces 0.4 mol CO₂, each monosaccharide molecule (CₓHᵧO

D. 4 ✅

This question asks for the pressure ratio (P_A / P_B) of two ideal gases, A and B, given that A has double the density and half the molar weight of B at the same temperature. To answer this question, we must recall our background knowledge of the ideal gas law.

The ideal gas law relates pressure (P), volume (V), number of moles (n), and temperature (T) for an ideal gas using the equation PV = nRT.

We can't directly solve for pressure using the ideal gas law with the given information. However, we can utilize the relationship between density (d) and number of moles (n) for a fixed volume (V):

PV = nRT
Rewriting: P = (n/V)RT

Since n = m/M, where m is the mass and M is the molar mass, we can substitute:

P = (m/MV)RT

We know that density d = m/V, so:

P = (d/M)RT → PM = dRT

We can now rewrite the ideal gas equation for each gas:

For gas A: P_A * M_A = d_A * R * T
For gas B: P_B * M_B = d_B * R * T

From the question, we are told that:

• d_A = 2 * d_B
• M_A = 1/2 * M_B

Substituting into the equation for gas A:

P_A * (1/2) * M_B = 2 * d_B * R * T

Divide both equations:

(P_A * (1/2) * M_B) / (P_B * M_B) = (2 * d_B * R * T) / (d_B * R * T)

Simplify:

P_A / P_B = 4

Therefore, the ratio of gas A's pressure to gas B's is 4.

A. They are stronger in CO₂ because they include carbon and oxygen. ✅

Due to its geometry, CO₂ exhibits stronger London dispersion forces than O₂ due to differences in molecular size and polarizability. CO₂ is a larger molecule than O₂, possessing more electrons. This increase in size and electron count makes CO₂ more polarizable, meaning its electron cloud can be more easily distorted. The greater polarizability results in stronger temporary dipoles and, consequently, stronger London dispersion forces. Thus, the combination of increased size, polarizability, and molecular structure accounts for the stronger London dispersion forces in CO₂.

B. The pressure of the gas produced in Reaction 2 will be larger. ✅

In reaction 1, MCO₃ (s) decomposes to MO (s) and CO₂ (g). CO₂ is relatively soluble in water. As the gas is produced, some will dissolve in the water, reducing the pressure of the free CO₂ gas in the tube. Meanwhile, in reaction 2, MxOy (s) decomposes to MxOy-z/2 (s) and z/2O₂ (g). O₂ is sparingly soluble in water. Most O₂ produced will remain in the gas phase, exerting a higher pressure than the CO₂ in Reaction 1.

✅ D. M1 and M2 must belong to different groups.

This question asks about the information that can be inferred from reactions 1 and 2 about the metals used. To answer this question, we must recall our background knowledge of the periodic table groups.

Group 1 elements are referred to as alkali metals because, when subjected to water, their oxides and hydroxides produce alkaline solutions.

Groups 3 through 12 elements are referred to as transition metals because they can form cations with varying charges.

Group 2 elements are known as alkaline earth metals because the oxides and hydroxides of these elements are naturally alkaline and can be found in the earth's crust.

Alkali metals (Group 1) are highly reactive due to their low ionization energies. They readily form ionic oxides (MO) that dissolve easily in water, generating hydroxides (MOH) and releasing hydrogen gas. Notably, these metals also readily form stable ionic carbonates (MCO₃).

Alkali earth metals (Group 2) are less reactive than Group 1 but still exhibit reactivity. Their ionic oxides (MO) are less soluble in water compared to Group 1 oxides and react slowly with water to form hydroxides. Similar to Group 1, they form stable ionic carbonates (MCO₃).

Transition metals encompass a diverse range of elements with varying reactivity depending on the specific metal. They can exhibit multiple oxidation states and form a variety of oxides (MxOy) with different properties. Some oxides are stable, while others decompose at high temperatures, releasing oxygen. Unlike Groups 1 and 2, transition metals generally do not form stable carbonates.

Group 14 (Carbon Group) also showcases varied reactivity within its elements. Carbon, for example, is relatively unreactive, while others like silicon are more so. They can form covalent oxides (like CO₂ for carbon) or oxides with a mix of ionic and covalent character (like SiO₂ for silicon). The stability of these oxides varies. Notably, carbon readily forms a stable covalent carbonate (CO₃²⁻), but other elements in the group typically do not.

Reaction 1 depicts the decomposition of a metal carbonate (MCO₃) into a metal oxide (MO) and carbon dioxide (CO₂). This type of reaction is characteristic of metals that form stable carbonates, which are typically Group 1 (alkali metals) and Group 2 (alkaline earth metals).

Reaction 2 shows the dissociation of a metal oxide (MxOy) into a lower oxide (MxOy-z/2) and oxygen gas (O₂). This behavior is not as specific to certain groups.

Analyzing the reactions can help us understand which group M1 and M2 might belong to on the periodic table.

Since Reaction 1 produces CO₂ from decomposing a metal carbonate (MCO₃), M1 is likely a good fit for Group 1 (alkali metals) or possibly Group 2 (alkaline earth metals). These groups readily form stable carbonates. However, M2 is unlikely to be a Group 1 element because Group 1 metals typically don't form higher oxides (MxOy) that decompose at room temperature, as seen in Reaction 2.

While M1 forming a carbonate suggests Group 1 or 2, it's not a perfect fit for Group 2 either. Dissociation of a metal oxide (Reaction 2) isn't exclusive to Group 2. Additionally, the presence of CO₂ from M1 further reduces the chance of M2 also belonging to Group 2.

Transition metals are a possibility for M2. Many can form oxides, and some decompose at higher temperatures, releasing oxygen, similar to Reaction 2. However, transition metals generally don't form stable carbonates like M1 seems to in Reaction 1.

Group 14 (carbon group) is a possibility for M1 or M2 with limitations. While some Group 14 elements form carbonates that decompose at high temperatures, the CO₂ production from M1 suggests it might be different from a Group 14 element (like carbon) that readily forms CO₂.

In conclusion, the information points towards M1 likely belonging to Group 1 (alkali metals) or Group 2 (alkaline earth metals) due to CO₂ production. M2, on the other hand, could be a transition metal or a Group 14 element, but it's less likely to be in the same group as M1 (not Group 1 or 2).

D. The experiment works because the buffering capacity of MOPS and Tris/HCl coincide. ✅

This question is asking about the expected result if the experiment is repeated with the MOPS buffer. To answer this question, we must refer to the passage and recall our background knowledge of buffers.

Enzymes often have a specific pH range where they function optimally. Fluctuations in pH can significantly affect enzyme activity. Buffers maintain a stable pH environment to optimize enzyme activity. The buffering capacity of a buffer solution refers to its ability to resist changes in pH upon the addition of acids or bases.

The third paragraph of the passage states, “Each APC variant (2 nm) was incubated with multiple concentrations of S-2366 in 20 mm 1 ml Tris/HCl (pKa = 8.1).” This means that the original experiment used a Tris/HCl buffer with a pKa of 8.1. This buffer effectively resists pH changes around this pKa value.

The proposed experiment uses a MOPS buffer with a pKa of 7.15, which is slightly lower than Tris/HCl.

Both MOPS and Tris/HCl are commonly used biological buffers. While their pKa values differ slightly, they both have sufficient buffering capacity around physiological pH (pH 7.4). As long as the MOPS buffer concentration is chosen appropriately to maintain the desired pH range (around 7.4), it should be able to buffer the solution similar to Tris/HCl adequately. The enzyme activity likely wouldn’t be significantly affected by this small change in pKa.

Therefore, the proposed experiment would work because the buffering capacity of MOPS and Tris/HCl coincide.

A. Increasing the distance “Ha” by 1 cm ✅

This question asks which action in a Heron’s fountain setup will increase the pressure at the bottom container. To answer this question, we must recall our background knowledge of hydrostatic pressure.

Hydrostatic pressure refers to the pressure exerted by a fluid at rest due to gravity. In Heron’s fountain, the pressure in container 3 is caused by the weight of the water column above it in container 2. The fluids in containers 1 and 3 are regarded as one water column as they are interconnected.

Ha represents the height difference between the water surfaces in containers 2 and 3. A larger Ha signifies a taller water column in container 2.

According to the principle of hydrostatic pressure, P = ρgd, where ρ is the density of the fluid, g is a constant, and d is the depth below the surface of the fluid. A taller water column exerts a greater force due to gravity, leading to higher pressure at the bottom.

Because of this, raising container 1 above the other container will raise the fluid’s surface height and consequently increase d in ρgd, which will raise the hydrostatic pressure that the fluid in container 3 experiences. This distance and hydrostatic pressure will grow with an increase in the terms Ha or Hw.

C. 2 ✅

Originally, both waves are in phase:

• Wave 1: amplitude = 2
• Wave 2: amplitude = 1
• Since they’re in phase, the amplitudes add:
  Resultant amplitude = 2 + 1 = 3

If the second wave is phase-shifted by π (i.e., 180°), it becomes perfectly out of phase with the first wave:

• Instead of adding, it now subtracts:
  Resultant amplitude = 2 − 1 = 1

Change in amplitude = 3 − 1 = 2

So the amplitude of the resultant wave decreases by 2 units.

B. Glycine ✅

A. zero, because excess charge within the conductor moves to the external surface. ✅

At equilibrium, the electric field is generally zero at any point within a conducting material. Resistive materials (generally non-metals) contain electrons that are mostly fixed in place, whereas conductive materials contain both fixed and free electrons. In a conductive material, the core electrons of atoms are typically fixed in place, while the outer electrons are typically free to move.

In conductive materials (e.g., metals such as copper), free electrons react directly to the forces exerted by external electric fields. In an initially uncharged solid, free electrons will be distributed evenly around the surface of the solid because they are repelled from one another, according to Coulomb’s law. Suppose an electric field is applied to a solid after establishing this static equilibrium. In that case, the free electrons will rearrange themselves on the conductor surface in the orientation needed to produce an electric field of zero within the solid.

(Choice B) Conductive materials contain both fixed and free electrons.

(Choice C) While excess charge accumulates on the surface, the electric field inside the conductor itself becomes zero.

(Choice D) Similar to B, conductive materials contain both fixed and free electrons.

B. 9.25 ✅

The buffer solution consists of ammonia (NH₃) and its conjugate acid, ammonium chloride (NH₄Cl).
The pKa of NH₄⁺ is 9.25, which means the buffer solution can resist changes in pH when small amounts of acid or base are added. This means that the environment contains:

NH₄⁺ = 0.35 M × 80 mL = 28 mmol
NH₃ = 0.30 M × 80 mL = 24 mmol

When NaOH is added, it reacts with the ammonium ions (NH₄⁺) in the buffer solution according to the following reaction:
NH₄⁺ (aq) + OH⁻ (aq) ⇌ NH₃ (aq) + H₂O (l)

This reaction consumes some of the OH⁻ ions added by the NaOH solution and produces NH₃, a weak base. The net effect is a smaller increase in pH compared to what would be observed in pure water.

This means that the new environment contains:
NH₄⁺ = 28 mmol – (0.10 M × 20 mL) = 26 mmol
NH₃ = 24 mmol + (0.10 M × 20 mL) = 26 mmol

We can then calculate the pH using the equation pH = pKa + log([NH₃]/[NH₄⁺])
pH = 9.25 + log(26/26) = 9.25

B. E(M) > E(L) > E(K) ✅

Electrical energy consumption depends on an appliance’s power rating and how long it is operated for.
The formula for calculating energy consumption is:

Energy Consumption (Wh) = Power (W) × Time (h)

We need to consider both power and runtime for each appliance to determine which consumes the most energy.

To calculate the estimated daily energy consumption (Wh) for each appliance:

• K: 1500 W × 0.2 h = 300 Wh
• L: 80 W × 7 h = 560 Wh
• M: 100 W × 6 h = 600 Wh

Since the question asks about the average electrical energy consumption over a month (assuming 30 days), we can multiply the daily consumption by 30:

• M: 600 Wh/day × 30 days = 18,000 Wh
• L: 560 Wh/day × 30 days = 16,800 Wh
• K: 300 Wh/day × 30 days = 9,000 Wh

Therefore, the order of energy consumption from highest to lowest is M > L > K, making answer choice B the correct answer.

B. Reduced number of action potentials ✅

B. HCl (g) ✅

This question asks which compound is most sensitive to increased pressure when dissolved in water. We must use our knowledge of Henry's Law to answer this question.

Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid at a constant temperature. In simpler terms, as the pressure of a gas above a liquid increases, more of that gas will dissolve in the liquid.

HCl, or hydrochloric acid, is a strong acid. This means when it dissolves in water, it readily breaks apart (dissociates) nearly entirely into individual ions: hydrogen (H⁺) and chloride (Cl⁻). Water is a polar molecule with a positive and a negative end. Because HCl separates into charged ions, these ions can interact strongly with the oppositely charged ends of water molecules through electrostatic forces (ionic bonds). These favorable interactions allow a large amount of HCl to dissolve in water.

Henry's Law applies to undissociated gases dissolving in liquids, stating that the amount of undissociated gas that dissolves in a liquid is proportional to the pressure of that gas above the liquid. After HCl dissociation in water, the ions are no longer acting as a gas, but rather as charged particles interacting with the water molecules.

Therefore, pressure changes primarily affect the dissociation equilibrium, leading to a perceived increase in solubility.

(Choice A) Carbon dioxide is a weak acid. It reacts with water to a small extent, forming carbonic acid (H₂CO₃) and bicarbonate (HCO₃⁻) ions. However, a significant portion of CO₂ remains as undissociated CO₂ molecules in the solution. These undissociated molecules are not as attracted to water as the separated ions in HCl. They can only form weaker hydrogen bonds with water molecules.

(Choice C) Methanol is liquid at room temperature. The solubility of liquids is independent of pressure. Moreover, methanol can only form hydrogen bonds with water molecules. These hydrogen bonds, although present, are weaker than the ionic attractions between H⁺ and Cl⁻ ions in HCl. Therefore, HCl would be more soluble in water.

(Choice D) Sodium chloride is an ionic solid, and its solubility in water is primarily driven by the ionic interactions and not significantly affected by pressure changes. NaCl is an ionic compound. In water, it separates into Na⁺ and Cl⁻ ions. However, these ions interact with water molecules through weaker ion-dipole attractions compared to the strong ionic bonds between H⁺ and Cl⁻ in dissociated HCl.

C. single-stranded DNA. ✅

In double-stranded DNA, the percentages of adenine and thymine would be equal (A = T) because they form a complementary base pair. According to the given composition, 18% of the bases are adenine, and 31% are thymine. This indicates that the microorganism does not follow the normal base pairing rules of DNA. Therefore, we can infer that this microorganism is single-stranded.

D. Four ✅

An ionizable group is a functional group in a molecule that can gain or lose a proton (H⁺) depending on the surrounding pH. The resulting electrically charged atom or molecule is called an ion. To determine the number of ionizable groups in an isolated phosphotyrosine molecule, we need to understand the chemical properties of the phosphotyrosine residue.

A phosphotyrosine molecule contains a phosphate group (PO₄³⁻) attached to a tyrosine amino acid molecule. The phosphate group carries a negative charge at physiological pH and can ionize to release a proton (H⁺). In the case of a phosphotyrosine residue, there are four potential ionizable groups: two OH groups from the phosphate group, an NH₂ group, and a carboxyl group of tyrosine.

A. catalytic receptor ✅

In the second paragraph, the passage states, "TGFβ binding to its receptor causes JAK2 to become phosphorylated via the transfer of a phosphate group from ATP." This indicates that the TGFβ receptor has catalytic activity, as it initiates a phosphorylation reaction upon ligand binding.

Upon cytokine binding to the receptor, the molecules of JAK2 are associated with the receptor, which phosphorylates it. STAT3 binds to the phosphorylation sites and is, in turn, phosphorylated at Tyr-705 by JAK2. Two phosphorylated STAT3 molecules form a dimer that translocates to the nucleus. In the nucleus, the STAT3 dimer identifies and engages with DNA and initiates transcription of its target genes in conjunction with the transcription machinery.

Therefore, we can infer that the TGFβ receptor is a tyrosine kinase receptor and, as a consequence, possesses catalytic activity, also known as an enzyme-linked receptor.

(Choice B) G-protein coupled receptors (GPCRs) are a different class of receptors that typically activate intracellular signaling pathways through interaction with G proteins.

(Choice C) Ligand-gated ion channels are receptors that, upon ligand binding, allow the passage of ions across the cell membrane.

(Choice D) Voltage-gated ion channels are activated by changes in membrane potential.

D. n.t. siRNA treated with TGFβ vs. SOCS3 siRNA treated with TGFβ ✅

SOCS3 knockdown means to decrease the expression or activity of the SOCS3 protein in a biological system. SOCS3 is a negative regulator of cytokine signaling pathways. It achieves this by binding to specific signaling proteins and preventing their activation.

The passage states:

“In vitro, TGFβ incubation is sufficient to induce an SSc-like phenotype in healthy fibroblasts.”

This statement establishes the rationale for using TGFβ treatment in the experimental design. It suggests that TGFβ is important for studying the effects of SOCS3 knockdown on fibroblasts in the context of SSc.

siRNA stands for small interfering RNA. It is a class of RNA molecules that plays a significant role in gene regulation and silencing specific genes. By using siRNA, researchers can target, silence, or knock down the expression of specific genes in vitro (in cell culture).

To compare the effects of SOCS3 knockdown specifically in the presence of TGFβ (the relevant condition for studying SSc), researchers must compare:

• n.t. siRNA treated with TGFβ vs. SOCS3 siRNA treated with TGFβ

This option allows for a direct comparison between the two groups, effectively demonstrating the effects of SOCS3 knockdown, making choice D the correct answer.

B. regulate the rate of water transport. ✅

Aquaporins, also called water channels, are channel proteins that form pores in the membranes of biological cells, mainly facilitating water transport between cells.

All biological membranes are water-permeable. Without aquaporins, water movement across the membrane is dramatically reduced. Therefore, water moves much more easily and rapidly through aquaporins than elsewhere. Cells that transport large amounts of water, like gut epithelial, cannot function without aquaporins.

Aquaporins are channels that can control water flow by opening or closing themselves. They can also be added or removed from a cell membrane very quickly. This means aquaporins can regulate the amount of water passing through cell membranes and epithelia. For example, in kidney tubules, the antidiuretic hormone stimulates the epithelial cells to insert more aquaporins into the plasma membrane, increasing the membrane's water permeability. This allows the kidneys to reabsorb more water when the body needs to retain it or excrete more water when there's excess.

In short, the diffusion of water through the phospholipids of a membrane is slow and cannot be regulated. Diffusion through aquaporins is fast and regulated.

C. Decreased total peripheral resistance ✅

Water homeostasis refers to the body's ability to maintain a stable and balanced water level within its cells and throughout the body. This balance plays a role in multiple physiological functions, including temperature regulation, transporting nutrients and waste products, and maintaining cell structure. When the body becomes dehydrated, the renin-angiotensin-aldosterone system (RAAS) and the sympathetic nervous system are activated to conserve water and restore balance.

The RAAS is a hormonal pathway activated in response to decreased blood volume. It stimulates the release of aldosterone, a hormone that promotes sodium and water reabsorption in the kidneys.

The sympathetic nervous system is one of the two main branches of the autonomic nervous system and is responsible for the body's "fight-or-flight" response. During dehydration, it constricts blood vessels, raising blood pressure and promoting water retention.

Total peripheral resistance refers to the resistance the blood vessels offers to blood flow. In dehydrated individuals, decreased plasma volume causes an increase in blood viscosity, which can lead to an increase in total peripheral resistance. Therefore, decreased total peripheral resistance is the least likely physiological change in significantly dehydrated athletes.

(Choice A) The parasympathetic nervous system is part of the nervous system that slows the heart, dilates blood vessels, decreases pupil size, increases digestive juices, and relaxes muscles in the gastrointestinal tract. When an individual is dehydrated, the decreased plasma volume increases blood viscosity, increasing cardiac output and blood pressure. This means there is typically an increase in sympathetic activity and a decrease in parasympathetic activity. Therefore, decreased parasympathetic activity is a likely physiological change in dehydrated athletes.

(Choice B) Dehydration leads to a decrease in overall body water content, which triggers the body to conserve water. The body accomplishes this by reducing urine output by promoting water reabsorption in the proximal convoluted tubules. So, decreased urine output is a likely physiological change in dehydrated athletes.

(Choice D) When the body becomes dehydrated, it may have difficulty regulating heat since heat regulation occurs through water excretion in sweat and increased blood flow to the skin. With reduced blood volume due to dehydration, less blood is available to be sent to the skin for cooling purposes. Therefore, decreased body temperature is a likely physiological change in dehydrated athletes, making it an incorrect answer.

B. Thin descending limb of Henle ✅

Osmolality, not to be confused with osmolarity, is the solution concentration expressed as the total number of solute particles per kilogram. The osmolality of the blood is typically 275-295 mOsmol/kg, which means it has a certain concentration of solutes. As the filtrate passes through different segments of the nephron, its osmolality changes due to the reabsorption or secretion of solutes.

The loop of Henle is a long U-shaped portion of the tubule that recovers water and sodium chloride from the glomerular filtrate. The loop of Henle has three segments with different characteristics that enable countercurrent multiplication.

The thin descending limb of Henle is passively permeable to water but not to solutes. As the filtrate descends through this segment, water is reabsorbed by osmosis, while solutes remain in the tubule, increasing the concentration of solutes in the filtrate. This concentration gradient is responsible for the high osmotic value of the filtrate in the thin descending limb of Henle compared to the blood.

(Choice D) The distal convoluted tubule is responsible for fine-tuning the reabsorption and secretion of specific ions, such as sodium, potassium, and hydrogen. There, the filtrate is further diluted; however, it does not significantly change the osmolality of the filtrate compared to the blood.

A. Only Gb3-transfected cells with endogenous Gb3 exhibited cellular injury. ✅

The passage's first paragraph states, “The function of the B pentamer is to bind to the host receptor globotriaosylceramide, Gb3, initiate endocytosis, and gain cellular entry.” This means that the B subunit of Shiga toxin (Stx) binds to the Gb3 receptor on the host cell surface. This binding initiates the toxin to enter the cell.

Gb3-transfected cells have undergone a procedure to introduce genes or genetic material that codes for the production of Gb3 molecules. This essentially equips the cell with the ability to produce its Gb3 receptors, even if it didn't have them naturally. Endogenous Gb3 refers to the Gb3 molecules that are naturally produced by the cell and are present on its surface. These are the "native" receptors. Exogenous Gb3 refers to the Gb3 molecules that are introduced from an external source, not produced by the cell itself.

In this case, the experiment aims to confirm that recipient cells must express the Stx receptor (Gb3) for toxin-positive vesicles to cause toxin-mediated inhibition of protein synthesis.

By only considering cells with endogenous Gb3, we can isolate the role of the natural receptor. If these cells show injury upon exposure to toxin-positive vesicles, it strongly suggests that the toxin requires the Gb3 receptor to enter and inhibit protein synthesis.

Therefore, by observing injury only in cells with naturally occurring Gb3 receptors, choice A provides the most compelling evidence that recipient cells need the specific Stx receptor for the toxin to inhibit protein synthesis.

B. N-glycosidase ✅

The Stx A subunit is known to have toxic effects on ribosomes. It inhibits protein synthesis by removing a single adenine residue from the 28S rRNA of the 60S, which disrupts the ribosome function.

N-glycosidases are a class of enzymes that cleave the glycosidic bond between a nitrogenous base and a sugar molecule. In the case of Stx A subunit toxicity, it acts as an N-glycosidase, removing the adenine residue from the ribosomal RNA. Therefore, an N-glycosidase enzyme can mimic the effects of Stx A subunit toxicity on ribosomes.

C. 2/3 ✅

An allele is a variant form of a gene. In autosomal recessive diseases, a dominant allele masks the expression of the recessive allele.

In this case, we can infer that the allele for phenylketonuria is recessive, denoted as c, while the dominant allele, denoted as C, does not express the disease. In other words, the normal phenotype is associated with the genotype CC and Cc (carrier). The disease phenotype is associated with the genotype cc.

Since the parents are carriers, they carry a dominant and recessive allele, denoted as Cc. The child inherits one allele from each parent. Therefore, the possible genotypes of the child are CC, Cc/cC, and cc.

In this case, the child has a normal phenotype, meaning they must have either the genotype CC or Cc. Two of the 3 possible genotypes (Cc, cC, and CC) that have a normal phenotype result in the child being a carrier. Therefore, the probability that the child is a carrier (Cc/cC) is ⅔.

C. 1/4✅

Let's denote the dominant allele as 'A' and the recessive allele as 'a'. According to the information given, the gene has two alleles, so we have 'A' and 'a'.

Let's assume that the frequency of the dominant allele 'A' is represented by 'p', and the frequency of the recessive allele 'a' is represented by 'q.' Since there are only two alleles, 'p' + 'q' = 1.

Now, we are given that the number of individuals with a heterozygous genotype is six times that of the homozygous recessive genotype. A heterozygous individual has two different alleles for a particular gene. A homozygous individual has two identical alleles for a particular gene. Therefore, we can infer that ‘Aa’ is 6 times the frequency of ‘aa.’

To solve the equation, let's represent the frequency of the heterozygous genotype (Aa) as '2pq' (since we have two alleles, 'A' and 'a') and the frequency of the homozygous recessive genotype (aa) as 'q²' (since both alleles must be 'a').

The question stem states that "the number of individuals with a heterozygous genotype is six times that of a homozygous recessive genotype."

Given that:
2pq = 6q²

Divide both sides by 2q to get:
p = 3q

Now we know that p = 3q, and we also know that p + q = 1. We can substitute the value of p in terms of q to get:
3q + q = 1
4q = 1
q = 1/4

D. DNA polymerase synthesis ✅

The passage states, “When pRB is hypophosphorylated, it binds to E2F1, suppressing the G1/S transition.” This means that when protein retinoblastoma has low phosphorylation levels, it inhibits the cell cycle from advancing from the G1 phase into the S phase, where DNA synthesis occurs. At the G1 phase, the cell prepares for DNA replication by making more proteins, organelles, and enzymes required for DNA replication, which includes DNA polymerase.

B. decreased testosterone levels. ✅

The passage explains that CAH patients experience impaired production of cortisol levels, which inhibits the feedback mechanism and causes an overproduction of androgens, such as testosterone.

The passage states, "This defect impairs the production of glucocorticoids and mineralocorticoids, leading to a bottleneck, creating a major backing up of precursor steroids overflowing into other steroid pathways. In female patients, this leads to excessive circulating adrenal-source androgen precursors (e.g., androstenedione)."

Therefore, we can interpret that by reversing cortisol levels back to normal levels and supplementing glucocorticoids through hormone replacement therapy, the negative feedback mechanism would suppress the activity of the pituitary gland. This would decrease testosterone levels and its precursors, such as 17α-Hydroxyprogesterone, androstenedione, and dehydroepiandrosterone.

D. Hemizygous ✅

The passage states, “AIS arises from a mutation in the androgen receptor (AR) located on the X chromosome.” Therefore, we can interpret AIS as an X-linked disorder. The passage states, “Key congenital disorders are 46, XY DSD, androgen insensitivity.” This means that AIS primarily affects individuals with XY karyotype, i.e., males. Using these two pieces of information, we can conclude that the mutated androgen receptor gene on the X chromosome is the only copy of that gene in males.

Hemizygous refers to the presence of only one copy of a particular gene in an individual, typically seen in males who have only one X chromosome. Therefore, the term "hemizygous" accurately describes the genetic status of males with AIS.

B. Lower FSH and LH ✅

This question asks about the effects of androgenic steroids on FSH and LH levels. Follicle-stimulating hormone (FSH) and Luteinizing hormone (LH) are gonadotropins produced by the pituitary gland.

Exogenous androgenic steroids would have identical effects as endogenous androgens. This is interpreted from the passage that states, “Exogenous testosterone exerts the same clinical effects on every known androgen-responsive tissue as endogenous testosterone, apart from spermatogenesis.”

Upon androgenic steroid administration, the exogenous testosterone is expected to suppress the production of FSH and LH through a negative feedback mechanism. The increased testosterone levels signal the hypothalamus and pituitary gland to decrease the release of FSH and LH. As a result, the circulating levels of both FSH and LH decrease.

C. n chromosomes. ✅

Since the question states nondisjunction happens in both meiotic divisions, the resulting daughter cells will likely have either an extra chromosome (n+1) or a missing one (n−1). Therefore, the least likely outcome is a daughter cell with the euploid chromosome number (n). This is because double nondisjunction inherently disrupts the normal chromosomal separation process, making it highly improbable for both meiotic divisions to incorrectly distribute chromosomes while maintaining the euploid number.

B. increased glucose concentrations. ✅

In paragraph 3, the passage states the enzymes that require biotin cofactor are:
Pyruvate carboxylase (Oxalacetate formation)
Propionyl-CoA carboxylase (Succinyl-CoA formation)
Methylcrotonyl-CoA carboxylase (Leucine metabolism)
Acetyl-CoA carboxylase (Malonyl-CoA formation)

Oxaloacetate is an intermediate of the citric acid cycle, where it reacts with acetyl-CoA, a molecule that provides an acetyl group into the citric acid cycle to be oxidized for energy production to form citrate. In biotinidase deficiency, oxaloacetate formation is decreased as pyruvate carboxylase is not activated by the biotin cofactor. Due to the decrease in oxaloacetate levels, the citric acid cycle flow decreases (Choice C). Subsequently, acetyl-CoA levels increase (Choice A) as they are not reacting with oxaloacetate.

Malonyl-CoA is the rate-determining metabolite in fatty acid synthesis. In biotinidase deficiency, malonyl-CoA formation is decreased as the biotin cofactor does not activate the enzyme acetyl-CoA carboxylase. This leads to decreased fatty acid synthesis (Choice D).

Succinyl-CoA plays a key role in the citric acid cycle. In biotinidase deficiency, succinyl-CoA formation decreases because the enzyme propionyl-CoA carboxylase cannot be activated by the biotin cofactor, leading to further disruption of the citric acid cycle.

Leucine is an essential amino acid for protein synthesis. In biotinidase deficiency, leucine metabolism is decreased as methylcrotonyl-CoA carboxylase cannot be activated by the biotin cofactor. This leads to increased levels of leucine.

Therefore, increased glucose concentrations (Choice B) is the least likely outcome among the given options as none of the carboxylases directly affects its levels.

B. mitochondria. ✅

Gluconeogenesis is the process by which the body synthesizes new glucose molecules from non-carbohydrate precursors. It provides glucose, the primary energy source for cells when dietary intake is insufficient. Gluconeogenesis can utilize various precursors, including amino acids, lactate produced by anaerobic respiration in muscles, and glycerol derived from triglycerides during fat breakdown. First, the starting materials (amino acids, lactate, or glycerol) are converted into pyruvate, a key intermediate. Pyruvate is then converted to oxaloacetate by pyruvate carboxylase, an enzyme requiring ATP and biotin. Oxaloacetate cannot freely cross the mitochondrial membrane. In the cytosol, oxaloacetate is converted to PEP by a series of enzymatic steps, utilizing energy from ATP and GTP. This is the committed step in gluconeogenesis, meaning once this step occurs, the pathway is committed to producing glucose. PEP is converted to glucose through a series of enzyme-mediated reactions, similar to the final steps of glycolysis but reversed.

A. introduce new carbon-carbon bonds. ✅

Carboxylases allow the production of new carbon-carbon bonds by introducing HCO₃⁻ or CO₂ into target molecules. In holocarboxylase synthetase deficiency, the enzymes are not activated by biotin and, therefore, cannot perform their function.

B. Log phase ✅

The passage’s second paragraph mentions that the S. aureus strain 17810R was "deprived of energy sources by starvation." This indicates that the control group in this experiment is the starved colony of S. aureus.

Also, the fourth paragraph states, "Researchers concluded that cells could maintain a balance between ATP utilized for cellular endergonic reactions and ATP re-synthesized through membrane potential generated by endogenous respiration. This allows the cells to maintain a high ATP pool during 30-minute observation." This means the researchers observed a high ATP pool during the 30-minute observation period.

ATP, or adenosine triphosphate, is a nucleotide that provides energy to drive and support many processes in living cells, including chemical synthesis. A high ATP pool indicates that the cells have sufficient energy reserves, typically associated with active growth and metabolism. The log phase, or the exponential phase, is characterized by rapid cell division and active metabolism, where the cells are actively synthesizing macromolecules, such as proteins and nucleic acids, and require a substantial amount of energy in the form of ATP.

D. Two cells formed after mitosis, and two cells formed after meiosis II ✅

In meiosis I, homologous chromosomes pair up. Homologous chromosomes are pairs of chromosomes, one from the mother and one from the father, that are similar in shape, size, and genetic content but are not identical. During prophase I, homologous chromosomes undergo crossing-over, where they exchange segments of genetic material. This crossing-over creates new combinations of genes, contributing to genetic diversity. During metaphase I, homologous chromosomes align at the metaphase plate. In anaphase I, the homologous chromosomes are pulled to opposite poles of the cell. This separation results in two non-identical daughter cells, each with a haploid set of chromosomes, as seen in telophase I and cytokinesis. These cells are not identical because of the independent assortment of maternal and paternal chromosomes in addition to the genetic recombination from crossing-over.

Meiosis II is similar to mitosis in that it involves the separation of sister chromatids. Sister chromatids are the identical copies of a chromosome created during DNA replication. In prophase II, chromosomes condense again, and during metaphase II, they align at the metaphase plate. Anaphase II involves the separation of sister chromatids to opposite poles. Finally, in telophase II and cytokinesis, each of the two cells from meiosis I divides again, resulting in four genetically distinct haploid cells. Meiosis II is actually very similar to mitosis, but the key difference between them is that the cells involved are haploid, not diploid, and the resulting cells are genetically unique due to the preceding cross over process in meiosis I.

Now, back to the question, it is stated that crossing over was prevented during meiosis, so which cells would have the same genetic compositions? We know that the cells after mitosis are definitely identical because mitosis produces two genetically identical daughter cells. Based on what we have explained above, meiosis II would produce two cells, which are only different because of the crossing-over process in meiosis I. So, it makes sense that if we prevent crossing over in meiosis I, then the two cells produced in meiosis II would actually be identical, too. Therefore, (Choice D) is the correct answer choice.

(Choice A) Even without crossing-over, the two cells formed after meiosis I are not genetically identical due to the independent assortment of maternal and paternal chromosomes.

(Choice B) These cells are not genetically identical because meiosis I produces genetically diverse cells, and meiosis II further separates the sister chromatids, maintaining the genetic diversity.

(Choice C) This answer is incomplete as it does not include the two cells produced in meiosis II.

D. 4 × 8 ✅

In a Punnett square, each gamete type formed by the crossed individuals is written at the beginning of a single column or row. Therefore, the size of the Punnett square we will use depends on the number of gametes that individuals will form.

To determine the minimum number of Punnett squares required, we must consider the different combinations of alleles for each genotype.

The first genotype, AaBBCCDdEE, has the following allele combinations:
Aa (2 possibilities), BB (1 possibility), CC (1 possibility), Dd (2 possibilities), and EE (1 possibility).
Multiplying these possibilities together gives us 2×1×1×2×1 = 4 different combinations for the first genotype.

The second genotype, AaBbccddEe, has the following allele combinations:
Aa (2 possibilities), Bb (2 possibilities), cc (1 possibility), dd (1 possibility), and Ee (2 possibilities).
Multiplying these possibilities together gives us 2×2×1×1×2 = 8 different combinations for the second genotype.

Therefore, the Punnett square grid should be 4×8.

C. Inductive Reasoning ✅

C. Analyzing the city map and eliminating the routes to narrow down the options ✅

C. If patients perceived an increase in self-efficacy in managing pain post-intervention ✅

According to Social Cognitive Theory (SCT), individuals who believe in their ability to perform a behavior are likelier to engage in and persist with that behavior. Self-efficacy refers to an individual's belief in their ability to execute behaviors necessary to achieve specific goals. In the context of chronic pain management, patients with higher perceived self-efficacy in managing their pain are more likely to adhere to their treatment plans. This adherence may include following recommendations from healthcare professionals, engaging in prescribed exercises or activities, and effectively managing pain-related symptoms. By feeling confident in their ability to cope with pain and adhere to treatment protocols, patients are more motivated to persist with their treatment plans, leading to better outcomes in managing chronic pain. Therefore, perceived self-efficacy in managing pain post-intervention would significantly predict treatment adherence among patients, as suggested by Social Cognitive Theory (SCT).

4o

B. Drive reduction theory ✅

Drive reduction theory states that organisms are motivated to take action in order to reduce drives, such as hunger, thirst, or in this case, anxiety. According to the researcher's hypothesis, individuals experiencing heightened anxiety leading to panic attacks during the intervention phase may increase their adherence to the post-discharge treatment plan as a means of reducing their anxiety levels. This aligns with the concept of drive reduction, where behaviors are motivated by the desire to alleviate internal tension or discomfort.

(Choice A) Instinct theory suggests that behavior is driven by innate, instinctual impulses. It proposes that certain behaviors are genetically programmed and do not require learning or experience. In the context of the hypothesis presented in the question, the instinct theory does not directly explain how worsening symptoms of anxiety leading to panic attacks would influence adherence to a treatment plan. It primarily focuses on behaviors that are instinctually driven rather than those influenced by psychological or emotional states. While anxiety and panic responses may have evolutionary roots, the hypothesis described in the passage focuses more on the role of anxiety as a drive that motivates behavior, rather than an instinctual response.

(Choice C) Incentive theory suggests that behavior is motivated by the anticipation of rewards or punishments associated with that behavior. It proposes that individuals are more likely to engage in behaviors that lead to positive outcomes or rewards and avoid behaviors associated with negative outcomes or punishments. While incentives may play a role in motivating behavior, the hypothesis in the question focuses more on the physiological response to anxiety and its potential impact on adherence to a treatment plan, rather than the anticipation of external rewards or punishments.

B. rationalization. ✅

Rationalization involves creating logical explanations or excuses to justify actions or beliefs. Blaming the teacher's grading style to explain exam failure instead of acknowledging the lack of preparation aligns with rationalization, as it's a way of justifying the failure without accepting personal responsibility.

D. Parkinson’s Disease ✅

GABA Agonists: These medications enhance the activity of gamma-aminobutyric acid (GABA), a neurotransmitter that inhibits brain activity. GABA agonists can have sedative, anxiolytic (anti-anxiety), and anticonvulsant (anti-seizure) effects. The question asks for the disorder that is least likely to be treated with GABA agonists, implying that while GABA agonists may be used in some of these disorders, one is notably less likely to be treated with these medications than the others.

GABA agonists are not the primary treatment for Parkinson's disease. Dopamine agonists, levodopa, and other medications targeting dopamine levels are more commonly used to manage motor symptoms in Parkinson's disease. Therefore, Parkinson's disease is the least likely candidate for GABA agonist treatment among the options provided.

D. City anonymity reduces the social influence on diet and activity, lessening conformity. ✅

Sociopsychological concepts refer to how societal and psychological factors influence individual behaviors and attitudes, including diet, physical activity, and overall lifestyle choices that can affect obesity rates. To answer this question, we need to evaluate how each statement connects sociopsychological theories or concepts with the observed differences in obesity rates between urban and rural areas. The focus should be on identifying the statement that either contradicts established sociopsychological theories or fails to adequately apply these theories to explain the urban-rural obesity rate disparity.

(Choice D) Social influence is a well-established concept in social psychology, emphasizing how people are influenced by the actions and attitudes of those around them, such as friends, family, and even strangers observed in their environment. Despite anonymity in cities, social influence through observation and interaction remains strong, with some arguing that the diverse range of people encountered in urban areas can intensify social comparison. Regarding obesity rates, adherence to societal norms regarding food and activity levels is likely a more influential factor than anonymity. Moreover, city anonymity could increase conformity to broader societal trends, including dietary habits and activity levels, driven by media and advertising rather than diminishing it. Urban residents might still feel pressure to conform to prevailing lifestyle norms, including healthy and unhealthy behaviors. The availability of a wide range of food options and lifestyle choices in urban areas means that social influence can manifest in complex ways, not necessarily lessened by anonymity.

C. Occupational segregation ✅

The passage mentions: "The decline in industries traditionally dominated by full-time male workers, such as utilities, mining, and manufacturing, led to drops in employment rates for men across all age groups between 1977 and 1997." We need to evaluate each of the answer choices to this piece of information:

Occupational segregation refers to the distribution of people across and within occupations based on demographic characteristics, most often gender. If certain industries, like utilities, mining, and manufacturing, were dominated by men, reducing occupational segregation might encourage a more diverse workforce in these sectors, potentially slowing their decline.

B. The number of relatives an individual supports ✅

Inclusive fitness = an individual’s total genetic success, which includes:

1. Direct fitness → the number of offspring you produce
2. Indirect fitness → helping close relatives (who share your genes) survive and reproduce

C. Native German speakers consistently rate objects with male-gendered articles as more similar to human male characteristics. ✅

B. Deindividuation ✅

In paragraph 4, a participant expresses a desire to emulate behaviors depicted in media portrayals, stating, "If a magazine would say, ‘bulimia ruined my life,’ I would read it just to get ideas... I wanted to figure out what a celebrity did because I need to do the same thing." This quote illustrates how the participant loses their sense of individual identity and adopts behaviors based on external influences, such as media messages or societal norms. This phenomenon aligns with the concept of deindividuation, where individuals in group settings experience a loss of self-awareness and engage in behaviors that align with group norms or external influences.

D. Selective serotonin reuptake inhibitors (SSRIs) ✅

To understand the question stem, we need to consider the coping mechanisms observed in the participants described in the passage. The passage discusses how individuals with anorexia turned to the media as a coping mechanism, seeking validation or solutions from media portrayals of body image and eating disorders. While anorexia is the primary focus of the study, it's essential to note that the coping mechanisms described, such as seeking validation or guidance, can also be indicative of individuals with obsessive-compulsive disorder (OCD). One participant explicitly mentions feeling the need to emulate behaviors associated with celebrity, indicating obsessive thoughts or behaviors.

SSRIs are commonly used in the treatment of OCD due to their effectiveness in reducing symptoms related to obsessive thoughts and compulsive behaviors. The coping mechanisms observed in the participants, such as seeking validation or solutions from media portrayals, align with the obsessive thought patterns characteristic of OCD. SSRIs work by increasing serotonin levels in the brain, which can help alleviate obsessive thoughts and reduce the urge to engage in compulsive behaviors. Given this understanding, the correct answer is selective serotonin reuptake inhibitors (SSRIs).

A. Correctly identifying prohibited items will increase over time. ✅

C. This disorder could be the obstructive type of apnea due to throat muscles or the central type due to brain dysfunction. ✅

When we examine the figure, the question stem and the y-axis of the graph allude to oxygen saturation. Only one sleep condition of the answer choices affects oxygen saturation: sleep apnea. "A-" means "in the absence of," and "-pnea" means "breathing." Therefore, apnea means "sleep without breathing." There are two types of sleep apnea we need to know about for the exam: obstructive sleep apnea and central sleep apnea.