This question asks to select the distance between the Accommodating Intraocular Lens (AIOL) and the retina that allows the patient to clearly see an object located at infinity. To answer this question, we must use our knowledge of geometrical optics.
The passage states “The the AIOL is a biconvex converging lens with a focal length of 3 cm.” Converging lenses refract parallel light rays passing through them, converging them at a single focal point. The focal length (f) of a lens is the distance between the lens and its focal point, where parallel light rays convergefor a convex/converging lens.
The human eye itself contains a lens that functions to converge light onto the retina. Objects we look at reflect light; that light enters our eye. Working with the cornea, and the ciliary muscles, the eye’s lens then focuses this light onto our retinas.
When light rays from far away enter the eye, they travel in parallel lines. The artificial lens must bend these rays so they meet exactly on the retina, the light-sensitive tissue at the back of the eye. The distance from the lens to the retina needs to be the same as the lens’s focal length, which is the distance where parallel light rays naturally converge after passing through the lens. If this distance isn’t right, the light won’t focus sharply on the retina, and the person’s vision will be blurry.
If the distance between the AIOL and the retina was 1.0 cm, that would be 2 cm less than the focal length of the AIOL. With a shorter distance between the AIOL and the retina, the light rays have not yet converged to a focus by the time they reach the retina. As a result of the focal point being located further back than the retina, the focus would be behind the retina. This is similar to hyperopia (farsightedness).
Like the first answer choice, if the distance between the AIOL and the retina was 2.0 cm, that would be 1 cm less than the focal length of the AIOL. With a shorter distance between the AIOL and the retina, the light rays have not yet converged to a focus by the time they reach the retina. As a result of the focal point being located further back than the retina, the focus would be behind the retina. Please refer to the image in answer choice A for a visual representation of this phenomena.
When an object is extremely far away (at “infinity” in optical terms), its light rays spread out so minimally that by the time they reach our eyes, they are practically parallel to each other. For someone with an artificial lens implant (AIOL) to see those objects clearly, the AIOL must bend these parallel rays so they converge on the retina.
This relationship can be mathematically expressed using the thin lens equation:1/do + 1/di = 1/f
where:
f = focal length of the AIOL
do = object distance (infinity, so 1/do≈0)
di= image distance (distance between AIOL and retina)
therefore, for an object at infinity:
1/f=1/di
di= f
Thus, the distance between the AIOL and the retina is equal to the focal length of the AIOL, which is f = 3.0 cm.
If the distance between the AIOL and the retina was 4.0 cm, that would be 1 cm greater than the focal length of the AIOL. With a longer distance between the AIOL and the retina, the light rays would converge to a focus before reaching the retina. As a result of the focal point being located in front of the retina, the light rays would begin to diverge again by the time they reach the retina, creating a blurred image. This is similar to myopia (nearsightedness).
This question asks us to calculate the nearest location of an object that can be seen clearly by a patient wearing the AIOL. We can utilize our general content knowledge about geometrical optics and passage information to answer this question.
The passage states in the second paragraph that “The AIOL moves away from the retina by as much as 5.0 mm due to the contraction of the eye’s focusing muscles” and that this “allows the patient to see clearly objects located nearby.” Based on this passage information, we can then assume that the AIOL will be moved 5.0 mm away from the retina to see an object nearby. We can also deduce, that for objects nearby, the AIOL will need to be in the farthest position from the retina.
The thin lens formula is given by:
1/o + 1/i = 1/f
Where: o = object distance (distance from the object to the lens) i = image distance (distance from the lens to the retina in this case), and f is the focal length of the lens.
This problem asks us to determine the nearest object distance that a patient with an AIOL can see clearly.
The image distance is given as i = 3.0 cm + 5.0 mm = 3.5 cm, and the focal length of the lens is f = 3.0 cm. To solve for the object distance using the thin lens formula, we start with the equation: 1/o + 1/i = 1/f. Our goal is to isolate the object distance o.
This answer choice could have resulted from a miscalculation. For a detailed explanation please reference above.
This answer choice could have resulted from a miscalculation. For a detailed explanation please reference above.
This answer choice could have resulted from a miscalculation. For a detailed explanation please reference above.
In this question we are asked to determine which polymer in Table 1 is best suited for the fabrication of an AIOL. We must use information from the passage as well as our knowledge of optics.
First, we should know the definitions of some terms mentioned in the passage.
The coefficient of thermal expansion measures how much a material expands or contracts when its temperature changes. In general, it quantifies the change in length, area, or volume of a material per degree of temperature change. In intraocular lenses, a low coefficient of thermal expansion ensures that the lens maintains its shape and focal properties despite small temperature variations.
In optics, the attenuation coefficient quantifies the rate at which light intensity decreases as it propagates through a medium due to absorption and scattering. It determines how much light is lost per unit distance in a material. For AIOLs, a low attenuation coefficient ensures maximum light transmission, improving vision clarity.
Index of Refraction is a dimensionless number that describes how light slows down and bends (refracts) when it enters a material from another medium, such as from air into water or glass. In intraocular lenses (AIOLs), an optimal refractive index ensures that light focuses correctly on the retina, restoring sharp vision.
The AIOL must maintain its volume within ±0.020% when subjected to a temperature change of ±2.0°C. For polymer 1, the relative volume change is calculated as 3×2.0×10−5 K−1×(±2.0∘C), resulting in ±1.2×10−4= ±0.012%. Similarly, for polymer 2, the relative volume change is 3×3.0×10−5K−1×(±2.0∘C), yielding ±1.8×10−4= ±0.018%. However, polymer 1 has a higher attenuation coefficient than polymer 2, meaning it transmits less light. Since both polymers have the same refractive index, a lens made from polymer 1 must have the same thickness as a lens made from polymer 2 in order to maintain a focal length of 3.0 cm.
We can refer to second paragraph in the passage where it is mentioned that “the AIOL must maintain its volume within ±0.020% when its temperature changes by ±2.0°C”.
From table 1 we see that the coefficient of thermal expansion of polymer 2 is 3.0 × 10−5 K−1 . then, we multiply this value by 3, where the factor 3 accounts for expansion in all three dimensions (length, width, and height) and finally, the result is multiplied by ±2.0 which is the temperature change in this case
For polymer 2, the relative volume change is 3 × 3.0 × 10−5 K−1 × (±2.0°C) = ±1.8 × 10−4 = ±0.018%.
When we calculate relative volume change for polymer 1 it would be 3 × 2.0 × 10−5 K−1 × (±2.0°C) = ±1.2 × 10−4 = ±0.012%. although polymer 1 has lower relative volume change than polymer 2, it is not preferred. Looking at the attenuation coefficient , we recognize that polymer 2 has a lower attenuation coefficient than polymer 1; thus it allows more light to pass through it than polymer 1 given that it has the same refractive index as polymer 1, and so the lens made of polymer 2 will have the same thickness as the lens made of polymer 1 in order to yield a focal length of 3.0 cm.
For polymer 3, the relative volume change is calculated as 3×4.0×10−5 K−1 ×(±2.0∘C), resulting in ±2.4×10−4 = ±0.024%, which is above the maximum allowable limit of ±0.020%.
Similarly, for polymer 4, the relative volume change is 3×5.0×10−5 K−1×(±2.0∘C), yielding ±3.0×10−4= ±0.030%, also exceeding the permitted limit of ±0.020%.
This question wants us to reason through the answer choices by using the process of elimination and understanding the relationship between frequency of a ultrasound waves (USW), amplitude, wavelength, and energy.
The question states the process of emulsification of an opaque eye lens utilizes USW waves that propagate at a speed of 1500 m/s, and a frequency of 30Mhz.
There is not enough information in the passage to determine the USW amplitude.
The question asks us to identify the correct characteristic of the ultrasound waves (USW) used in the emulsification of an opaque eye lens, given a propagation speed of 1500 m/s and a frequency of 30 MHz. To determine the wavelength, we apply the fundamental wave equation: wavelength equals velocity divided by frequency (λ = v/f). Plugging in the provided values, we divide the wave velocity of 1500 m/s by the frequency of 30 million hertz (30 x 106 Hz). This calculation yields a wavelength of 0.00005 meters, which is equivalent to 0.05 millimeters.
There is not enough information in the passage to determine the USW energy because the USW amplitude is unknown.
USW frequency range is from 20 kHz (20×103 Hz= 0.02 MHz) to over 1 GHz (2×109 Hz= 1000 MHz). The statement “The USW cannot have frequencies other than 30 MHz.” is incorrect because ultrasound waves of different frequencies could still be used for emulsification, provided they meet the necessary wavelength requirement as long as within the accepted USW frequency range.
This question asks about the number of neutrons in the metalloid atom present in the compound in the polymers used for the manufacturing of AIOLs. To answer this question, we must use the information provided by the passage as well as our knowledge of atomic nuclei and the periodic table.
This is incorrect because it describes the mass number of silicon. The mass number is the sum of protons and neutrons and does not represent the number of neutrons alone.
(Choice B) is a miscalculation. This exceeds the actual number of neutrons by 6.
(Choice C) is also a miscalculation. This represents the number of neutrons plus 4.
In the second paragraph of the passage, it is mentioned that the AIOL is made from a SiO2-based polymer. In the question, it is required to know the number of neutrons in the metalloid atom. We need to recall that metalloids are elements that exhibit properties intermediate between those of metals and nonmetals, and they include boron, silicon, arsenic, tellurium, antimony, and germanium. The compound SiO₂, or silicon dioxide, contains the element silicon, which is a metalloid. Therefore, we can understand that the metalloid referred to in the question is silicon (Si).
To get the number of neutrons, we need to use the periodic table. In the periodic table, we can find:
The mass number (A) of an atom, which is the total number of protons and neutrons in its nucleus.
The atomic number (Z) of an atom, which is the number of protons in its nucleus.
The number of neutrons will not be written directly, so we have to calculate it.
To determine the number of neutrons in a silicon atom:
The most common isotope of silicon is silicon-28 (28Si).
The atomic number (Z) of silicon is 14, which means that it has 14 protons.
The mass number (A) of silicon-28 is 28.
The number of neutrons (N) is calculated as:
N= A−Z= 28−14= 14
Thus, the number of neutrons in the silicon atom present in the SiO₂-based polymers is 14 (Choice D).
In this question we are asked to determine the maximum magnitude of the work done by the focusing muscles on the AIOL. We can utilize our understanding of work and passage information to answer this question.
This is the result of the calculation 2.0 N × 5.0 mm = 1.0 × 10−2 J. There is a mistake in unit conversion. N should be 2.0 μN = 2.0×10−6 N and d should be 5.0×10−3 m. Therefore, we get in A the wrong answer (1.0 × 10–2 J instead of (1.0 × 10–8 J)
This is the result of the calculation 2 × 2.0 μN × 5.0 m = 2.0 × 10−5 J.
Here, we have a mistake in displacement unit conversion. d should be 5.0×10−3 m not 5.0 m. In addition, the force and distance are multiplied by 2 which is not possible since cosθ never equals 2.
The passage states the AIOL moves away from the retina “by as much as 5.0 mm” due to the contraction of the “focusing muscles of the eye”. To determine the maximum magnitude of the work done by the focusing muscles on the AIOL, we use the work formula:
W=F⋅ d⋅ cos(θ)
where:
W is the work done,
F = 2.0 μN = 2.0×10−6 N (force on the AIOL),
d = 5.0 mm = 5.0×10−3 m (maximum displacement of AIOL),
θ=0° since the force is in the same direction as displacement, so cos(0°) = 1.
This is the result of the calculation 2 × 2.0 μN × 5.0 μm = 2.0 × 10−11 J.
We have again a mistake in displacement unit conversion as it should be 5.0×10−3 m. In addition, the force and distance are multiplied by 2 which is not possible since cosθ never equals 2.
This question asks about the ground state electron configuration of the metalloid atom in the polymers used for the manufacturing of AIOL. To answer this question, we must understand the principles of electron configuration and use the supplied periodic table.
The ground state electron configuration of an atom describes the arrangement of its electrons in the lowest possible energy levels, which is the most stable state of the atom. To determine the ground state configuration, we need to know the atomic number of the element, which tells us how many electrons are present in a neutral atom.
In the second paragraph of the passage, it is mentioned that the AIOL is made from a SiO₂-based polymer. In the question, it is required to know the ground state electron configuration of the metalloid atom. We need to recall that metalloids are elements that exhibit properties intermediate between those of metals and nonmetals, and they include boron, silicon, arsenic, tellurium, antimony, and germanium. The compound SiO₂, or silicon dioxide, contains the element silicon, which is a metalloid. Therefore, we can understand that the metalloid referred to in the question is silicon (Si) for which we need to get the electron configuration.
First, we should get the atomic number of silicon from the periodic table. The atomic number is 14, and we know that the number of electrons equals the number of protons in a neutral atom. So silicon has an atomic number of 14, meaning it has 14 protons and therefore 14 electrons in its neutral ground state. These 14 electrons will be arranged in the lowest energy configuration possible, which represents the ground state electron configuration of silicon.
Second, we determine the electron configuration. Electrons fill orbitals in a special order following the sequence: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, and so on, based on increasing energy levels. The filling follows the Aufbau principle, where we place electrons into the lowest available energy orbitals first before moving to higher ones. This means we begin filling from 1s, then move to 2s, followed by 2p, and continue in that order. We also apply the Pauli exclusion principle, which states that each orbital can hold a maximum of two electrons with opposite spins, and Hund’s rule, which states that electrons will fill degenerate orbitals (orbitals of the same energy, such as the three 2p orbitals) singly with parallel spins before pairing up. By applying these rules systematically, we arrive at the correct ground state configuration for Si.
The ground-state electron configuration of Si is:
1s2 2s2 2p6 3s2 3p2
It can also be written in short form as [Ne] 3s2 3p2 because the ground-state electron configuration of Ne is 1s2 2s2 2p6.
Therefore, we can assume that (Choice A) is the correct configuration.
This is the ground state electron configuration of P, which is not present in the compound on which the polymers are based.
This is the ground state electron configuration of S, which is not present in the compound on which the polymers are based.
This is the ground state electron configuration of Cl, which is not present in the compound on which the polymers are based.
The question asks about the type of chromatographic purification of Ngb-H64Q-CCC that was performed in the second purification step. To answer this question, we must refer to the passage and understand some basic principles of the different types of chromatography.
In the first paragraph of the passage, it is mentioned that “researchers purified crude Ngb-H64Q-CCC by (1) running the protein solution through a molecular weight-cutoff gel filtration column, (2) passing the resulting protein through a column filled with porous cellulose beads covalently attached to poly-L-lysine, which selectively binds bacterial endotoxin”.
Affinity chromatography (Choice A) is a biochemical separation technique used to purify specific molecules based on their highly specific interactions with a ligand attached to a stationary phase, usually a column. The principle of affinity chromatography relies on the specific binding between a target molecule and a ligand.
The ligand is immobilized on a solid support, such as agarose beads in a chromatography column, and when a mixture passes through, only the target molecule binds to the ligand, while other molecules are washed away. The target molecule is then eluted using a special buffer that disrupts the interaction.
This suggests that the removal of this impurity is a form of affinity chromatography because the poly-L-lysine is covalently attached to the porous cellulose beads and serves as a specific ligand that binds bacterial endotoxin. This selective binding between the poly-L-lysine ligand and the endotoxin is the defining feature of affinity chromatography. The target in this case is the endotoxin, and it is retained in the column due to its specific interaction with the immobilized ligand. This mechanism of purification, which relies on a specific biochemical interaction, confirms that the second purification step is correctly identified as affinity chromatography.
Hydrophobic interaction chromatography occurs in step 3, not step 2.
Hydrophobic Interaction Chromatography (HIC) is a type of liquid chromatography used to separate and purify biomolecules based on their hydrophobicity. It relies on hydrophobic interactions between proteins (or other molecules) and a hydrophobic stationary phase under high salt conditions.
More hydrophobic molecules bind more strongly to the column, while less hydrophobic ones pass through. As the salt concentration is gradually reduced, proteins elute based on their hydrophobicity—less hydrophobic proteins elute first, and more hydrophobic ones elute later.
Normal phase chromatography involves a polar stationary phase, like silica gel. This type of chromatography is not described in the passage. Normal Phase Chromatography is a type of liquid chromatography in which the stationary phase is polar, and the mobile phase is nonpolar. It is primarily used to separate compounds based on their polarity, where more polar compounds interact more strongly with the stationary phase and elute later, while less polar compounds elute first.
Size-exclusion chromatography occurs in step 1, not step 2. Size Exclusion Chromatography, also known as gel filtration chromatography, is a technique for separating molecules based on their size (molecular weight) by passing them through a column filled with porous beads. Larger molecules elute first, while smaller molecules take longer to pass through.
This question asks about the partial pressure of CO(g) at the lethal level used by the researchers. To answer the question, we should recall our knowledge of gas laws and extract the relevant information from the passage.
We are asked to determine the partial pressure of carbon monoxide in a gas mixture when CO is present at a concentration of 3.0% by volume.
Avogadro’s Law states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules. For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant.
This means that the fraction of a specific gas in a mixture by volume is the same as its mole fraction in the gas mixture. Since CO makes up 3.0% by volume, it also constitutes 3.0% of the total number of moles of gas in the system.
The partial pressure (Pi) of a gas in a mixture is given by Dalton’s Law of Partial Pressures, which states:
Ptotal = P1 + P2 + P3 + … + Pn
This result is based on 1.5% CO(g) by volume, not 3.0%. Therefore, answer A (11 mmHg) is approximately half the correct answer of B (23 mm Hg).
Each gas in the mixture exerts a pressure proportional to its mole fraction. The partial pressure of an individual gas is calculated as:
PCO = XCO × Ptotal
where:
PCO is the partial pressure of CO,
XCO is the mole fraction of CO,
Ptotal is the total pressure of the gas mixture.
Given that:
CO is 3.0% by volume, so its mole fraction is:
XCO = 3.0 / 100 = 0.030
The total pressure of the gas mixture is 760 mmHg.
Applying the formula for partial pressure:
PCO = Ptotal × XCO
PCO = (760 mmHg) × (0.030)
PCO = 22.8 mmHg
Rounding to 2 significant figures (since 3.0% has 2 significant figures):
PCO = 23 mmHg
This result is based on 4.5% CO(g) by volume, not 3.0%. As a result, answer C (34 mmHg) is approximately 1.5 times answer B (23 mmHg).
This result is based on 6.0% CO(g) by volume, not 3.0%. Consequently, answer C (46 mmHg) is the double of answer B (23 mmHg).
This question asks about the region of the electromagnetic spectrum to which the laser radiation described in the passage belongs. To answer this question, we should know the ranges of the different regions of the electromagnetic spectrum.
In air, X-rays have wavelengths in the range of 0.001–1 nm. The wavelength of the laser used in the experiment is 532 nm and is out of this range.
In air, ultraviolet radiation has wavelengths in the range of 1–400 nm. The wavelength of the laser used in the experiment is 532 nm and is out of this range.
In the second paragraph of the passage, it is stated that “Laser pulses at 532 nm wavelength (in air)”. Accordingly, the wavelength of the laser used in the experiment is 532 nm. Since 532 nm falls within the visible light range (400–700 nm) in air, this laser radiation belongs to the visible spectrum.
In air, infrared radiation has wavelengths in the range of 750–25,000 nm. The wavelength of the laser used in the experiment is 532 nm and is out of this range.
This question is based on Figure 1, where we should determine how much lower is the lactate concentration in Ngb variant-treated blood versus albumin-treated blood 40 minutes after lethal CO(g) exposure. To answer this question, we should understand Figure 1 and be able to extract information from the graph.
This value is the lactate concentration of albumin-treated blood after 40 min, not the difference between this value and that of Ngb variant-treated blood.
This value is the lactate concentration of Ngb variant-treated blood after 40 min, not the difference between albumin-treated blood and Ngb variant-treated blood.
This value is the lactate concentration of Ngb variant-treated blood after 20 min, not the difference between the lactate concentrations of albumin-treated blood and Ngb variant-treated blood at 40 min.
At 40 min, we can see that albumin-treated blood has a lactate concentration of 13 mM, whereas Ngb variant-treated blood has a lactate concentration of 9 mM.
Thus, we calculate the difference between the 2 values:
13 mM – 9 mM = 4 mM
This means that the lactate concentration in Ngb variant-treated blood is 4 mM lower versus albumin-treated blood at 40 min.
The question asks us to determine the association constant (KA) for CO binding to the Ngb-H64Q-CCC hemoprotein variant.
Before we start with the calculation, let us define KA and Kd
The dissociation constant Kd represents the equilibrium constant for the dissociation of a complex. A lower Kd value indicates higher binding affinity (stronger binding).
The association constant KA is the equilibrium constant for the formation of the complex. A higher KA means stronger binding affinity.
From the passage, we are given the M-value:
M = KA(CO) / KA(O2) = 1.0 × 104
In the question, the dissociation constant for O2 is given:
Kd(O2) = 2.5 × 10−8 M
Thus, we can now calculate KA(O2), which is the association constant for O2. KA(O2) is inversely proportional to Kd(O2):
KA(O2) = 1 / Kd(O2)
We substitute now the given Kd(O2):
KA(O2) = 1 / (2.5 × 10−8 M) = 4.0 × 107 M–1
As we know that M = KA(CO) / KA(O2), we can rearrange to solve for KA(CO):
KA(CO) = M × KA(O2)
Substituting the values:
KA(CO) = (1.0 × 104) × (4.0 × 107 M-1)
KA(CO) = 4.0 × 1011 M-1
This is KA(O2) or 1/Kd(O2).
This is the M-value given M–1 units.
This is Kd(O2) × the Ngb variant concentration in the lethal CO dose experiments.
The question asks about the technique that would most likely discover fractions that contained low concentrations of intact Ngb variant during its purification. The question requires information from the passage as well as our knowledge of hemoprotein properties and detection techniques.
It is important that we pay attention to the fact that the question specifically emphasizes detecting low concentrations of intact Ngb variant. Low concentrations would not be visibly red, even though the hemoprotein is inherently colored. So, visual inspection alone would not be reliable at low concentrations, and this should direct us toward instrument-based techniques.
Hemoprotein solutions are red, not colorless. Low concentrations of a red-colored hemoprotein may appear nearly colorless to the human eye, but that does not mean the hemoprotein is absent. Detecting faint color changes by eye is not reliable or sensitive, especially when the goal is to detect intact protein in trace amounts.
This is incorrect because although hemoproteins like Ngb-H64Q-CCC are red, relying on visible red coloration as a detection method is not sensitive enough for detecting low concentrations. Fractions containing small but significant amounts of protein may not appear visibly red and would be missed entirely using this approach. This method lacks the precision and sensitivity required in purification workflows.
This is incorrect because red-colored solutions do not absorb red light—they reflect or transmit it. A substance appears red because it absorbs the complementary color, which is green light.
Spectrophotometry is a technique that allows us to measure how much light is absorbed at specific wavelengths, even at very low concentrations. We know from the passage that Ngb-H64Q-CCC is a red-colored hemoprotein as mentioned in the first sentence. While it is true that the solution appears red, that color is not the light being absorbed—it is the light being reflected or transmitted. The important concept here is that a red-colored substance appears red because it absorbs light in the complementary part of the visible spectrum, which is green light, not red light.
So when we use a visible spectrophotometer, we do not measure the absorbance at red wavelengths (around 620–750 nm). Instead, we measure the absorbance in the green region, roughly 500–570 nm, because that is where red substances absorb light most strongly.
This is why, in detecting low concentrations of a red hemoprotein, we set the spectrophotometer to a wavelength in the green range—the region where maximum absorbance occurs, which makes D the correct answer choice. The higher the absorbance at these green wavelengths, the greater the concentration of the red hemoprotein.
This question asks us to determine the approximate isoelectric point (pI) of the C-terminal amide-capped tripeptide Asp–Ala–Cys–NH2. The isoelectric point is the pH at which the peptide has no net charge. At this point, the molecule is in its zwitterionic form, meaning it contains both positive and negative charges but is overall neutral. To calculate the pI, we need to understand which groups in the peptide are ionizable and how their charge states change with pH.
It is also important to clarify what is meant by an amide-capped C-terminus. In an uncapped peptide, the C-terminal residue ends in a carboxyl group (–COOH), which is ionizable and typically negatively charged at physiological pH. However, when the C-terminus is capped with an amide group (–NH2), it means that the –OH group of the carboxyl is replaced with –NH2, forming an amide functional group (–CONH2). This functional group is neutral and non-ionizable, which means that it does not contribute to the net charge and is not used in the isoelectric point calculation. Now we can look at the relevant ionizable groups in Asp–Ala–Cys–NH2. The N-terminus is a primary amine with a pKa of 9.8.
It is positively charged when protonated and neutral when deprotonated. Aspartic acid contains a side chain carboxylic acid group with a pKa of 3.9. This group is neutral when protonated and negatively charged when deprotonated. Cysteine contains a thiol side chain with a pKa of 8.4, and this group is neutral when protonated and negatively charged when deprotonated. Alanine has no ionizable side chain, so it does not affect the net charge. At very low pH, all ionizable groups are protonated. The N-terminus is +1, the Asp side chain is neutral, and the Cys side chain is neutral, giving a total charge of +1. As we increase the pH past 3.9, the Asp side chain loses a proton and becomes negatively charged, so the overall charge becomes 0. If we continue to raise the pH past 8.4, the Cys side chain also deprotonates and becomes negatively charged, resulting in a net charge of –1. At pH values above 9.8, the N-terminus deprotonates and becomes neutral, and both side chains remain deprotonated and negatively charged, giving a net charge of –2. The isoelectric point is the pH at which the net charge is zero. This occurs between the pKa values that bracket the neutral form, which in this case is between the deprotonation of the Asp side chain (3.9) and the deprotonation of the Cys side chain (8.4).
9.8 is the pKa of the N-terminus. At this pH, the N-terminus is 50% protonated, so we approximate its contribution as +0.5. Both the Asp and Cys side chains are deprotonated, contributing –1 each. This gives a net charge of +0.5 – 1 – 1 = –1.5, which is clearly not the isoelectric point.
This results from incorrectly averaging the pKa of the N-terminus and the Asp side chain. However, the N-terminus remains protonated at the point where the net charge is zero, so it should not be used in the calculation.
We calculate the pI as the average of the two pKa values that flank the zwitterionic form: (3.9 + 8.4) / 2 = 6.15, which rounds to approximately 6.1. This is the pH at which the peptide has no net charge.
3.9 is the pKa of the Asp side chain. At this pH, the Asp side chain is half-ionized, contributing –0.5. The N-terminus is fully protonated (+1), and the Cys side chain remains neutral. This gives a net charge of +0.5, meaning the molecule is still positively charged and not at its isoelectric point.
This question asks about the maximum number of full α-helical turns in a protein monomer composed of 4,428 amino acid residues.
To determine the maximum number of full α-helical turns in a protein monomer composed of 4,428 amino acid residues, we use the structural property that each full turn of an α-helix contains approximately 3.6 amino acid residues.
(Choice A) This calculation assumes 4.1 amino acid residues per helical turn, which corresponds to a π-helix, not an α-helix.
By dividing the total number of residues by the number of residues per helical turn, we calculate the total number of possible helical turns as 4,428 divided by 3.6, which equals approximately 1,230. Therefore, the maximum number of full α-helical turns in a protein monomer composed of 4,428 amino acid residues is approximately 1,230.
(Choice C) This result is based on 3 amino acid residues per helical turn, which is for a 310-helix, not an α-helix.
(Choice D) This result is based on 2.7 amino acid residues per helical turn. There are no protein helices with this ratio.
The question asks about the approximate molecular weight in kDa of a protein composed of 90 amino acid residues. We need to use our knowledge of proteins and amino acids to answer this question. To estimate the molecular weight (MW) of a protein composed of 90 amino acid residues, we use the approximation that the average molecular weight of a single amino acid residue in a protein is 110 Da (Daltons).
This result is based on 80 Da per amino acid residue, not 110.
This result is based on 90 Da per amino acid residue, not 110.
This result is based on 100 Da per amino acid residue, not 110.
The total molecular weight is calculated as: MW = 90 × 110 = 9,900 Da = 9.9 kDa.
This question asks about the type of turn in a protein that has a hydrogen bond between the i and i + 5 end residues of the turn. To answer this question, we must understand the principles of protein secondary structure, specifically turns, which are short sequences of amino acids that allow a polypeptide chain to reverse direction. Turns are stabilized by intramolecular hydrogen bonds and play a role in protein folding. Different types of turns are classified based on the number of residues involved and the positioning of the hydrogen bond donor and acceptor within the sequence.
A π-turn is characterized by a hydrogen bond between the carbonyl oxygen of residue i and the amide hydrogen of residue i + 5. This means that π-turns involve six residues, making them the longest type of turn. These turns are less common but can contribute to tight and complex folding patterns in proteins.
A γ-turn has a hydrogen bond between the i and i + 2 residues, meaning that it is a much tighter turn involving only three residues. Because γ-turns form such compact structures, they differ significantly from π-turns, which involve six residues.
A β-turn has a hydrogen bond between the i and i + 3 residues, meaning that it consists of four residues and is a more common turn found in protein structures. β-turns are frequently observed in globular proteins and are critical for connecting β-strands in β-sheets.
An α-turn has a hydrogen bond between the i and i + 4 residues, involving five residues in total. While α-turns contribute to protein folding, they are distinct from π-turns, which involve one additional residue and a longer loop structure.
This question asks about the percentage of the Vmax attained for a single-substrate enzyme following Michaelis–Menten kinetics at [S]0 = 4 × KM.
To answer the question, we need to recall our information about enzyme kinetics and remember some important terms related to enzymatic activity. Vmax is the maximum reaction rate that an enzyme-catalyzed reaction can achieve when the substrate concentration is saturating (i.e., all enzyme active sites are occupied by the substrate). It represents the point where adding more substrate does not increase the reaction rate because the enzyme is working at full capacity. KM is the substrate concentration at which the reaction rate is half of Vmax. It is a measure of the enzyme’s affinity for its substrate:
• Low KM → High affinity (enzyme binds substrate tightly, needs less substrate to reach half of Vmax).
• High KM → Low affinity (enzyme binds substrate weakly, requires more substrate to reach half of Vmax).
To determine the percentage of the maximum velocity (Vmax) attained by an enzyme following Michaelis–Menten kinetics at a substrate concentration of [S]0=4KM, we use the Michaelis–Menten equation.
This rate is for [S]0 = 5 × KM.
Using the Michaelis–Menten equation: V0 = Vmax [S] / (KM+[S])
If [S]0=5 × KM, we substitute: V0=Vmax (5KM) / (KM+ 5KM) V0= 5 VmaxKM / 6KM V0 =5 Vmax / 6 V0 = 0.833 Vmax
Thus, the percentage of Vmax attained at [S]0=5×KM would be 83%.
The Michaelis–Menten equation describes the relationship between the initial velocity (V0) of an enzyme-catalyzed reaction and the substrate concentration ([S]): V0 = Vmax [S] / (KM+[S])
where:
• Vmax is the maximum reaction velocity.
• KM is the Michaelis constant.
• [S] is the initial substrate concentration.
Given that [S]0=4× KM, we substitute this into the equation: V0=Vmax (4KM) / (KM+ 4KM) V0= 4 VmaxKM / 5KM V0 =4 Vmax / 5 V0 =0.8 Vmax
Thus, the percentage of Vmax attained at [S]0=4×KM is 80%.
At a substrate concentration of 4 times the Michaelis constant (4KM), the reaction velocity (V) reaches 80% of the maximum velocity (Vmax).
This rate is for [S]0 = 3 × KM.
Using the Michaelis–Menten equation: V0 = Vmax [S] / (KM+[S])
If [S]0=3 × KM, we substitute: V0=Vmax (3KM) / (KM+ 3KM) V0= 3 VmaxKM / 4KM V0 =3Vmax / 4 V0 = 0.75 Vmax
Thus, the percentage of Vmax attained at [S]0=3×KM would be 75%.
This rate is for [S]0 = 2 × KM.
Using the Michaelis–Menten equation: V0 = Vmax [S] / (KM+[S])
If [S]0=2 × KM, we substitute: V0=Vmax (2KM) / (KM+ 2KM) V0= 2 VmaxKM / 3KM V0 =2 Vmax / 3 V0 = 0.667 Vmax
Thus, the percentage of Vmax attained at [S]0=2×KM would be 66.7%.
This question asks about the gap between the atomic energy levels that correspond to the emission of therapeutic high-energy X-rays.
The problem involves high-energy X-rays emitted due to transitions between atomic energy levels. According to quantum mechanics, when an electron moves from a higher energy level to a lower one, it emits a photon whose energy (E) is equal to the difference between the two energy levels.
To determine the energy gap between atomic energy levels that correspond to the emission of therapeutic high-energy X-rays, we use the fundamental equation from quantum mechanics that relates energy to frequency:
E = hf
where:
E is the energy of the photon (Joules),
h is Planck’s constant (6.6 × 10−34 J·s),
f is the frequency of the X-ray photons (1.5 × 1017 Hz, given in the passage).
This is consistent with calculating the energy gap as (6.6 – 1.5) × 10–14 J without scientific basis.
This implies the X-rays’ frequency is 1.0 × 1019 Hz.
This is consistent with calculating the energy gap as (6.6 + 1.5) × 10–16 J without scientific basis.
First, we begin by substituting the given values into the equation:
E = (6.6 × 10−34 J·s) × (1.5 × 1017 Hz)
E = 9.9 × 10-17 J
This value (9.9 × 10-17 J) represents the energy difference between the two atomic energy levels involved in the emission of the X-ray.
When an electron jumps down from a higher energy level to a lower one, it emits an X-ray photon with this energy.
This question asks about what best approximates the typical number of electrons generated per second by the X-ray detectors that receive the transmitted low-energy X-rays.
To determine the typical number of electrons generated per second by the X-ray detectors, we use the relationship between electric current and charge flow.
We are given:
Electric current from the detectors: I = 15 μA = 15 × 10−6 A
Elementary charge (charge of one electron): e = 1.6 × 10−19 C
From the definition of electric current:
I = Total charge per second / Time (1 s)
Since the charge is carried by electrons, the total charge per second is given by:
Q = ne
where:
Q is the total charge (Coulombs),
n is the number of electrons per second,
e is the elementary charge.
Rearranging for n:
n = Q / e = I / e
This implies the typical current intensity is 8 × 1012 × 1.6 × 10−19 C/s = 1.3 μA instead of 15 μA.
Current (I) = 15 μA = 15 × 10−6 C/s
Elementary charge (e) = 1.6 × 10−19 C
We use the formula:
n = I / e
Substitute the values:
n = (15 × 10−6 C/s) ÷ (1.6 × 10−19 C)
First, estimate the numeric division:
We know that 15 ÷ 1.5 = 10
Since 1.6 is slightly greater than 1.5,
15 ÷ 1.6 ≈ slightly less than 10 → approximately 9.4
Now apply exponent rules:
10−6 ÷ 10−19 = 1013
So:
n ≈ 9.4 × 1013 electrons per second
This implies the typical current intensity is 7 × 1014 × 1.6 × 10−19 C/s = 112 μA instead of 15 μA.
This implies the typical current intensity is 5 × 1015 × 1.6 × 10−19 C/s = 8 mA instead of 15 μA.
This question asks about the chemical classification of the element detected in the left ventricle when the absorbed dose per unit time exceeded 0.04 Gy/s.
The problem describes a study on cardiovascular injury caused by repeated exposure to high-energy X-rays. Researchers used computed tomography (CT) scans to analyze heart tissue, detecting calcium-containing plaques in the left ventricle of patients subjected to an absorbed dose exceeding 0.04 Gy/s.
To determine the chemical classification of the detected element, we need to analyze the element associated with calcium-containing plaques and how X-ray absorption is influenced by the atomic number of elements.
The plaques detected in the heart contain calcium, which suggests that calcium (Ca) is the primary element involved. These calcium deposits are common in cardiovascular diseases and occur due to calcium phosphate (Ca3(PO4)2) buildup in artery walls.
X-ray absorption depends on atomic number (Z); elements with higher atomic numbers absorb more X-rays due to their greater electron density. Calcium (Z = 20) absorbs X-rays more efficiently than elements like carbon (Z = 6) or oxygen (Z = 8), making it easily detectable in CT scans.
Since calcium absorbs more X-rays, the detectors receive fewer photons in calcium-rich regions, allowing the CT scan to highlight areas of calcification. Elements are classified into different chemical categories based on their properties.
Calcium is not an alkali metal because it is Group 2, not Group 1, of the periodic table.
Calcium (Ca) belongs to the group of alkaline earth metals, which are found in Group 2 of the periodic table. Calcium is classified as an alkaline earth metal because it has two valence electrons in the outermost shell, readily forms Ca2+ ions, and combines with phosphate (PO43-) to form calcium phosphate, a key component of bone and arterial plaque.
Therefore, the chemical classification of the element detected in the left ventricle is alkaline earth metal, specifically calcium (Ca), which belongs to Group 2 of the periodic table.
The passage does not state any of the transition metals (Groups 3–12 elements in the periodic table) as having occurred in the left ventricle when the high-energy X-ray absorbed dose per unit time exceeded 0.04 Gy/s.
No halogen occurred in the left ventricle when the high-energy X-ray absorbed dose per unit time exceeded 0.04 Gy/s, according to the passage. Halogens are in Group 17 of the periodic table, whereas calcium, the element detected in the left ventricle, is in Group 2.
The question asks about the physical effect which constitutes the basis on which the X-ray detectors for CT scan operate.
The Doppler effect manifests as a change in the observed frequency of a wave emitted by a source as a result of the relative motion between the source and the observed.
Mass–energy conversion refers to the effect that manifests as nuclear fission or fusion.
The passage describes how detectors absorb transmitted X-rays and generate an electric current, which is a direct application of the photoelectric effect.
The photoelectric effect is a phenomenon in which electrons are released from the surface of a metal when exposed to light. These emitted electrons are referred to as photoelectrons denoted by e−. The current produced as a result of the ejected electrons is called photoelectric current.
Both the emission of photoelectrons and their kinetic energy depend on the frequency of the incident light. The process by which photoelectrons are ejected from the metal’s surface due to exposure to light is known as photoemission.
This effect occurs because electrons in the metal’s surface absorb energy from the incoming light. When this absorbed energy is sufficient, the electrons can break free from the attractive forces holding them to the metal’s nuclei.
The Venturi effect refers to the measurement of volume flow rate or flow speed of a fluid based on the difference of static pressure in a tube with variable diameter.
The question asks about the effect of a temperature increase on the dimensions of the opening through which the X-rays for CT scan pass.
A temperature increase will cause thermal expansion of the copper plate, leading to an increase in both the width (D = 1.0 μm) and length (L = 10.0 μm) of the opening through which the X-rays pass. Since the expansion is proportional to the initial dimension, the longer dimension (L) will expand more than the shorter dimension (D).
The linear increase in the length of the opening is ΔL = Lα∆T and in the width is ΔD = Dα∆T.
ΔL = Lα∆T
where:
ΔL is the change in length due to temperature,
α is the coefficient of linear expansion of copper (1.7 × 10−5 °C−1),
L is the initial length of the dimension,
ΔT is the temperature change.
ΔD = Dα∆T
where:
ΔD is the change in length due to temperature,
α is the coefficient of linear expansion of copper (1.7 × 10−5 °C−1),
D is the initial length of the dimension,
ΔT is the temperature change.
We can see that ΔL > ΔD as L > D. Since L for the length (10.0 μm) is ten times larger than that for the width (1.0 μm), the absolute expansion in length will be ten times greater than the expansion in width, assuming uniform heating.
This implies the length is smaller than the width before the temperature increase, whereas the passage states that width D = 1.0 μm and length L = 10.0 μm.
This implies the length is equal to the width before the temperature increase, whereas the passage states that width D = 1.0 μm and length L = 10.0 μm.
Both L and D increase because of thermal expansion (the tendency of matter to increase in length, area or volume, changing its size and density, in response to an increase in temperature.
This question asks about the minimum radiation energy absorbed in two seconds by 20 grams of heart tissue that is associated with the formation of plaques.
To determine the minimum radiation energy absorbed in two seconds by 20 grams (0.020 kg) of heart tissue, we use the definition of Gray (Gy):
1 Gy = 1 J/kg
Given that the absorbed dose rate is 0.04 Gy/s, the time duration is 2 seconds, and the mass of the heart tissue is 0.020 kg, we first calculate the total absorbed dose using the formula:
Total Dose = (Dose rate) × (Time)
Substituting the values, we get:
(0.04 Gy/s) × (2 s) = 0.08 Gy
Since 1 Gy corresponds to 1 J/kg, the total energy absorbed (E) is given by:
E = Total Dose × Mass
E = (0.08 J/kg) × (0.020 kg) = 0.0016 J = 1.6 × 10−3 J
Therefore, the minimum radiation energy absorbed in two seconds by 20 grams of heart tissue is 1.6 × 10−3 J (1.6 mJ).
This implies the exposure time is 26 s instead of 20 s.
This is the energy absorbed by 17 grams of tissue.
Given that the absorbed dose rate is 0.04 Gy/s, the time duration is 2 seconds, and the mass of the heart tissue is 0.020 kg, we first calculate the total absorbed dose using the formula:
(0.04 Gy/s) × (2 s) = 0.08 Gy.
E = (0.08 J/kg) × (0.020 kg) = 0.0016 J = 1.6 × 10−3 J.
Therefore, the minimum radiation energy absorbed in two seconds by 20 grams of heart tissue is 1.6 × 10−3 J (1.6 mJ).
This implies the plaques were formed in the left ventricle when the absorbed dose per unit time exceeded 0.04 Gy/s.
The question asks about the classification of neutral atoms of the chemical element found in plaques in the left ventricle based on its magnetic properties.
As mentioned in the paragraph, the chemical element found in plaques is calcium. To classify calcium based on its magnetic properties, we need to analyze its electronic structure and how it behaves in the presence of a magnetic field. Calcium (Ca, atomic number 20) has the ground-state electronic configuration: Ca:[Ar]4s2,
where [Ar] represents the electron configuration of argon (1s22s22p63s23p6):
The two outermost electrons occupy the 4s orbital. The magnetic behavior of an atom is determined by the presence or absence of unpaired electrons. Diamagnetic atoms have all paired electrons, meaning they do not generate a net magnetic moment and are repelled by a magnetic field, while paramagnetic atoms have one or more unpaired electrons that generate a net magnetic moment and are attracted to a magnetic field. Since calcium in its ground state ([Ar]4s2) has all paired electrons, it is diamagnetic under normal conditions.
Calcium has the electronic structure [Ar] 4s2. The presence of the empty d orbital allows an electron from the last occupied s orbital to acquire energy in the presence of a magnetic field and occupy the empty d orbital. This unpairs the last electrons and causes paramagnetic properties as opposed to diamagnetic.
Fe, Ni, and Co are the primary ferromagnetic materials of which none are found in the plaques.
Calcium has magnetic properties when exposed to an external magnetic field. Calcium, in its ground state, is generally diamagnetic because all of its electrons are paired. However, when exposed to an external magnetic field, it can exhibit paramagnetic properties due to electronic excitation. This occurs when one of the paired electrons in the 4s orbital gains energy and moves to the empty 3d orbital, resulting in an excited-state configuration of [Ar]4s13d1.
The question suggests that an external magnetic field can excite an electron from the 4s2 orbital to an empty 3d orbital, leading to the configuration [Ar]4s13d1. In this excited state, calcium has two unpaired electrons: one in the 4s orbital and one in the 3d orbital.
The presence of unpaired electrons in this configuration means that calcium exhibits paramagnetic behavior in this state. Paramagnetism arises because unpaired electrons have intrinsic magnetic moments that align with an external magnetic field, causing attraction. In conclusion, in its ground state ([Ar]4s2), calcium is diamagnetic, but in an excited state ([Ar]4s13d1), it has unpaired electrons and becomes paramagnetic. Since the question considers the effect of an external magnetic field, which can excite the electron to the empty 3d orbital, the calcium in plaques is classified as paramagnetic under these conditions.
This question is asking which amino acid stabilizes the [Fe4S4] and [NiFe4S4] clusters in CODH. To answer this, we need to recall our information about amino acids and their side chains. We need to extract information from the passage to find the description of this interaction.
In the second paragraph of the passage, it is stated that “These clusters are held in place within the protein by complexation of four metal centers to thiolate ligands of amino acid residues.”
This tells us that the coordination involves thiolate ligands, which are negatively charged sulfur atoms formed when a thiol group (-SH) loses a proton. A thiolate (–S–) group can act as a strong ligand and bind metal ions such as iron and nickel and stabilize metal clusters within proteins. Therefore, we need to identify an amino acid that contains a thiol group in its side chain and is capable of forming a thiolate.
Cysteine (Cys, C) is an amino acid characterized by its –CH2–SH side chain.
It is unique among amino acids in its ability to deprotonate, forming a thiolate (-S–) group from its thiol (-SH) side chain. This property enables cysteine to strongly interact with metal ions, which plays a role in stabilizing metal clusters.
In addition, the two cysteine (C) residues labeled as C526 and C333 are included in figure 1.
We can, therefore, infer that Choice A is correct.
Glutamate (Glu, E) is an amino acid with a –(CH2)3CO2– side chain. It features a carboxylate (-CO2–) functional group but lacks a thiol (-SH) group. This answer is, therefore, incorrect.
Lysine (Lys, K) possesses a –(CH2)4NH2 side chain, which includes an aliphatic amine (-NH2) functional group. However, it does not contain a thiol (-SH) group. This answer is, therefore, incorrect.
Methionine (Met, M) contains a –CH2–CH2–S–CH3 side chain with a thioether (-S-CH3) functional group. While sulfur is present, it is part of a thioether linkage rather than a free thiol (-SH) and can not form a thiolate. This answer is, therefore, incorrect.
The question asks us to determine the equilibrium constant (K) for Reaction 1. To answer this, we need to apply basic principles of chemical equilibrium and reaction kinetics using information provided in the passage.
The equilibrium constant, K, quantifies the ratio of products to reactants at equilibrium. It is given by the general expression:
K=[products] / [reactants]
However, the equilibrium constant can also be expressed as the ratio of the rate constants of the forward (kf) and backward (kb) reactions:
K=kf / kb
where:
kf is the rate constant for the forward reaction (CO2 reduction to CO and H2O),
kb is the rate constant for the reverse reaction (CO oxidation back to CO2).
From the first paragraph in the passage, we are given the rate constants:
Forward reaction: kf=12 s−1 (CO2 → CO + H2O)
Backward reaction: kb= 40,000 s−1 (CO + H2O → CO2)
We just need to use the following formula and check for the correct answer among the given choices.
K=kf / kb
If we make the calculation correctly, we will substitute as follows:
K=12 / 40,000
K=0.0003
This is the answer given in Choice A. Accordingly, we can say that Choice A is correct.
While the numerical value may be correct, this choice includes units, which is inappropriate. Equilibrium constants are dimensionless quantities because they are ratios.
This represents the reciprocal of the equilibrium constant for Reaction 1, as it was determined using K=kb / kf instead of the correct formula, K=kf / kb. That expression does not apply here, as it reverses the roles of product and reactant.
This value is the inverse of the actual equilibrium constant for Reaction 1, and equilibrium constants should always be expressed as dimensionless quantities.
The question requires that we determine the volume of CO2 gas captured by a 1 μg (0.015 nmol) CODH crystal under standard conditions, assuming ideal gas behavior. In order to answer this question, we need to use information from the passage as well as our understanding of the ideal gas law.
In order to calculate the volume of CO2 gas captured by the enzyme, we need to first determine the number of moles of CO2 molecules bound by the CODH crystal. Then, we use the ideal gas law to convert the amount of CO2 (in moles) into a gas volume under standard conditions.
There is one important aspect mentioned in the second paragraph of the passage, which states:
“In the CODH from C. hydrogenoformans in E. coli, the enzyme structure includes three [Fe4S4] clusters and two [NiFe4S4] clusters.” This tells us that each CODH enzyme contains exactly two [NiFe4S4] clusters, and these are the relevant metal centers for CO2 binding. Later in the passage, it is stated that:
“The active sites in these states exhibit a 1:1 ratio of Ni/CO2 and Ni/NCO–,”
which means that each Ni center binds exactly one CO2 molecule. Therefore, each CODH enzyme molecule binds two CO2 molecules, corresponding to the two nickel centers present.
Let’s make the required calculations accordingly.
Given that the total amount of CODH in the crystal is 1.5 × 10-11 mol (0.015 nmol), the total number of CO2 molecules bound can be determined by multiplying this by two, resulting in 3.0 × 10-11 mol CO2.
To determine the volume of CO2 gas captured by the CODH crystal under standard conditions, we use the ideal gas law: PV = nRT. Solving for volume, we get:
V = (nRT) / P
At Standard temperature and pressure (STP), P = 1 atm (standard pressure), n = 1 mol, R = 0.08206 L·atm/mol·K (ideal gas constant), and T = 273 K (standard temperature).
To calculate the volume of this amount of gas under standard temperature and pressure (STP), we use the fact that one mole of an ideal gas occupies 22.4 L at STP.
Now, using the ideal gas law and the fact that one mole of gas occupies 22.4 L at standard conditions, the volume of CO2 is calculated as follows:
V = (22.4 L/mol) × (3.0 × 10-11 mol) = 67.2 × 10-11 L = 6.72 × 10-10 L = 6.72 × 10-7 mL = 6.72 × 10-4 μL = 0.672 nL.
This value would be obtained if only one Ni center per CODH was assumed to bind CO2, rather than both. Since the passage states there are two [NiFe4S4] clusters per enzyme, each binding one CO2 molecule, this answer underestimates the number of moles of CO2 by a factor of 2. This answer is incorrect.
If we correctly calculate using the correct molar amount of CODH (1.5 × 10-11 mol), the 2:1 CO2-to-enzyme ratio (due to two Ni centers per CODH), and the molar volume of an ideal gas at STP (22.4 L/mol).
The correct volume is:
V=0.672 nL
This answer is correct.
This value comes from using a molar volume of 24.2 L/mol instead of 22.4 L/mol, which corresponds to room temperature (298 K) rather than standard temperature. This overestimates the volume and is incorrect.
This value results from assuming that all five [Fe4S4] clusters in CODH contain a Ni atom, whereas in reality, only two of the five clusters are [NiFe4S4] clusters. This answer is incorrect.
This question asks about the substance that must be added to 100 mL of 0.1 M isocyanic acid (HNCO) to ensure that 99.9% of it exists as the conjugate base (NCO–). To answer this question, we need to use the passage and apply acid–base equilibrium principles.
In the second paragraph of the passage, it stated that “At pH 8, the pKa value of isocyanic acid (HNCO) is 3.5.” This means that when the pH = 3.5, the concentrations of HNCO and its conjugate base NCO– are equal. But we are told to achieve 99.9% conjugate base, which corresponds to a [A–]/[HA] ratio of 999:1. We can apply the Henderson–Hasselbalch equation:
Since we require [NCO−] / [HNCO]=999, we get:
pH=3.5+log(999)≈3.5+3=6.5
To shift the equilibrium toward 99.9% NCO–, the pH must be raised significantly above the pKa of 3.5—specifically, to about pH 6.5 or higher. This can only be achieved by adding a strong base that effectively removes protons from HNCO and generates its conjugate base.
Adding a strong acid will lower the pH of the solution. Since the pKa of isocyanic acid is 3.5, lowering the pH will increase the concentration of the protonated form (HNCO) and reduce the formation of the conjugate base (NCO–). This directly opposes the condition described in the question, which requires that 99.9% of the isocyanic acid be in the conjugate base form.
The pH needs to be significantly higher than the pKa of isocyanic acid to drive the equilibrium toward the conjugate base. Adding a strong base raises the pH and results in the deprotonation of HNCO, forming NCO–. At approximately pH 6.5, 99.9% of the acid will exist as the conjugate base. This satisfies the condition given in the question, so this is the correct answer.
A weak acid like acetic acid has a pKa of around 4.6, which is slightly higher than that of isocyanic acid. However, adding a weak acid will not raise the pH and is more likely to lower it. Lowering the pH shifts the equilibrium toward the protonated form of isocyanic acid, not the conjugate base. Therefore, this choice does not result in 99.9% of HNCO being present as NCO–.
A buffer at pH 3.0 is below the pKa of isocyanic acid. At this pH, the majority of the isocyanic acid will remain in its protonated form. This prevents conversion to the conjugate base and does not meet the requirement that 99.9% of the acid be deprotonated.
This question asks about the heterocycle present in the side chains of residues 93 and 261, which helps stabilize the carbonite ion (CO22-) coordinated to cluster C in CODH. To answer this, we need to interpret Figure 1 and use our knowledge of amino acid structures and heterocycles.
In the figure, residues labeled H93 and H261 are clearly involved in interacting with the bound CO2 species.
A heterocycle is a ring structure that includes atoms other than carbon in the ring, most commonly nitrogen in biological molecules. By examining the side chains of these residues, we see a five-membered ring with two nitrogen atoms, which identifies the ring as an imidazole. The imidazole ring, with nitrogens at positions 1 and 3, is a five-membered aromatic heterocycle found in the side chain of histidine. Histidine is an amino acid commonly involved in metal coordination and enzymatic catalysis. In CODH, unprotonated imidazole can act as an N-donor ligand to Fe2+ in the NiFe4S4 cluster, which stabilizes the bound CO22- ion.
Protonated imidazole serves as a strong hydrogen bond donor, stabilizing one of the oxygen atoms in carbonite (CO22-), ensuring that the molecule remains bound in the active site. This interaction prevents the premature release of CO22- and facilitates its reduction in the enzyme.
Pyridine is a six-membered aromatic ring containing one nitrogen atom, but it does not occur in the side chain of any natural amino acid. It cannot be the heterocycle observed at residues 93 and 261 and, therefore, cannot be responsible for stabilizing the bound CO22-.
Indole is the bicyclic heterocycle found in tryptophan. While it contains a nitrogen atom, its geometry and electron distribution make it less suitable for metal coordination. Tryptophan is not involved at these residues, and indole is not the ring shown in the figure.
This is the correct answer. The five-membered ring with two nitrogen atoms seen at H93 and H261 is an imidazole, characteristic of histidine. It stabilizes CO22- through metal binding and hydrogen bonding, consistent with both the structure and function described in the passage.
Furan is a five-membered aromatic ring with an oxygen atom. It is not present in any standard amino acid side chain and does not participate in metal coordination in biological systems, making it irrelevant in the context of this enzyme.
This question asks us to identify which carboxylic acid is formed after the diol tautomerizes along the reduction path from CO2 to CO. To answer this, we need to use the passage and our understanding of organic intermediates and tautomerization reactions.
The passage states that “The bound CO22- (carbon dioxide dianion, or carbonite) and H2NCO2– (carbamoyl dianion) fragments suggest the initial formation of dihydroxycarbene (HO–C–OH) upon CO22- protonation, which tautomerizes into a carboxylic acid before CO release.” This is the part of the passage that guides us on the intermediate transformation. The question refers specifically to the tautomerization of the diol, and the diol mentioned is HO–C–OH, which is dihydroxycarbene.
Tautomerization refers to the intramolecular rearrangement of bonds, typically involving proton transfer and formation of a new functional group. In this case, tautomerization of dihydroxycarbene (HO–C–OH) involves the conversion of one hydroxyl group into a carbonyl, yielding a carboxylic acid.
So, the tautomerization of HO–C–OH leads to the formation of formic acid (HCOOH). This is consistent with the one-carbon structure and matches the passage’s indication that this intermediate is then further reduced to carbon monoxide (CO).
The reaction pathway is:
HO–C–OH → HCOOH → CO
Thus, the carboxylic acid formed after tautomerization is formic acid.
Acetic acid contains two carbon atoms, with the structure CH3COOH. The intermediate described in the passage, dihydroxycarbene (HO–C–OH), contains only one carbon atom, so it cannot form a two-carbon carboxylic acid. Acetic acid is not a plausible product of this tautomerization step. This answer is incorrect.
Carbamic acid has the structure H2N–COOH and results from a different pathway involving an amine group. The passage does mention H2NCO2– (carbamoyl dianion), but that refers to a different pathway. The tautomerization described in this question starts from the diol (HO–C–OH), not from a nitrogen-containing species. Therefore, carbamic acid is not the correct product of the tautomerization being asked about.
Chloroformic acid (ClCOOH) is a halogen-substituted carboxylic acid and requires the presence of a chlorine atom. There is no indication in the passage of any chlorine-containing species or substituents. It is unrelated to the CO2 reduction mechanism and the intermediates described, so this answer is not correct.
Formic acid (HCOOH) is a one-carbon carboxylic acid and is logically formed from the tautomerization of dihydroxycarbene (HO–C–OH), which also contains one carbon. This transformation is explicitly described in the passage, and formic acid would then be further reduced to CO, completing the pathway. This is the correct answer.
The question requires that we determine which class of enzymes lysozymes belong to. To answer this, we need to use our understanding of enzyme classification and the specific biochemical function of lysozyme.
Lysozyme is classified as a glycosidase, which is a type of hydrolase enzyme. Glycosidases catalyze the hydrolysis of glycosidic bonds—the covalent linkages between carbohydrate units or between a sugar and another molecule. These enzymes break down complex carbohydrates by cleaving these bonds with the addition of water.
In the case of lysozyme, they break the components of the peptidoglycan polymer found in bacterial cell walls. This cleavage weakens the structural integrity of the bacterial wall, leading to cell lysis and bacterial death. Due to this antimicrobial function, lysozyme is commonly found in tears, saliva, mucus, and egg whites, where it serves as a natural defense mechanism against bacterial infection.
This is the correct answer. Hydrolases are enzymes that catalyze bond cleavage via the addition of water. Lysozyme is a glycosidic hydrolase that breaks glycosidic bonds in bacterial peptidoglycan. Therefore, lysozyme falls squarely into this enzyme class.
Isomerases catalyze structural rearrangements within a molecule, converting it from one isomer to another. Lysozyme does not change the configuration or structure of a molecule without breaking bonds. Instead, it cleaves glycosidic bonds, so this classification is incorrect.
A lyase catalyzes the addition of a molecule to a substrate to break a multiple bond, or the elimination of a molecule from a substrate to create a multiple bond. This class of enzymes operates without the use of water or redox cofactors. In contrast, lysozyme catalyzes the hydrolysis of a glycosidic bond between sugar residues in the bacterial cell wall, using water as a reactant to break the bond. There is no formation or cleavage of double bonds involved, and the reaction mechanism is not consistent with the function of a lyase. Therefore, lysozyme cannot be classified as a lyase.
Oxidoreductases catalyze redox reactions, transferring electrons between molecules. Lysozyme does not participate in electron transfer; it breaks glycosidic bonds via water-mediated hydrolysis. Thus, this class does not apply.
This question asks us to identify which enzyme catalyzes a reaction that directly activates or deactivates enzyme activity through covalent enzyme modification. To solve this, we need to recall how enzymes are regulated. Enzyme regulation can occur through non-covalent interactions, such as allosteric binding or feedback inhibition, or through covalent modification, which involves the addition or removal of functional groups like phosphate, acetyl, or methyl groups.
Covalent modification can lead to the activation or inactivation of the enzyme being modified. A common and well-characterized type of covalent modification is phosphorylation, where a phosphate group is added to specific amino acid residues—usually serine, threonine, or tyrosine—on the target enzyme. This phosphorylation event is catalyzed by a specific class of enzymes.
Let us now look at the answer choices to determine which one fits this role.
Adenylyl cyclase catalyzes the conversion of ATP into cyclic AMP (cAMP), which serves as a second messenger in signaling pathways. While cAMP can regulate enzyme activity, it does so indirectly by activating proteins like protein kinase A (PKA) rather than modifying enzymes itself. Since adenylyl cyclase does not directly phosphorylate or modify enzymes, it is not the correct answer.
ATP synthase catalyzes the formation of ATP from ADP and inorganic phosphate (Pi) during oxidative phosphorylation. ATP is an energy carrier and can influence enzyme activity, but this effect is not due to covalent modification—it is instead through allosteric regulation or ATP-dependent phosphorylation by kinases.
Glycogen phosphorylase catalyzes the breakdown of glycogen by adding a phosphate group to release glucose-1-phosphate. However, it does not covalently modify enzymes; rather, it modifies glycogen itself. Glycogen phosphorylase is regulated by phosphorylation, but it does not catalyze phosphorylation of other enzymes, meaning it is not responsible for direct enzyme activation or deactivation.
This is the correct answer. Protein kinases catalyze the transfer of a phosphate group from ATP to specific amino acid residues (usually serine, threonine, or tyrosine) on target proteins, including enzymes. This covalent phosphorylation can activate or deactivate enzyme activity depending on the target and the cellular context. This is a direct regulatory mechanism, making protein kinase the only enzyme in this list that fits the description in the question.
This question asks about the magnitude of vector S, where Vector A has the reciprocally orthogonal components Ax = +3, Ay = +4, and vector B has Bx = –6, By = –8.
To find the magnitude of vector S, which is the sum of vectors A and B, we apply vector addition: the x-components and y-components of the two vectors are added separately to get the components of the resultant vector.
Vector A has components Ax = +3 and Ay = +4, while vector B has components Bx = -6 and By = -8.
By adding the corresponding components, we find that:
Sx = Ax + Bx = 3 + (–6) = –3
Sy = Ay + By = 4 + (–8) = –4
Finally, to calculate the magnitude of S, we use the Pythagorean theorem:
|S| = √(Sx² + Sy²) = √[ (–3)² + (–4)² ] = √(9 + 16) = √25 = 5.
This is the correct result from applying vector addition and using the Pythagorean theorem to calculate the magnitude of the resultant vector.
This is the result of the calculation −(Bx + By + Ax + Ay) = −(−6 − 8 + 3 + 4) = 7. This is an incorrect calculation.
This is the result of the calculation AxBx − AyBy = −18 − (−32) = 14. This is an incorrect calculation.
This is the result of the calculation Ax + Ay − (Bx + By) = 7 − (−14) = 21. This is an incorrect calculation.
This question asks about the propagation speed of the transverse wave that propagates along the x-axis.
To determine the propagation speed of the transverse wave given by the equation: y(x, t) = 24 × sin[2π(x / 8 − t / 2)] we analyze its wave characteristics.
A traveling wave moving in the positive x-direction can be expressed in the standard form:
y(x, t) = A × sin[2π(x / λ − t / T)]
where:
A is the amplitude (24 cm in this case)
λ is the wavelength (distance over which the wave repeats)
T is the period (time for one complete oscillation)
v is the wave speed, defined as v = λ / T
By comparing the given equation with the standard wave equation, we identify:
λ = 8 cm (from the term x / 8)
T = 2 s (from the term t / 2)
Using the wave speed formula: v = λ / T = 8 cm ÷ 2 s = 4 cm/s
This is the wave amplitude to which units of speed are associated. The amplitude of the wave is 24 cm, which represents the maximum displacement of the wave from its equilibrium position. However, speed is a measure of how fast the wave propagates, and its units should be cm/s, not just cm. The wave speed is calculated using the wavelength and period, not the amplitude.
This is consistent with the calculation: 24 cm ÷ 2 s = 12 cm/s.
The wave speed formula is: v = λ / T
where λ = 8 cm, T = 2 s
Substituting in: v = 8 cm ÷ 2 s = 4 cm/s
The given calculation, 24 cm ÷ 2 s = 12 cm/s, is incorrect because 24 cm is the amplitude, not the wavelength. Amplitude does not determine wave speed.
To determine the propagation speed of the transverse wave given by the equation: y(x, t) = 24 × sin[2π(x / 8 − t / 2)] we analyze its wave characteristics.
A traveling wave moving in the positive x-direction can be expressed in the standard form: y(x, t) = A × sin[2π(x / λ − t / T)]
where: A is the amplitude (24 cm in this case), λ is the wavelength (distance over which the wave repeats), T is the period (time for one complete oscillation), v is the wave speed, defined as v = λ / T
By comparing the given equation with the standard wave equation, we identify: λ = 8 cm (from the term x / 8), T = 2 s (from the term t / 2)
Using the wave speed formula: v = 8 cm ÷ 2 s = 4 cm/s
This is consistent with the calculation: 12 cm ÷ 4 s = 3 cm/s
The correct formula for wave speed is: v = λ / T
The numbers 12 cm and 4 s do not correspond to the given wavelength (8 cm) or period (2 s), so this calculation does not match the correct wave speed formula.
The correct calculation should be: 8 cm ÷ 2 s = 4 cm/s, not 12 cm ÷ 4 s.
This question asks about the compound which is formed instead of Compound 2 if NaBH4 is also added in Step 1, shown in Figure 1.
To answer this question, we must understand the chemical transformation shown in the reaction in Figure 1 of the passage and our knowledge of the reactivity of NaBH4.
In this case, the reaction catalyzed by Fub7 involves the formation of an iminium ion intermediate that forms from the condensation between the primary amine and the aldehyde from the pyridoxal moiety. Normally, this intermediate would proceed to form Compound 2. However, if sodium borohydride (NaBH4) is added, it acts as a reducing agent, specifically reducing the iminium ion to a secondary amine before the reaction can progress to Compound 2. This reduction leads to the formation of an alternative product instead of Compound 2. This means the structure formed in the presence of NaBH4 would have the imine reduced to an amine, but all other features (such as the ring structure and substitution) would remain the same. So we are looking for a compound that is identical to Compound 2 except for the reduction of the imine double bond to a single bond with a protonated nitrogen.
This structure contains the amine from the reduced iminium ion, but the remaining alkyl chain that contained the carboxylate has been removed. The loss of the alkyl chain that included the carboxylate suggests that the reaction has not proceeded correctly with NaBH4 as expected.
This is the correct answer. It contains the same carbon framework as Compound 2 but shows the imine nitrogen reduced to an amine (N–H). This is consistent with the selective reduction by NaBH4. The rest of the molecule, including the carboxylate and phosphate substituents, remains unchanged. This matches the expected product formed when the imine is trapped before conjugation.
This option assumes that the NaBH4 would not reduce the iminium ion, which is not the case. The NaBH4 efficiently reduces the iminium ion, yielding a secondary amine.
The NaBH4 would reduce only one of the double bonds, not both. The iminium ion would be reduced, leaving the terminal double bond.
This question asks about the vitamin that is a cofactor for Fub7 to form Compound 2.
To answer this question, we must recall our knowledge of vitamins and understand the principles of enzyme cofactors and their roles in enzymatic reactions. Many enzymes require cofactors—non-protein molecules that assist in catalysis. One important cofactor is pyridoxal phosphate (PLP), which is derived from vitamin B6 and is essential for the function of many enzymes, particularly those involved in amino acid metabolism.
The passage states that Fub7 is a PLP-dependent enzyme. PLP is the active form of vitamin B6 and serves as a coenzyme in various biochemical reactions, including transamination, decarboxylation, and racemization of amino acids. Since Fub7 requires PLP to catalyze the formation of Compound 2, vitamin B6 must be its precursor.
Vitamin A is not a precursor for any coenzyme. Instead, it is primarily involved in vision, immune function, and cell growth. Unlike vitamin B6, vitamin A does not participate in enzymatic reactions as a cofactor.
Vitamin B5, also known as pantothenic acid, is a precursor for coenzyme A (CoA). CoA is important for fatty acid metabolism and the citric acid cycle but does not play a role in the PLP-dependent enzymatic reaction catalyzed by Fub7.
This is the correct answer. Vitamin B6 is the precursor of pyridoxal phosphate (PLP), the required cofactor for Fub7. PLP is central to many reactions involving amino acids and forms covalent intermediates that facilitate bond rearrangements. Since Fub7 uses PLP to catalyze the formation of Compound 2, the correct vitamin cofactor is vitamin B6.
Vitamin C, or ascorbic acid, functions as an antioxidant and plays a role in collagen synthesis and immune function. However, it is not a precursor for any enzymatic cofactor like PLP. Unlike vitamin B6, it does not contribute to enzymatic reactions in the same way.
This question asks us to determine the structural relationship between Compound 4 and the compound shown in the image. To answer this, we need to use the structure of Compound 4 from the figure in the passage as well as our understanding of isomerism and stereochemistry.
Isomers are compounds with the same molecular formula but different structures. There are different classes of isomers.
First, constitutional isomers have different connectivity between atoms—they differ in the order in which atoms are bonded.
Second, conformational isomers have the same connectivity and the same stereochemistry but differ due to rotation about single bonds—these differences do not involve breaking bonds.
The third major category is stereoisomers, which have the same connectivity but differ in the spatial arrangement of atoms. Stereochemistry shows how atoms are arranged around chiral centers, which are tetrahedral carbon atoms bonded to four different groups. Stereoisomers can be further divided into enantiomers and diastereomers.
Enantiomers are mirror images that are non-superimposable and differ at all chiral centers. Diastereomers also differ at chiral centers, but they are not mirror images and differ at some, but not all stereocenters.
Conformational isomers differ by rotation about a single σ bond. They are different spatial arrangements of the same molecule that can interconvert without breaking bonds—for example, staggered vs. eclipsed conformations. Compound 4 and the compound shown are not related by single-bond rotation; they have different configurations at chiral centers, which means that bonds must be broken to interconvert. Therefore, they are not conformational isomers.
Constitutional isomers have the same molecular formula but differ in the connectivity of atoms. That means the atoms are bonded in a different sequence. The structures in question both compounds have the same bonding pattern and identical functional groups in the same positions. The difference lies only in the spatial arrangement around the chiral centers, which means they have the same connectivity. Therefore, they are not constitutional isomers.
Diastereomers are stereoisomers that are not mirror images and differ at some, but not all, stereocenters. This occurs when there are two or more chiral centers, and only one of them is inverted. In this case, both stereocenters are inverted—Compound 4 has (2S,5R,) and the compound shown has (2R,5S). Since they differ at all stereocenters and are mirror images, they do not qualify as diastereomers.
Compound 4 and the given structure contain two stereocenters, and the absolute configurations at both centers are inverted. If a compound has two chiral centers and both of them switch configuration (e.g., from (2S,5R) to (2R,5S)), the two compounds are enantiomers. Enantiomers are non-superimposable mirror images that differ in configuration at all stereocenters. Based on the fact that compound 4 is (2S,5R), and the structure shown is (2R,5S), both stereocenters are inverted. That satisfies the definition of enantiomers. This is the correct answer.
This question asks about the number of electrons that are transferred when going from Compound 5 to Compound 6.
To answer this question, we must use our knowledge of the principles of oxidation and reduction in organic chemistry.
In general, oxidation is defined as the loss of electrons, while reduction is the gain of electrons. In organic chemistry, specifically, we detect oxidation by the formation of additional bonds to oxygen or the loss of hydrogen atoms, whereas reduction is detected by the opposite—losing bonds to oxygen or gaining hydrogen atoms.
In the transformation from Compound 5 to Compound 6, the most noticeable structural change is the introduction of two new double bonds in the ring.
This choice would imply that Compound 5 was reduced to Compound 6. However, this is incorrect because Compound 6 contains more double bonds and has lost hydrogen atoms. The formation of double bonds and the loss of hydrogen atoms are characteristic of oxidation, not reduction. If Compound 6 had been formed by reduction, we would expect the addition of hydrogen atoms and the removal of double bonds instead.
This is the correct answer. The formation of double bonds typically occurs through oxidation, where electrons are removed from the molecule. Since each π bond involves two electrons, the formation of two new double bonds corresponds to the loss of four electrons. Therefore, the total number of electrons transferred during the conversion from Compound 5 to Compound 6 is four, which is consistent with a four-electron oxidation process.
This choice suggests that Compound 5 undergoes reduction, but as explained above, this does not match the structural changes observed. The formation of new double bonds and the apparent loss of hydrogen atoms clearly indicate an oxidation process. Therefore, this choice is incorrect.
This choice proposes that only two electrons are lost, but that would correspond to the formation of a single double bond. In the actual transformation, two new double bonds are formed, which would require four electrons to be lost. Thus, this underestimates the extent of the oxidation and does not account for the full transformation observed from Compound 5 to Compound 6.
This question asks for the IUPAC name of the product formed when Compound 10 undergoes the Fub8-catalyzed reaction. To answer it, we need to carefully read the passage and use Table 1, which provides the structure of the R group for each substrate. The reaction catalyzed by Fub8 converts a carboxylic acid to an aldehyde, and from the table, we see that Compound 10 has an R group of HC≡C(CH2)4⁻. This means the carbon backbone contains a total of seven carbons when we include the aldehyde carbon: one carbon from the aldehyde (C1), followed by four CH2 units (C2–C5), and then two additional carbons from the alkyne (C6 and C7), with a triple bond between C6 and C7.
Using IUPAC nomenclature rules, we begin numbering from the aldehyde carbon, which must be carbon 1, since the aldehyde functional group takes naming priority. This gives us a seven-carbon chain, so the base name is hept-. The triple bond is between carbon 6 and 7, so we indicate the position of the triple bond with -6-yne, and the suffix for the aldehyde is -al.
Hept-6-enal is incorrect because the infix -en- refers to a double bond, not a triple bond. The R group in Compound 10 contains a triple bond, so it should be named with -yne, not -ene.
This is the correct answer. Hept-6-ynal indicates a seven-carbon chain (hept-) with a triple bond beginning at carbon 6 (-6-yn-) and an aldehyde group at carbon 1 (-al). This matches the structure of the product formed when Compound 10 undergoes Fub8-catalyzed conversion to the aldehyde.
Hex-5-ynal is incorrect because it refers to a six-carbon chain (hex-), which is too short for the actual seven-carbon structure of Compound 10. This neglects the carbon atom that carries the aldehyde group. It also incorrectly places the triple bond at position 5 instead of 6.
Oct-7-ynal is incorrect because it refers to an eight-carbon chain (oct-), which is one carbon too long. It also places the triple bond at position 7, which would place it at the very end of the chain and is inconsistent with the given R group.
This question asks about the compound in Table 1, which provides the highest catalytic efficiency for Fub8, assuming product release is rate determining. To understand this question, we must understand the principles of enzyme kinetics, specifically catalytic efficiency, which is determined by the ratio kcat/KM.
The catalytic efficiency of an enzyme describes how effectively it converts a substrate into a product. The higher the kcat/KM value, the more efficiently the enzyme processes the substrate. The kcat (turnover number) represents how many substrate molecules are converted to product per unit time by one enzyme molecule, and the KM (Michaelis constant) indicates the substrate concentration at which the reaction proceeds at half of its maximum rate. A low KM means high binding affinity, and a high kcat means fast turnover. Together, a high kcat/KM ratio reflects an enzyme that binds well and reacts quickly. Since product release is rate-determining, we are focusing on how fast and efficiently the substrate is processed up to the point of product formation.
From Table 1, we calculate the catalytic efficiency (kcat/KM) for each substrate:
Compound 7: 0.5 / 0.03 = 16,700 M−1s−1
Compound 8: 0.012 / 0.03 = 400 M−1s−1
Compound 9: 0.019 / 0.016 = 1,188 M−1s−1
Compound 10: 0.026 / 0.19 = 137 M−1s−1
Compound 7 has kcat = 0.5 s−1 and KM = 0.03 mM, giving a kcat/KM value of 16,700 M−1s−1. This is the highest catalytic efficiency among the compounds listed, indicating that Compound 7 is the most effective substrate for Fub8.
Compound 8 has kcat = 0.012 s−1 and KM = 0.03 mM, giving a kcat/KM value of 400 M−1s−1. This value is significantly lower than that of Compound 7, making it a less efficient substrate.
Compound 9 has kcat = 0.019 s−1 and KM = 0.016 mM, giving a kcat/KM value of 1,188 M−1s−1. Although this is the second-highest value, it is still much lower than the value for Compound 7.
Compound 10 has kcat = 0.026 s−1 and KM = 0.19 mM, giving a kcat/KM value of 137 M−1s−1. This is the lowest catalytic efficiency among the compounds and indicates the least favorable substrate for Fub8.
This question asks us to determine the electron configuration of the metal ion used to obtain the 1H NMR spectrum. To answer this, we must first identify the metal ion mentioned in the passage and then apply the principles of electron configurations for ions in transition metals.
In the first paragraph of the passage, it is stated: “Analysis of the 1H NMR spectrum for the Cu2+–Compound 2 complex indicated that the enol proton at 13.58 ppm is absent…” This tells us directly that the NMR spectrum in question involves Cu2+, so we are being asked for the electron configuration of the copper(II) ion.
Copper has an atomic number of 29 as per the supplied periodic table. Its ground-state electron configuration is not what we might initially expect from a strict application of the Aufbau principle. According to the Aufbau order, electrons should fill 4s before 3d, which would predict [Ar] 4s2 3d9.
However, copper actually adopts the configuration [Ar] 4s1 3d10 because a fully filled 3d subshell provides additional stability, making it energetically more favorable than a half-filled 3d9 with a full 4s2.
When forming Cu2+, copper loses two electrons. Electrons are always removed first from the highest principal quantum number, not necessarily from the last orbital filled. In this case, the 4s electron (n = 4) is removed before any 3d electrons (n = 3), even though 4s was filled first. So the first electron is removed from 4s1, and the second is removed from the 3d10 subshell.
This is correct. Cu2+ has lost one electron from the 4s orbital and one from the 3d orbital, resulting in [Ar] 3d9, which matches the electron configuration expected for copper(II).
This configuration refers to a different block of the periodic table. It is not associated with copper, which is a 3d transition metal.
This configuration, [Ar] 4s13d8, would imply that two electrons were removed from the 3d subshell without removing the 4s electron first. That contradicts the rule that electrons are lost first from the highest principal energy level, which in this case is the 4s orbital.
This configuration corresponds to elements in the 4d block with a 5s orbital, such as ruthenium or rhodium, not to copper. Cu is part of the 3d transition metals and does not have a 5s orbital occupied in its ground state.
This question asks about the amount of Co2+–Compound 2 that was added to the agar plate. To answer this, we need to use information from the passage and apply a basic unit conversion.
The passage states, “The test solutions were prepared by dissolving the compounds in DMSO (100 μg/mL) and adding 20 μL to an agar plate…” This tells us that the solution had a concentration of 100 μg/mL, and the volume applied to the agar plate was only 20 μL, which is equal to 0.020 mL.
To calculate the mass added, we use the formula:
mass = volume × concentration
= 0.020 mL × 100 μg/mL
Note that the value 100 μg/mL means that in every 1 mL of solution, there are 100 micrograms of compound. So when only 0.020 mL is added, it contains a proportionally smaller amount.
This value, 0.02 μg, is incorrect. It likely results from a miscalculation, such as a conversion error.
This value, 0.2 μg, contains an error in converting between metric units or may result from using 2 μL instead of 20 μL.
This is the correct answer. The passage states that 20 μL (which is 0.020 mL) of a 100 μg/mL solution was applied to the agar plate. Using the equation:
mass = volume × concentration = 0.020 mL × 100 μg/mL = 2 μg.
This value, 20 μg, likely results from mistaking the volume 20 μL for the mass, assuming that 20 μL directly equals 20 μg without using the correct concentration. However, at 100 μg/mL, 20 μL only corresponds to 2 μg, not 20 μg.
This question asks for the reagent that is needed to produce the compound shown from Compound 1 in the presence of pyridine. To answer the question, we need to compare the structure of Compound 1 with the structure of the product and also use our knowledge of acylation reactions in organic chemistry.
Acylation reactions are transformations that introduce an acyl group (RCO–) onto a nucleophile, such as an alcohol or amine. In this case, the phenol –OH groups in Compound 1 are the nucleophiles that will react. The typical reagents used for this purpose are acid chlorides (RCOCl) and acid anhydrides, which are good acyl donors because of their reactive carbonyl groups and good leaving groups (Cl– or –OCOR).
The passage tells us that pyridine is present. Pyridine is a mild base and nucleophilic catalyst that is commonly used to neutralize HCl, the byproduct of reactions involving acid chlorides. Its presence confirms that an acid chloride is probably being used. Additionally, the reference to a neutralization workup suggests that excess acid or base is being removed at the end.
Therefore, we are looking among the answer choices for a reagent that is an acyl donor, has a reactive carbonyl group, and can acylate the phenol groups in Compound 1 in the presence of pyridine.
Methyl chloride is not an acylating agent. It lacks a carbonyl group, which is required for transferring an acyl (RCO–) group. It is simply an alkyl halide (CH3–Cl) and cannot form the product shown in the question. This means it cannot produce the transformation shown, so this choice is incorrect.
This is correct. Acetyl chloride (CH3COCl) is an acid chloride. It contains a reactive carbonyl group and a chloride leaving group, allowing it to acylate phenols effectively. In the presence of pyridine, the phenol –OH groups in Compound 1 react with acetyl chloride to form acetate esters, and pyridine neutralizes the HCl byproduct. This yields the desired product shown in the question, where both phenolic –OH groups have been converted into acetyl esters (–OCOCH3).
This is incorrect. The molecular formula corresponds to acetone (CH3COCH3), which is a ketone, not an acid chloride. Ketones are not acylating agents because they lack a leaving group and cannot directly transfer an acyl group to phenol.
This is incorrect. This molecule is chloroacetone, which contains both a halogen and a carbonyl. If reacted with a phenol, it would yield an undesired product—specifically an –OCH2COCH3 group, not the product shown in the question. Therefore, it is not suitable for this transformation.
This question asks about the region of the electromagnetic spectrum in which Compound 2 absorbs.
To answer this question, we must refer to the passage and use our information about the principles of the electromagnetic spectrum and how different wavelengths of light correspond to different regions of absorption.
The electromagnetic spectrum consists of various types of electromagnetic radiation, categorized based on wavelength or frequency. When a compound absorbs light at a specific wavelength, it typically falls within one of these defined regions.
The infrared region of the electromagnetic spectrum extends from 750 nm to 0.01 cm. Infrared absorption is typically associated with vibrational transitions of molecules, often studied using infrared spectroscopy (IR spectroscopy). Since Compound 2 absorbs at 423 nm, it is outside the infrared range and does not correspond to vibrational transitions.
The visible region of the electromagnetic spectrum spans wavelengths from 400 nm to 750 nm in air. According to the first paragraph in the passage, Compound 2 exhibits a strong absorption band at 423 nm, which falls within this range. Absorption in the visible region means that Compound 2 interacts with light that the human eye can perceive, which may result in a visible color. This answer correctly identifies the spectral region.
The ultraviolet (UV) region of the electromagnetic spectrum ranges from 100 nm to 400 nm. UV absorption typically involves electronic transitions of molecules, often observed in UV-Vis spectroscopy. While compounds with UV absorption often have conjugated systems, the absorption of Compound 2 at 423 nm places it just beyond the UV range into the visible spectrum.
The X-ray region spans 0.1 nm to 10 nm. X-ray absorption is associated with high-energy transitions of core electrons in atoms, typically used in techniques such as X-ray crystallography. Since Compound 2 absorbs at 423 nm, which is far outside the X-ray region, this answer is not applicable.
This question asks us to identify which metal–Compound 2 complex is most effective at inhibiting the growth of B. subtilis. To answer this, we need to examine Table 1 in the passage and understand what is meant by “zone of inhibition.”
The second paragraph of the passage explains that “To test the antimicrobial activity of the metal–Compound 2 complexes, test solutions were prepared by dissolving the compounds in DMSO (100 μg/mL) and adding 20 μL to an agar plate loaded with the bacterium Bacillus subtilis (B. subtilis). The zone of inhibition was measured and compared to the known antibiotic drug amikacin to gauge antimicrobial activity. A blank test showed that DMSO did not inhibit bacterial growth.”
We can understand from this that the zone of inhibition refers to an area on the agar plate where the bacteria could not grow, and the larger this zone is, the more effective the compound must be at stopping bacterial growth. The name “zone of inhibition” also helps us interpret this because it suggests a region where bacterial growth is being inhibited. Since DMSO by itself had no effect on bacterial growth, we know that the observed inhibition is due to the metal–Compound 2 complex. That means we can directly compare the sizes of the inhibition zones in Table 1 to determine which compound was the most effective. The greater the inhibition zone im mm, the more effective it is.
This complex had a 13 mm zone of inhibition. While this indicates some level of antimicrobial activity, it is less effective than the Zn2+ complex, which produced a significantly larger inhibition zone.
This complex had a 15 mm zone of inhibition, which is slightly more effective than Co2+–Compound 2, but it is still less effective than Zn2+–Compound 2, which exhibited a 23 mm zone.
This complex showed 0 mm zone of inhibition, which means it exhibited no detectable antimicrobial activity against B. subtilis. It was the least effective among the tested metal complexes.
According to Table 1, the Zn2+–Compound 2 complex produced the largest zone of inhibition, measuring 23 mm. This indicates it was the most effective antimicrobial agent among all the tested complexes.
This question asks about the E/Z designations for Compound 1.
To answer this question, we must refer to the structure in Figure 1 and use our knowledge of the principles of E/Z (cis/trans) isomerism in alkenes. The E/Z system is used to describe the relative positions of the highest-priority substituents around a double bond using Cahn-Ingold-Prelog priority rules. If the highest-priority groups on each carbon of the double bond are on opposite sides, the alkene is assigned the E (entgegen, “opposite”) designation. If the highest-priority groups are on the same side, the alkene is assigned the Z (zusammen, “together”) designation.
(1Z, 6Z) is incorrect because this would mean that both double bonds have their highest-priority substituents on the same side, which is not the case for Compound 1. Instead, in Compound 1, these substituents are on opposite sides, meaning the double bonds should be E, not Z.
(1Z, 6E) is incorrect because while the double bond between C6 and C7 is correctly assigned E, the double bond between C1 and C2 should also be E, not Z. The highest-priority substituents at C1 and C2 are on opposite sides, requiring the E designation.
(1E, 6Z) is incorrect because while the double bond between C1 and C2 is correctly assigned E, the double bond between C6 and C7 should also be E, not Z. The highest-priority substituents at C6 and C7 are on opposite sides, requiring the E designation.
Both double bonds in Compound 1—between C1 and C2, and C6 and C7—have their highest-priority substituents on opposite sides, giving them the E designation.
Therefore, this answer is correct.
This question asks about the approximate chemical shift of the CH3 hydrogen atoms in Compound 1. To answer this question, we must refer to the structure in Figure 1 and use our knowledge of the principles of proton nuclear magnetic resonance (1H NMR) spectroscopy.
NMR is an analytical technique used to determine the electronic environment of hydrogen atoms in a molecule. It helps us learn more about the chemical structure and functional group positioning, which allows us to distinguish among different types of hydrogen atoms based on their surroundings.
A chemical shift (δ, in ppm) in NMR tells us where a particular hydrogen atom appears on the spectrum. The position depends on how shielded or deshielded the hydrogen is by surrounding electrons. Shielded means that the hydrogen nucleus is surrounded by a relatively high electron density, protecting it from the external magnetic field and causing it to appear upfield (at a lower δ). Deshielded means that electron density is pulled away from the nucleus, exposing it more to the magnetic field and shifting it downfield (at a higher δ). Factors such as electronegativity, hybridization, conjugation, and hydrogen bonding all affect the degree of shielding.
In this case, we are asked about the chemical shift of the CH3 (methyl) hydrogen atoms in Compound 1, which is curcumin. From the structure, we can see that the CH3 groups are part of methoxy (–OCH3) substituents, meaning each methyl group is directly bonded to an oxygen atom. Oxygen is highly electronegative and pulls electron density away from the hydrogen atoms on the CH3 group. This deshielding effect causes the chemical shift to move downfield. In a typical alkyl group not attached to oxygen, the CH3 signal would appear around 1.0 ppm, but with the deshielding effect of oxygen, we should expect it to appear somewhere between 3.0 and 4.0 ppm, more or less.
This region corresponds to highly deshielded protons, typically found in phenolic –OH groups, especially when they are involved in hydrogen bonding. In those cases, the chemical shift can be as high as 9–10 ppm. The CH3 protons in a methoxy group do not experience this level of deshielding, so they do not appear in this region. This choice is incorrect.
This region is typical for vinyl protons (–CH=CH–), which are found next to double bonds and are influenced by conjugation and π-electron delocalization. These protons typically appear around 5–6 ppm. The CH3 protons in Compound 1 are not part of a conjugated double bond system, so they do not resonate in this region. This choice is incorrect.
This is correct. The CH3 groups in curcumin are part of methoxy (–OCH3) groups, meaning each methyl group is directly attached to an electronegative oxygen atom. This oxygen causes a deshielding effect, shifting the resonance downfield. In a normal alkyl group, the methyl signal would appear around 1.0 ppm, but due to the presence of the oxygen, the shift moves to around 3.5 ppm. This falls within the 3.0–4.0 ppm range.
This region corresponds to CH3 groups in simple alkyl chains, where the protons are attached to non-electronegative atoms, such as carbon. In such cases, the hydrogen atoms are well-shielded and appear upfield, typically around 1.0 ppm. However, since the methyl protons in curcumin are bonded to oxygen, they are deshielded and appear downfield, not in this region.
This question asks us to identify the modification to a Venturi meter that would increase the maximum measurable air flow speed at the entrance. To solve this, we need to understand the relationship between the Venturi meter’s components, the Bernoulli’s principle, and the continuity equation.
The Venturi meter measures fluid flow speed by measuring the pressure difference between the wider entrance and the narrower throat. As air flows through the Venturi meter and encounters the narrower throat, its speed increases. According to Bernoulli’s principle, this increase in speed leads to a decrease in pressure within the narrower section.
We would expect the liquid in the manometer tube connected to the narrower section to rise, indicating this pressure drop. The maximum measurable flow speed is limited by the height the liquid rises in the manometer tube, which is in turn influenced by the density of the liquid and the geometry of the meter.
We need to analyze each answer choice to determine which modification allows for a greater pressure difference and thus a higher measurable flow speed.
Shortening the manometer tube decreases the maximum measurable speed by preventing the liquid from rising in tube 2 to the height corresponding to the speed measured. This limits the measurable pressure difference and therefore the maximum flow speed. Thus, it is not the correct answer.
This is the correct answer. Increasing the manometer liquid density increases the flow speed range, meaning it increases the flow speed at the entrance that can be measured. Here’s a breakdown of how we arrive at the equation and its implications:
Continuity Equation: A1v1 = A2v2, where A1 and v1 are the cross-sectional area and velocity at the wider entrance, and A2 and v2 are the cross-sectional area and velocity at the narrower throat.
Bernoulli’s Equation: P1 + ½ρairv12 = P2 + ½ρairv22, where P1 and P2 are the pressures at the wider entrance and narrower throat, respectively, and ρair is the density of the air.
Hydrostatic Pressure Measurement: P1 – P2 = ρliquidgh, where ρliquid is the density of the liquid in the manometer, g is the acceleration due to gravity, and h is the height difference between the liquid levels in the manometer tubes.
Rearrange Bernoulli’s equation to solve for P1 – P2: P1 – P2 = ½ρairv12[(A1/A2)2 – 1].
Substitute P1 – P2 = ρliquidgh into the rearranged Bernoulli equation: ρliquidgh = ½ρairv12[(A1/A2)2 – 1].
Solve for v1: v1 = √(2ρliquidgh/ρair[(A1/A2)2 – 1]).
From this equation, we can see that increasing the liquid density ρliquid directly increases the maximum flow speed v1 that can be measured. Hence, the correct answer is B.
Increasing the amount of manometer liquid (i.e., the volume) decreases the flow speed at which the liquid in tube 2 can enter the narrow section of the Venturi meter. Therefore, this is not the correct answer.
The maximum value of distance h, set by the maximum flow speed at the entrance, cannot exceed the length of the liquid column in the manometer. Decreasing the amount of manometer liquid decreases the maximum flow speed at the entrance that can be measured. Thus, this is not the correct answer.
This question asks us to identify which object, a large mass or a small mass, has a larger kinetic energy when both objects have the same momentum. To solve this, we need to understand the relationship between momentum, kinetic energy, and mass. Momentum is given by p = mv, and kinetic energy is given by KE = ½mv². We are given that the momenta of the two objects are equal, and we need to relate this information to their respective kinetic energies.
The small mass object has the larger kinetic energy. We know that momentum is defined as p = mv, where p is momentum, m is mass, and v is speed, and kinetic energy is defined as KE = ½mv², where KE is kinetic energy.
Since the two objects have the same momentum, we can express their momenta as p1 = m1v1 and p2 = m2v2, and these are equal.
We can rearrange the kinetic energy equation to express it in terms of momentum by substituting v = p/m into KE = ½mv², resulting in KE = ½m(p/m)², which simplifies to KE = p²/(2m).
Now, if we compare two objects with the same momentum, one with a smaller mass m1 and one with a larger mass m2, we have kinetic energies KE1 = p1²/(2m1) and KE2 = p2²/(2m2).
Since m1 is smaller than m2, and p1 equals p2, then KE1 must be larger than KE2. Therefore, the smaller mass object has the larger kinetic energy.
The large mass object does not have the larger kinetic energy. This is incorrect because, as shown in the main solution, the kinetic energy is inversely proportional to the mass when the momentum is constant.
While speed is a component of kinetic energy, it is not the determining factor in this scenario. The problem states that the momenta are equal, which constrains the relationship between speed and mass. Therefore, this is not the correct answer.
The kinetic energies of the two objects are not the same. As shown in the main solution, the kinetic energy is inversely proportional to the mass when momentum is constant. Therefore, this is not the correct answer.
The mechanical advantage (MA) is defined as the ratio of the resistance force to the effort force. In this scenario, the resistance force is the component of the object’s weight acting parallel to the inclined plane, plus the force due to kinetic friction.
The component of the weight acting parallel to the plane is given by mg·sin(θ), where m is the mass of the object and g is the acceleration due to gravity. The force due to kinetic friction is μmg·cos(θ), where μ is the coefficient of kinetic friction.
Since the object moves at a constant speed, the applied effort force must equal the sum of these resistance forces. Therefore, the applied force is F = mg·sin(θ) + μmg·cos(θ).
The mechanical advantage is then calculated as MA = mg/(mg·sin(θ) + μmg·cos(θ)), which simplifies to MA = 1/(sin(θ) + μcos(θ)).
This expression, MA = μ + sin(θ)/cos(θ), suggests that the mechanical advantage increases linearly with the coefficient of kinetic friction, which is physically incorrect. The presence of friction should decrease the mechanical advantage, not increase it.
The expression MA = μsin(θ) + cos(θ) implies an incorrect relationship between the angle of the incline and the mechanical advantage. It incorrectly suggests that the normal force directly influences the mechanical advantage in this manner, which is not physically accurate.
The expression MA = μsin(θ)/(cos(θ)) suggests that a frictionless inclined plane (μ = 0) would have a mechanical advantage of zero, which contradicts the fundamental principles of inclined planes. Even without friction, an inclined plane provides a mechanical advantage due to the reduction in the required force.
The expression MA = 1/(sin(θ) + μcos(θ)) is the correct representation of the mechanical advantage. It accurately reflects the inverse relationship between the applied force and the mechanical advantage, and correctly incorporates the effects of both the angle of the incline and the kinetic friction.
This question asks us to identify the property that makes mercury more suitable than water for use in clinical capillary thermometers, based on the provided table. To solve this, we must consider the requirements for a liquid used in a thermometer.
A thermometer relies on the consistent and measurable expansion of a liquid with temperature changes, and the liquid must also move reliably through the capillary tube. The table provides values for density, thermal expansion coefficient, viscosity, and surface tension for both water and mercury.
We need to analyze each property in the context of these requirements, comparing the values for water and mercury to determine which property is most significant for the liquid’s suitability in a capillary thermometer.
Density is not the primary reason mercury is preferred. While mercury has a significantly higher density (13500 kg/m3) compared to water (1000 kg/m3), density itself does not directly contribute to the accurate measurement of temperature in a capillary thermometer. The key factor is the liquid’s response to temperature changes, not its mass per unit volume.
Viscosity is not relevant to the suitability of mercury in a clinical capillary thermometer. Viscosity is a fluid’s resistance to flow. In a thermometer, the liquid must *not* flow freely; it must move only in response to thermal expansion. If the liquid were to flow easily, it would be difficult to accurately measure its position in the capillary tube, as it would move due to gravity or other external forces rather than solely due to temperature changes. Therefore, viscosity is not a determining factor in mercury’s suitability for this application.
Surface tension is the property that makes mercury more suitable. Mercury has a much higher surface tension (500 × 10−3 N/m) than water (72 × 10−3 N/m). Capillary thermometers rely on thermal expansion and the capillary effect. The capillary effect is enhanced by higher surface tension. This higher surface tension allows mercury to form a more distinct and cohesive meniscus, which is crucial for accurate reading in a narrow capillary tube. The distinct meniscus created by the higher surface tension makes it easier to accurately measure the height of the mercury column, which corresponds to the temperature.
The thermal expansion coefficient is the fractional change in volume per degree change in temperature. While this is an important property for thermometers, both water (2.1 × 10−4 K−1) and mercury (1.8 × 10−4 K−1) have suitable thermal expansion coefficients. The difference in their thermal expansion coefficients is not the primary reason mercury is preferred.
This question asks about the molar mass of CaM after its reaction with one equivalent of Compound 1.
To understand this question, we must understand the principles of mass spectrometry (MS) and protein bioconjugation, specifically how the modification of calmodulin (CaM) with Compound 1 affects its molar mass. Mass spectrometry measures the mass-to-charge ratio (m/z) of ions, and when the charge (z) is normalized to +1, the detected values correspond directly to the molecular weights of the different species present in the sample.
This value represents the unmodified CaM, which has not undergone any reaction with Compound 1. Since the problem explicitly states that CaM reacts with one equivalent of Compound 1, its mass must increase, making this choice incorrect.
In this case, unmodified CaM has a known molar mass of 16,621 g/mol, and the reaction with one equivalent of Compound 1 results in the addition of an adduct with a mass of 96 g/mol (C4H4N2O). Therefore, the molar mass of CaM after reaction with one equivalent of Compound 1 is 16,621 g/mol + 96 g/mol = 16,717 g/mol.
This number corresponds to the m/z value (with z normalized to +1) for the second peak in the mass-spectrum of CaM modified in the presence of calcium ions (Figure 2), which refers to the protein modified at its two methionine residues. Note that this adduct also causes the faint second peak in the mass-spectrum of CaM modified in the absence of calcium ions (Figure 2).
This value corresponds to the m/z ratio (with z normalized to +1) for the second peak in the mass spectrum of CaM modified in the presence of calcium ions (Figure 2), indicating modification at two methionine residues. This adduct also contributes to a faint second peak in the mass spectrum of CaM modified in the absence of calcium ions (Figure 2).
This value corresponds to the m/z ratio (with z normalized to +1) for the third peak in the mass spectrum of CaM modified in the presence of calcium ions (Figure 2), indicating that the protein has been modified at three methionine residues.
This question asks about the amino acid in CaM that is expected to be responsible for binding the ion that triggers conformational changes in the protein.
To understand this question, we must understand the principles of calcium ion coordination in proteins and the role of specific amino acid side chains in stabilizing metal ion binding. Calcium ions (Ca2+) are positively charged and typically interact with negatively charged or polar amino acid residues that can provide lone pairs of electrons for coordination. In calmodulin (CaM), calcium binding induces conformational changes, which allow the protein to interact with its target molecules.
The side chain of arginine contains a positively charged guanidinium group at physiological pH. Since like charges repel, arginine would not effectively bind a positively charged calcium ion and instead may interfere with calcium coordination.
Aspartate (option B) is the correct answer because it has a negatively charged carboxyl (-COO–) group in its side chain. This carboxylate group can engage in electrostatic interactions with Ca2+, stabilizing its binding. Additionally, calcium ions tend to coordinate with oxygen-containing ligands, and the oxygen atoms in aspartate’s carboxyl group make it well-suited for calcium coordination.
Leucine is a non-polar, hydrophobic amino acid with an aliphatic side chain. Leucine does not have functional groups capable of directly coordinating with calcium ions, and it does not contribute to calcium-dependent conformational changes in CaM.
Phenylalanine, like leucine, is a hydrophobic amino acid. Its aromatic benzyl side chain does not participate in ionic or electrostatic interactions with calcium ions, making it an unlikely candidate for calcium binding.
This question asks about the best description of the ion that induces conformational changes in CaM.
To understand this question, we must understand the principles of atomic structure, ion formation, and the classification of elements based on their valency and charge. Atoms achieve stability by either gaining or losing electrons to form ions, and the number of electrons lost or gained determines the ion’s charge (valency). Metals, which are found on the left side of the periodic table, tend to lose electrons and form positively charged ions (cations), whereas nonmetals typically gain electrons and form negatively charged ions (anions). The valency of an ion refers to the number of electrons lost or gained, which determines the charge of the ion.
A monovalent cation has a charge of +1, meaning it has lost only one electron. Examples include Na+ (sodium ion) and K+ (potassium ion). However, calcium forms a +2 cation (Ca2+), making it divalent rather than monovalent.
Calcium is a metal, and metals do not form anions. An anion is a negatively charged ion that results from the gain of electrons, which is typical of nonmetals (e.g., Cl–, F–, Br–). Since calcium loses electrons, it cannot be an anion. Calcium ion is positively charged, which classifies it as a cation.
Calcium (Ca) is a metal that loses two valence electrons to form the Ca2+ ion. This makes it a divalent metal cation, meaning it has a +2 charge and is classified as a metal ion. In calmodulin (CaM), the binding of these divalent calcium ions induces conformational changes by interacting with negatively charged amino acid residues.
Calcium is not an anion; it is a cation. A divalent anion would have a -2 charge, such as sulfate (SO42-) or oxide (O2-). Calcium, however, forms a +2 charged ion (Ca2+), making it a divalent cation rather than an anion.
This question asks about the reason why the protocol for CaM labeling by Compound 1 includes a step where Compound 4 is added to the reaction mixture.
To understand this question, we must understand the principles of chemical reactivity, functional group selectivity, and reaction quenching in bioconjugation. In a chemical reaction, excess reagents are often used to drive the reaction to completion, but any unreacted reagent must be removed or neutralized to prevent unwanted side reactions.
Compound 4 (N-acetylmethionine) features an acetyl group linked to the α-amino group of methionine through an amide bond, preventing it from functioning as an acetylating agent.
According to the passage, the exposure of methionine residues on the surface of CaM is induced by the binding of calcium ions.
In the case of methionine-based protein bioconjugation, Compound 1 reacts selectively with accessible methionine residues in CaM. Once all exposed methionine residues are modified, any remaining Compound 1 must be neutralized to prevent non-specific reactions that could interfere with downstream applications.
Compound 4 (N-acetylmethionine) is a functional analog of methionine, meaning it shares similar chemical properties but does not participate in the same interactions. Once all accessible methionine residues in CaM have reacted with Compound 1, Compound 4 serves as a competitive quencher by reacting with any remaining Compound 1 in the solution. This ensures that no excess Compound 1 is available to modify other unintended sites, maintaining the specificity of the labeling reaction.
Compound 4, identified as N-acetylmethionine, serves as a functional analog of the protein’s methionine residues and, according to the reaction scheme in the passage, is not expected to react with benzaldehyde.
This question asks about the elemental classification of the catalyst for the conversion of Compound 2 to Compound 3 in Figure 1.
To understand this question, we must understand the principles of the periodic table’s classification of elements, particularly the properties and grouping of metals. The periodic table organizes elements based on their atomic number, electron configuration, and chemical properties.
Elements are divided into different categories, including transition metals, metalloids, alkali metals, and alkaline earth metals. The classification of an element determines its chemical reactivity and its role in catalysis. In the reaction from Compound 2 to Compound 3, the catalyst used is copper(I).
Copper is in Group 11 (Group IB) of the periodic table, which is in the d-block of the periodic table (Groups 3 to 12) containing transition metals.
Transition metals are defined as elements that have partially filled d orbitals, which allow them to exhibit multiple oxidation states and serve as effective catalysts in chemical reactions.
Metalloids include elements such as boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), and astatine (At). These elements exhibit properties intermediate between metals and nonmetals and are not typically used as catalysts in redox reactions. Copper, on the other hand, is fully metallic and does not belong to this category.
Alkaline earth metals belong to Group 2 (formerly Group IIA) of the periodic table, which includes beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). These metals have two valence electrons and typically form +2 cations. Unlike transition metals, they do not have partially filled d orbitals. Copper, being in Group 11, does not fit this classification.
Alkali metals are found in Group 1 (formerly Group IA) of the periodic table, which includes lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr). These elements have a single valence electron and readily form +1 cations, making them highly reactive. Copper, positioned in Group 11, does not belong to this group.
This question asks about the number of stereoisomers that Compound 1 has.
To understand this question, we must understand the concept of chirality and stereoisomerism. Stereoisomers are molecules that have the same molecular formula and connectivity but differ in the spatial arrangement of their atoms. The number of possible stereoisomers of a molecule is determined by the number of chiral centers, which are typically sp³-hybridized carbon atoms with four different substituents. The maximum number of stereoisomers for a given molecule can be calculated using the 2n rule, where n is the number of chiral centers.
Compound 1 has one chiral center. Since the number of stereoisomers is determined by the equation 2n, and n = 1 (one chiral center), the total number of stereoisomers for Compound 1 is 2. These two stereoisomers exist as non-superimposable mirror images, also known as enantiomers.
Three stereoisomers occur only when there are two chiral centers in a symmetrical molecule that results in a pair of enantiomers and an additional meso compound. A meso compound contains an internal plane of symmetry, making it optically inactive. However, since Compound 1 has only one chiral center, a meso form is not possible, and three stereoisomers cannot exist.
A compound with two chiral centers typically has four stereoisomers, but Compound 1 contains only one sp³-hybridized carbon with four different substituents, making it a single chiral center.
A compound with three chiral centers would have eight stereoisomers, but Compound 1 has only one sp³-hybridized carbon with four distinct substituents, meaning it has just one chiral center.
This question asks about the compound which is most appropriate to accomplish the synthesis in Figure 2. To understand this question, we must understand the principles of acid-base reactions, specifically the deprotonation of carboxylic acids.
Sodium bromide (NaBr) is a neutral salt that dissociates into Na⁺ and Br⁻ in solution. Bromide (Br⁻) is the conjugate base of hydrobromic acid (HBr), which is a strong acid. Because strong acids have very weak conjugate bases, Br⁻ is not a strong enough base to deprotonate a carboxylic acid effectively.
Sodium chloride (NaCl), is a neutral salt that dissociates into Na⁺ and Cl⁻ in solution. Chloride (Cl⁻) is the conjugate base of hydrochloric acid (HCl), that is a strong acid. Since Cl⁻ is an extremely weak base, it cannot remove the proton from a carboxylic acid.
The deprotonation of carboxylic acid and its conversion into its carboxylate salt requires a strong base. Sodium hydroxide (NaOH) is a strong base capable of deprotonating the carboxylic acid functional group in Compound 2 in an acid–base reaction to form the corresponding carboxylate salt. This reaction is important in the final step of the synthesis shown in Figure 2, where the presence of sodium (Na⁺) in the product indicates that NaOH was the reagent used.
Sodium iodide (NaI), also falls into the same category. Iodide (I⁻) is the conjugate base of hydroiodic acid (HI), which is an even stronger acid than HBr and HCl. Since HI is a strong acid, its conjugate base (I⁻) is weak and cannot act as a strong enough base to deprotonate a carboxylic acid.
This question asks about the inhibition mechanism exhibited by Compound 2 against HKT.
To understand this question, we must understand the principles of enzyme kinetics and how different types of inhibitors affect enzyme activity, particularly in the context of the Lineweaver–Burk plot, which is a graphical representation of the reciprocal of enzyme velocity versus the reciprocal of substrate concentration. The inhibition mechanism of Compound 2 against 3-Hydroxykynurenine transaminase (HKT) is determined by analyzing changes in kinetic parameters such as Vmax (maximum velocity) and KM Michaelis constant), which describe the enzyme’s efficiency and affinity for its substrate.
In competitive inhibition, an inhibitor competes with the substrate for binding to the enzyme’s active site. This results in an apparent increase in KM (lower substrate affinity) because a higher concentration of substrate is required to reach half of Vmax. However, the Vmax remains unchanged since increasing substrate concentration can overcome the inhibition. In a Lineweaver–Burk plot, competitive inhibition is identified by multiple lines that intersect at the y-axis (same 1/Vmax) but have different x-intercepts. In this case, since the lines do not share the same y-intercept, competitive inhibition is ruled out.
Mixed inhibition occurs when an inhibitor can bind to both the free enzyme and the enzyme-substrate complex, but with different affinities. This results in changes to both KM and Vmax, and the lines in a Lineweaver–Burk plot will not share the same slope, x-intercept, or y-intercept. In the given study, the inhibitory profile shows lines intersecting above the x-axis, indicating that both KM and Vmax are altered, which is a characteristic of mixed inhibition.
Noncompetitive inhibition occurs when an inhibitor binds equally well to both the free enzyme and the enzyme-substrate complex at a site distinct from the active site. This leads to a decrease in Vmax, but KM remains unchanged since substrate binding affinity is not affected. In a Lineweaver–Burk plot, noncompetitive inhibition is characterized by multiple lines sharing the same x-intercept but having different y-intercepts. Since the plot in this case shows different x-intercepts, noncompetitive inhibition is ruled out.
Uncompetitive inhibition occurs when an inhibitor binds exclusively to the enzyme-substrate complex, reducing both KM and Vmax while keeping their ratio (Vmax/KM) constant. This results in parallel lines in a Lineweaver–Burk plot, meaning that all lines have the same slope. Since the plot in this study does not display parallel lines, uncompetitive inhibition is incorrect.
This question asks about the type of column chromatography that was mentioned in the passage.
To understand this question, we must understand the principles of column chromatography, which is a widely used technique for purifying proteins based on their physical and chemical properties. Different types of chromatography separate proteins based on size, charge, hydrophobicity, or specific interactions with ligands.
Size exclusion chromatography separates molecules based on their size. This method utilizes a porous gel matrix that acts as a molecular sieve, allowing smaller molecules to enter the pores and be delayed, while larger molecules pass through more quickly. Gel filtration resins do not require NiSO4.
Antigen-antibody immunoaffinity chromatography is a highly specific technique that involves a resin functionalized with antibodies or antigens that selectively bind to a target protein. Since the passage does not mention the use of an antibody or antigen-functionalized resin, and instead describes a resin charged with Ni²⁺, this method is not applicable. The specificity in this case arises from the interaction between histidine residues in the His-tag and the metal ions rather than an antigen-antibody interaction.
Hydrophobic interaction chromatography separates proteins based on their hydrophobicity. This technique typically uses a resin with hydrophobic ligands that interact with the hydrophobic regions of proteins under high salt conditions. However, the passage describes an ionic resin charged with Ni²⁺, which does not primarily interact with hydrophobic side chains of the protein but rather binds specifically to histidine residues in His-tagged proteins.
In the passage, His-tagged 3-Hydroxykynurenine transaminase (HKT) was purified using a resin charged with NiSO₄, which indicates that a specific type of affinity chromatography was used.
Immobilized metal affinity chromatography (IMAC) relies on the strong and selective interaction between histidine residues in His-tagged proteins and metal ions immobilized on a resin. The passage specifies that the resin was charged with Ni²⁺, which is a characteristic of IMAC. In this method, the histidine residues in the His-tag form coordination bonds with Ni²⁺, allowing selective binding of the target protein to the resin. Unbound proteins are washed away, and the His-tagged protein is then eluted using imidazole or another competing ligand.
This question asks about the Kb of 3-HK.To understand this question, we must understand the principles of acid-base equilibrium and how pKa, pKb, Ka, and Kb are related.
The acid dissociation constant, Ka, measures the strength of an acid in solution and is defined as Ka = [H⁺][A⁻] / [HA]. The pKa is the negative logarithm of Ka, given by pKa = -log(Ka).
Similarly, the base dissociation constant, Kb, measures the strength of a base in solution and is defined as Kb = [OH⁻][BH⁺] / [B], with pKb given by pKb = -log(Kb).
There is a fundamental relationship between Ka and Kb, given by the equation Ka × Kb = Kw, where Kw is the ionization constant of water, equal to 1.0 × 10⁻¹⁴ at 25°C. Taking the negative logarithm on both sides, we get pKa + pKb = 14.
0.1 represents the Ka (Ka = 10⁻¹) of 3-HK rather than Kb. The question specifically asks for Kb, which requires additional calculation.
13 represents pKb (pKb = 14 – pKa), not Kb. Since Kb is calculated as 10⁻ᵖᴷᵇ, the correct answer should be 10⁻¹³, not 13.
Applying these concepts to the problem, we are given that 3-HK has a pKa of 1.0. Using the equation pKa + pKb = 14, we solve for pKb: pKb = 14 – 1 = 13. Now, we calculate Kb using Kb = 10⁻ᵖᴷᵇ, which gives Kb = 10⁻¹³.
This value corresponds to Kw, the ionization constant of water, not the base dissociation constant of 3-HK. While Ka and Kb are related through Kw, the actual Kb for 3-HK is 10⁻¹³, making choice C the correct answer.
This question asks about the IC50 of Compound 2 in nanomolar.
To understand this question, we must understand the principles of unit conversion, particularly in the context of molarity, which is a measure of concentration. Molarity (M) refers to the number of moles of a substance per liter of solution, and it has different subunits: millimolar (mM), micromolar (μM), and nanomolar (nM). The key relationships between these units are: 1 M = 1000 mM, 1 mM = 1000 μM, and 1 μM = 1000 nM.
This is the value of 72 μM expressed in molar (M). Since 1 μM = 10⁻⁶ M, we can convert 72 μM to M by 72 × 10⁻⁶ M = 0.000072 M. This value is in molar units, not nanomolar.
0.072 represents 72 μM converted to millimolar (mM). Since 1 mM = 1000 μM, we can convert 72 μM to mM as 72 × 10⁻³ mM = 0.072 mM. This is also not the correct conversion to nanomolar.
7200 represents 7.2 μM converted to nanomolar. Since 1 μM = 1000 nM, converting 7.2 μM to nM gives 7.2 μM × 1000 = 7200 nM. However, the IC50 value given in the passage is 72 μM, not 7.2 μM.
In this problem, the IC50 of Compound 2 is given as 72 μM. To convert this to nanomolar (nM), we use the relationship 1 μM = 1000 nM. Multiplying by 1000, we get 72 μM × 1000 = 72,000 nM.
This question asks about the number of hydrogen bond donors in XA.To understand this question, we must understand the principles of hydrogen bonding and how functional groups contribute to hydrogen bond donation.
A hydrogen bond donor is an atom or functional group capable of donating a hydrogen atom that is covalently bonded to a highly electronegative atom such as oxygen, nitrogen, or fluorine.
In the case of XA, the chemical structure provided in Figure 1 reveals that there are three hydroxyl (-OH) groups, one of which belongs to a carboxyl (-COOH) functional group. These hydroxyl groups contain hydrogen atoms directly bonded to oxygen, making them capable of donating hydrogen bonds to an acceptor.
The structure of 3-HK, not XA, contains four hydrogen bond-donating atoms.
XA contains five hydrogen bond-accepting groups with lone pairs capable of forming hydrogen bonds, but it has only three atoms that can donate hydrogen bonds.
Six hydrogen bond donor sites correspond to the structure of 3-HK, not XA. The additional hydrogen bond donors in 3-HK arise from an extra functional group that is absent in XA.
The correct answer is B
This question asks about the value of the molar heat capacity of the gas during a thermodynamic process in which pressure doubles and volume halves.
To understand this question, we must understand the principles of thermodynamics, specifically the first law of thermodynamics and the concept of molar heat capacity. The first law of thermodynamics states that the change in internal energy (ΔU) of a system is given by ΔU=Q−W, where Q is the heat added to the system and W is the work done by the system. The molar heat capacity (C) of a gas during a process is defined as the heat exchanged per mole per unit temperature change: C=Q/ΔT, where ΔT is the temperature change between the initial and final states.
In this problem, the gas undergoes a process where the pressure doubles and the volume halves. Using the ideal gas law, PV=nRT, for one mole of gas (n=1), we can analyze the temperature change. Let the initial state have pressure Pi, volume Vi, and temperature Ti. The final state has pressure = 2Pi and volume Vf =1/2 Vi. Applying the ideal gas law for both states, we get Pi Vi= R Ti and Pf Vf= (2Pi)×(Vi/2) =Pi Vi=RTf. Since Pf Vf = Pi Vi, it follows that Tf =Ti, meaning the temperature remains constant during the process (ΔT=0).
Now, using the definition of heat capacity, C=Q/ΔT, and since ΔT=0, the denominator is zero. For any nonzero heat (Q≠0), dividing by zero leads to an infinite molar heat capacity.
According to the first law of thermodynamics, this would suggest that the gas temperature remains constant throughout the process. However, this is incorrect, as the gas experiences intermediate states where its temperature first increases and then decreases.
The molar heat capacity can be determined using the first law of thermodynamics and the ideal gas law. Since the temperature does not change and heat is exchanged, the heat capacity is not undetermined but rather infinite.
This suggests that, according to the first law of thermodynamics, the final temperature differs from the initial temperature. However, based on the ideal gas law, the final and initial temperatures are equal since the product of pressure and volume remains unchanged between the two states.
This question asks us to identify the daughter nucleus produced when Radium-223 (Radium with a mass number of 223 and an atomic number of 88) undergoes alpha decay. To solve this, we need to understand the process of alpha decay. Alpha decay involves the emission of an alpha particle, which is a helium nucleus. A helium nucleus consists of 2 protons and 2 neutrons. This means the parent nucleus loses 4 nucleons (2 protons and 2 neutrons), resulting in a decrease of 4 in the mass number and a decrease of 2 in the atomic number. Therefore, the daughter nucleus will have a mass number that is 4 less than the parent’s mass number, and an atomic number that is 2 less than the parent’s atomic number. We can then compare this to the answer options.
Additionally, to understand why the other options are not the result of alpha decay, we can consider what type of decay would lead to them. Option B, Francium-223, would require beta minus decay, where a neutron is converted into a proton. Option C, Radium-223, would indicate no decay at all. Option D, Thorium-227, would be the result of beta plus decay, where a proton is converted into a neutron.
This answer choice is the result of alpha decay. In alpha decay, the parent nucleus Radium-223 experiences the following changes:
* Loss of 4 from the mass number: 223 – 4 = 219
* Loss of 2 from the atomic number: 88 – 2 = 86 The element with atomic number 86 is Radon. Therefore, the daughter nucleus is Radon-219.
This answer choice, Francium-223, would be the result of beta minus decay, where a neutron is converted into a proton, increasing the atomic number by 1 but leaving the mass number unchanged. Alpha decay decreases the mass number, so this is not the resulting daughter nucleus.
This answer choice, Radium-223, is the parent nucleus; it cannot be the daughter nucleus of its own decay. No change in the nucleus indicates no decay at all.
This answer choice, Thorium-227, would be the result of beta plus decay, where a proton is converted into a neutron, decreasing the atomic number by one and increasing the mass number. Alpha decay decreases the mass number, so this is not the resulting daughter nucleus.
This question involves the photoelectric effect. 7eV photons eject electrons (photoelectrons) from a 5eV work function metal sphere. The sphere becomes positively charged, opposing further emission. We need the max number of photoelectrons emitted before the sphere’s charge stops emission.
This is not the number of photoelectrons. This does not accurately reflect the relationship between the photon energy, work function, and the electrostatic potential on the sphere.
This is not the number of photoelectrons. This does not accurately reflect the relationship between the photon energy, work function, and the electrostatic potential on the sphere.
Emitted electrons have max 2eV kinetic energy. The sphere’s charge creates a potential energy barrier. Emission stops when N x e x 4.8 x 10^-8 V ≤ 2 eV. N ≈ 4 x 10^7.
This implies a 33cm radius sphere. This is not the number of photoelectrons. This does not accurately reflect the relationship between the photon energy, work function, and the electrostatic potential on the sphere.
In this problem, we are analyzing the conditions under which the hollow cube can float, and how the density of the liquid in which it floats affects our ability to read the markings on the cube. To determine the markings, we’ll need to be able to see enough of the cube above the liquid’s surface to make an accurate reading. This requires the cube to float partially submerged. To understand this, we need to consider buoyancy, which is the upward force exerted by a fluid that opposes the weight of an immersed object. Archimedes’ principle states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
If the liquid’s density is too low, the buoyant force will be insufficient to support the cube’s weight, and it will sink, making the markings unreadable. If the liquid’s density is too high, the cube will float higher, but we are looking for the minimum density at which the cube can still float partially, allowing us to read the markings. We need to find the specific density where the buoyant force just balances the cube’s weight, which will correspond to the minimum measurable density.
This option suggests a density below what the cube’s own effective density can allow for. Since the cube floats half-submerged in oil of 800 kg/m³, its average density is 400 kg/m³. It cannot float in a liquid of lesser density, thus making it impossible to read any markings on the cube.
This choice indicates a density where the cube would completely sink. With a density of 200 kg/m³, the buoyant force would be insufficient to counteract the weight of the cube, which has an effective density of 400 kg/m³. Consequently, the cube submerges fully, preventing any reading of the marked edges.
This density matches the cube’s effective density, calculated from its floating behavior in oil. The cube floats when the buoyant force equals its weight. Any liquid with a density lower than 400 kg/m³ will not provide enough buoyant force for the cube to float partially, rendering the markings unreadable.
This selection represents a scenario where the cube floats, but at a submersion level of 2/3. This is not the minimum density at which the cube can float partially, as it can float at a lower density, namely the cube’s effective density of 400 kg/m³. Therefore, this isn’t the minimum density that allows for a reading.
This question asks about the ratio of [OH–] to [H+] in the buffer described in the passage. To understand this question, we must understand the principles of pH, pOH, and the relationship between hydrogen ion [H+] and hydroxide ion [OH–] concentrations in aqueous solutions.
The pH of a solution is defined as:
pH = −log[H+],
while the pOH is defined as
pOH = −log[OH–].
These values are related by the equation pH + pOH = 14.
These values are related by the equation pH + pOH = 14.
Since the problem states that the buffer has a pH of 5.0, we can determine the corresponding [H+] and [OH–] concentrations. The hydrogen ion concentration is given by [H+] = 10⁻⁵ M. Using the relationship between [H+] and [OH–] in water, given by the ion-product constant for water (Kw = 10⁻¹⁴ at 25°C), we calculate [OH–] = Kw / [H+] = 10⁻¹⁴ / 10⁻⁵ = 10⁻⁹ M. Now, we find the ratio of [OH–] to [H+]: [OH–] / [H+] = 10⁻⁹ M / 10⁻⁵ M = 10⁻⁴:1.
Correct. The calculated ratio is 10⁻⁴:1.
This ratio corresponds to a pH of 6.0. At pH 6.0, the hydrogen ion concentration is [H+] = 10⁻⁶ M, and using [OH–] = 10⁻¹⁴ / [H+], we find [OH–] = 10⁻⁸ M. The ratio of [OH–] to [H+] is then 10⁻⁸ / 10⁻⁶ = 10⁻²:1, which does not match the given buffer’s pH of 5.0.
This ratio occurs at pH of 8.0. At pH 8.0, the hydrogen ion concentration is [H+] = 10⁻⁸ M, and [OH–] = 10⁻⁶ M. The ratio of [OH–] to [H+] is then 10⁻⁶ / 10⁻⁸ = 10²:1, which is incorrect for a pH 5.0 buffer.
This ratio occurs at pH of 9.0. At pH 9.0, the hydrogen ion concentration is [H+] = 10⁻⁹ M, and [OH–] = 10⁻⁵ M. The ratio of [OH–] to [H+] is then 10⁻⁵ / 10⁻⁹ = 10⁴:1, which is incorrect for the given pH.
To understand this question, we must understand the principles of complementary base pairing in DNA.
In double-stranded DNA (dsDNA), each base pairs with its complementary base according to Chargaff’s rules: adenine (A) pairs with thymine (T) through two hydrogen bonds, and cytosine (C) pairs with guanine (G) through three hydrogen bonds. DNA strands are also antiparallel, meaning one strand runs in the 5′ to 3′ direction while the complementary strand runs in the 3′ to 5′ direction. The given DNA oligomer in the passage is 5′–CTAGCGCGCTAG–3′. To determine its complementary sequence, we need to replace each base with its complementary counterpart and then reverse the order to maintain the antiparallel orientation. This results in the sequence 3′–GATCGCGCGATC–5′. When rewritten in the standard 5′ to 3′ direction, the complementary sequence is 5′–CTAGCGCGCTAG–3′.
This sequence mismatches bases, incorrectly pairing C with A instead of G and T with G instead of A. It does not reflect proper complementary base pairing rules.
This sequence pairs the A’s with T’s and the G’s with C’s correctly but fails to account for the DNA binding to this sequence, which will be flipped, with the 3′ of the complementary sequence binding to the 5′ of this sequence. The complementary sequence must be antiparallel to the original strand, meaning the 3′ end of one strand aligns with the 5′ end of the other. The correct complementary strand should be written in the 5′ to 3′ direction as 5′–CTAGCGCGCTAG–3′, not 5′–GATCGCGCGATC–3′.
This sequence incorrectly pairs C with A instead of G and T with G instead of A, making it an incorrect complementary sequence.
This question asks about the amino acid residue in Protein X which would most likely form electrostatic interactions with the DNA backbone in the minor groove.
To understand this question, we must understand the principles of electrostatic interactions between proteins and DNA. DNA is a negatively charged molecule due to the phosphate groups in its backbone. Proteins that bind to DNA often contain positively charged amino acids, such as arginine (R) and lysine (K), which can form favorable electrostatic interactions with the negatively charged phosphate groups.
The passage indicates that in the case of Protein X, only β-sheet residues bind to DNA. From the figure we can see that Arg24 (R24), a positively charged amino acid, is located in the β3 region. This amino acid interacts with the minor groove of the DNA, and therefore it can form bonds with the negatively charged phosphate backbone.
While Lys63 (K63) carries a positive charge, it is situated outside the β-sheet region of the protein, which is responsible for interacting with the phosphate backbone.
Although Lys52 (K52) is positively charged, it is found in the α-helical segment of Protein X, which does not participate in electrostatic interactions with the phosphate backbone.
Arg24 (R24), located in the β3 sheet, is positively charged and positioned to form electrostatic interactions with the negatively charged DNA backbone in the minor groove.
Despite being positioned within a β-sheet region, Glu11 (E11) carries a negative charge, preventing it from forming electrostatic interactions with the negatively charged phosphate backbone.
This question asks about the technique researchers could have used to most selectively bind Protein X on a column and elute it in a purified form.
To understand this question, we must understand the principles of protein purification techniques, specifically chromatography methods used to isolate a target protein based on its unique biochemical properties. Affinity chromatography is a powerful purification method that selectively binds proteins with high specificity by using a ligand immobilized on the column matrix.
Since Protein X binds specifically to double-stranded (ds) DNA, the most effective purification method would involve using a column with covalently bound dsDNA oligomer as the stationary phase. As the crude protein mixture passes through, only Protein X will strongly interact with the immobilized DNA, while other proteins will elute out. Subsequently, a NaCl gradient can be applied to disrupt electrostatic interactions between Protein X and the DNA, effectively eluting Protein X in a purified form. This ensures high specificity and selectivity in the purification process.
While cation exchange will effectively separate Protein X from other proteins with different net charges, it will not specifically bind Protein X to the column resin.
Although hydrophobic interaction chromatography can be used to purify Protein X, it will not specifically bind Protein X to the column.
Although gel-filtration or size-exclusion chromatography separates proteins by molecular weight, it will not specifically bind Protein X to the column.
This method uses immobilized dsDNA to selectively retain Protein X via specific interactions, allowing it to be eluted later using a salt gradient.
This question asks about the equilibrium constant expression for Protein X binding to the dsDNA oligomer (dsDNA).
To answer this question, we must use the information given to us by the passage and understand the principles of chemical equilibrium and the proper formulation of an equilibrium constant expression. The equilibrium constant expression is derived from the balanced chemical equation, where the concentration of products is placed in the numerator and the concentration of reactants is placed in the denominator, each raised to the power of their stoichiometric coefficients.
When we go through the passage, we can see that “studies found that two Protein X molecules can bind to each molecule of the double-stranded (ds) DNA oligomer. “
Accordingly, based on the passage, the binding reaction is given as:
2(Protein X) + 1dsDNA ⇌ (Protein X)₂(dsDNA)
The general form of the equilibrium constant (K) expression is:
K = [products] / [reactants]
Applying this to the binding reaction:
K = [(Protein X)₂(dsDNA)] / ([Protein X]² [dsDNA])
Since two Protein X molecules bind to one dsDNA molecule, the numerator should be (Protein X)₂(dsDNA), and the concentration of [Protein X] in the denominator must be squared.
This is incorrect because the product (Protein X)₂(dsDNA) is missing from the numerator and the stoichiometry of the Protein X reactant is not accounted for.
This is the correct expression: K = [(Protein X)₂(dsDNA)] / ([Protein X]² [dsDNA])
This expression incorrectly interprets the binding reaction as Protein X + 2dsDNA ⇌ (Protein X)(dsDNA)₂, which does not match the given reaction mechanism.
This question asks about the highest level of protein structure present in Protein X in the absence of DNA.
To understand this question, we must understand the principles of protein structure, which is classified into four levels: primary, secondary, tertiary, and quaternary. These structural levels describe how a protein is organized, from its amino acid sequence to the complex three-dimensional folding of multiple polypeptide chains.
The passage states that Protein X, in the absence of DNA, consists of multiple β-sheets, an α-helix, and a p-loop. These elements represent secondary structural features. However, the interaction between these secondary structures results in a three-dimensional conformation, which corresponds to the tertiary structure of the protein. Therefore, the interaction of multiple types of secondary structures of Protein X implies that this protein has tertiary structure.
The primary structure of a protein refers to its linear sequence of amino acids. While Figure 1 shows the sequence of Protein X, the passage describes additional levels of structural organization.
Although Figure 1 depicts the secondary structures within Protein X, the interaction of various secondary structure elements indicates that the protein possesses tertiary structure.
The interaction of multiple secondary structural elements—such as β-sheets, α-helices, and loops—results in a unique three-dimensional fold, consistent with tertiary structure.
The quaternary structure of a protein occurs when multiple polypeptide chains (subunits) assemble into a functional complex. The passage explicitly states that Protein X is monomeric.
This question asks about the number of non-canonical base-paired guanines found in the total stack of the Spinach RNA aptamer. To understand this question, we must understand the principles of RNA secondary and tertiary structure, specifically the formation of G-quadruplexes and non-canonical base pairing. Unlike standard Watson-Crick base pairing, where guanine pairs with cytosine, guanine-rich sequences can form a G-quadruplex, a stable structure where four guanines form a G-tetrad through Hoogsteen hydrogen bonding. These G-tetrads can stack on top of each other, stabilized by monovalent cations such as K⁺. The passage states that Spinach RNA contains two stacked G-quadruplexes, each consisting of a G-tetrad. The guanines forming these tetrads are: First G-tetrad (G23, G27, G59, G54) and Second G-tetrad (G22, G26, G61, G57). Each G-tetrad consists of four guanine bases that hydrogen bond with each other in a non-canonical manner. Since there are two G-tetrads stacked together, the total number of guanines involved in non-canonical base pairing is 4 guanines per G-tetrad × 2 G-tetrads = 8 non-canonical base-paired guanines.
The total stacked structure consists of two tetramers, each containing four guanine bases hydrogen-bonded together, resulting in a total of eight non-canonical base pairs rather than two.
Figure 1 depicts four non-canonical base pairs, but since there are two stacked G-quadruplexes, the total number of guanine-guanine pairings is eight, not four.
Figure 1 illustrates four guanines forming hydrogen bonds in a single G-quadruplex, and with the second G-quadruplex positioned beneath it, the total number of non-canonical base pairs amounts to eight.
This question asks about the approximate percentage of fluorine by mass in Compound 1. To understand this question, we must understand the principles of calculating the mass percentage of an element in a compound, which is determined using the formula: Mass % of Element = (Total mass of element in the compound / Molar mass of compound) × 100. Given that Compound 1 has a molar mass of 251.22 g/mol and contains two fluorine (F) atoms, we first determine the total mass of fluorine in the compound: 2 × 19 g/mol = 38 g/mol. Next, we apply the mass percentage formula: (38 g/mol / 251.22 g/mol) × 100 = 15.1%. Thus, the correct answer is C (15%) because it closely matches our calculated value.
This corresponds to the percentage of hydrogen in Compound 1. Hydrogen has a molar mass of 1 g/mol, and since Compound 1 contains 10 hydrogen atoms, their total mass is 10 g/mol. Using the mass percentage formula: (10 / 251.22) × 100 = 3.98% ≈ 4%, this confirms that option A represents the mass percentage of hydrogen, not fluorine.
This represents the percentage of nitrogen in Compound 1. Since nitrogen has a molar mass of 14 g/mol and there are two nitrogen atoms in the compound, their total mass is 28 g/mol. The mass percentage calculation is: (28 / 251.22) × 100 = 11.1% ≈ 11%, confirming that option B corresponds to nitrogen rather than fluorine.
This corresponds to the percentage of carbon in Compound 1. Carbon has a molar mass of 12 g/mol, and since the compound contains 12 carbon atoms, the total carbon mass is 144 g/mol. The mass percentage calculation is: (144 / 251.22) × 100 = 57.3% ≈ 57%, confirming that option D represents the carbon content, not fluorine.
This question asks about the approximate ratio of Compound 1 and its conjugate acid at pH 6.5.
To understand this question, we must understand the principles of acid-base equilibria and the Henderson–Hasselbalch equation, which describes the relationship between the pH of a solution, the pKa of an acid, and the ratio of its conjugate base to its protonated form.
The equation is written as pH = pKa + log([A–]/[HA]), where [A–] represents the concentration of the deprotonated (anion) form of the compound, and [HA] represents the concentration of its conjugate acid (protonated form).
In this case, Compound 1 has a pKa of 5.5, and the question asks us to determine the ratio of its deprotonated form to its protonated form at a pH of 6.5.
Using the Henderson–Hasselbalch equation:
6.5 = 5.5 + log([A–]/[HA])
Subtracting 5.5 from both sides:
1 = log([A–]/[HA])
Converting from log to exponential form:
10¹ = [A–]/[HA]
So the ratio is 10:1.
If the ratio of Compound 1 ([A–]) to its conjugate acid ([HA]) were 100:1, applying the Henderson–Hasselbalch equation would yield:
pH = 5.5 + log(100) = 5.5 + 2 = 7.5, which exceeds the given pH of 6.5. This is an incorrect calculation.
This is consistent with the calculated ratio using the Henderson–Hasselbalch equation for pH 6.5 and pKa 5.5, giving 10:1.
For the ([A–]/[HA]) ratio to be 1:1, the pH must be equal to the pKa. However, since the pKa is 5.5 and the pH is 6.5, this condition is not met. This is an incorrect calculation.
If the pH were 4.5, the Henderson–Hasselbalch equation becomes:
4.5 = 5.5 + log([A–]/[HA])
–1 = log([A–]/[HA])
So [A–]/[HA] = 10⁻¹ = 0.1 or 1:10. This is an incorrect calculation.
This question asks about the number of Lewis bases that coordinate the K⁺ ion associated with the G4-tetrads in Figure 1.
To understand this question, we must understand the principles of metal ion coordination in G-quadruplex structures.
A G-quadruplex (G4) is a secondary structure formed by guanine-rich sequences of RNA or DNA. These structures consist of stacked guanine tetrads (G-tetrads), where four guanines are arranged in a square planar configuration and stabilized by Hoogsteen hydrogen bonding. The central cavity of the G-quadruplex is negatively charged due to the carbonyl oxygens of the guanines, which enables selective coordination of monovalent metal cations such as K⁺ or Na⁺.
The passage indicates that in the case of Spinach RNA, two G4-tetrads are stacked together, creating a larger central cavity. A single K⁺ ion is positioned in the middle of these two tetrads.
The K⁺ ion is a Lewis acid (an electron-pair acceptor) that interacts with the Lewis bases (electron-pair donors). The Lewis bases in this case are the carbonyl oxygens from the guanines in both tetrads.
Since one K⁺ ion is in the center of the two stacked G4-tetrads, it interacts with the four carbonyl oxygens from the top tetrad and the four carbonyl oxygens from the bottom tetrad, making a total of 8 coordinating Lewis bases.
The structural diagram in the passage illustrates that the metal cation, acting as the Lewis acid, is positioned at the center of the tetrad. Given the tetrad’s symmetry, there is no justification for the metal ion to coordinate exclusively with only two guanines, which serve as the Lewis bases. This is an incorrect calculation.
This response assumes either that each G4-tetrad contains its own K⁺ ion at the center or that only one G4-tetrad has a K⁺ ion in its center. The first scenario would result in repulsion between the two K⁺ ions, while the second scenario is not feasible since the K⁺ ion is too large to fit within a single tetrad’s center. This is an incorrect calculation.
The structural diagram in the passage depicts the metal cation located in the center of the tetrad. Due to the tetrad’s symmetry, the number of Lewis bases coordinating with the metal cation must be either four (above the center of one tetrad) or eight (within the cavity between two stacked tetrads). Four is incorrect here. This is an incorrect calculation.
This is consistent with the fact that four guanines from each of the two stacked tetrads coordinate with the single K⁺ ion, totaling eight Lewis bases.
This question asks about the conditions that are required for the correct positioning of Compound 1 in the G4-tetrad as shown in Figure 2.
To answer this question, we must understand the principles of stereochemistry, particularly the E/Z isomerism of alkenes, and how the spatial arrangement of Compound 1 affects its binding to the G4-tetrad of the Spinach RNA aptamer.
The E/Z notation describes the relative positions of the highest-priority groups attached to each carbon in a double bond. In the case of anionic Compound 1, the fluorophore contains an alkene moiety with substituents of differing priorities.
According to the Cahn–Ingold–Prelog priority rules, the highest-priority groups on the alkene carbons are the nitrogen (N) and the phenyl carbon (C). The E (entgegen) designation indicates that these high-priority groups are positioned on opposite sides of the double bond, whereas the Z (zusammen) designation would mean that they are on the same side.
The correct alignment of anionic Compound 1, as depicted in Figure 2, requires E-stereochemistry so that the fluorophore can engage in hydrogen bonding with G28. The highest-priority groups on the alkene carbons—nitrogen (N) and the phenyl carbon (C)—are positioned opposite each other (“entgegen,” E) rather than on the same side (“zusammen,” Z).
Compound 1 does not contain any sp³-hybridized carbon atoms with four distinct substituents, meaning it lacks stereogenic centers and cannot have an (R)-configuration. This is an incorrect interpretation.
Since Compound 1 has no sp³-hybridized carbon atoms with four unique substituents, it does not possess stereogenic centers and therefore cannot have an (S)-configuration. This is an incorrect interpretation.
This is consistent with the stereochemical requirements shown in Figure 2. The fluorophore must adopt the E-configuration for correct binding to G28.
Figure 2 indicates that the alkene in anionic Compound 1 must adopt the E-configuration to properly engage in hydrogen bonding with G28, ensuring stable binding within the RNA structure. The Z-configuration would misalign the interacting groups. This is an incorrect interpretation.
This question asks about the color of the emitted fluorescence light upon activation of Compound 1 by the Spinach RNA aptamer.
To answer this question, we must understand the principles of fluorescence and the relationship between the wavelength of emitted light and its perceived color.
Fluorescence occurs when a molecule absorbs light at a specific excitation wavelength and then emits light at a longer wavelength. In this case, the Spinach RNA aptamer binds to the fluorophore Compound 1, activating its fluorescence.
According to the passage, Compound 1 has an excitation maximum at 468 nm and an emission maximum at 501 nm in air. The color of emitted light is determined by its wavelength, and visible light is categorized into different color ranges.
Green light spans wavelengths from approximately 490 nm to 570 nm, while other colors such as yellow, orange, and red have emission ranges at higher wavelengths.
The emission wavelength of Compound 1 is 501 nm, which falls within the green region of the visible spectrum (490 nm to 570 nm). This means that when Spinach RNA activates the fluorophore, it fluoresces with green light.
This is consistent with the emission wavelength of 501 nm, which falls within the green light range.
Orange light falls within the wavelength range of 580–620 nm. A substance that absorbs blue light would reflect orange light, but 501 nm is outside this range. This is an incorrect interpretation.
Red light spans wavelengths from 650–750 nm, significantly higher than the emission of 501 nm. This is an incorrect interpretation.
Yellow light is in the range of 570–585 nm. The emission at 501 nm is outside this range. This is an incorrect interpretation.
If we recall, a charged particle moving within a magnetic field experiences a force known as the Lorentz force. This force is given by the equation F = qvB, where ‘q’ is the charge of the particle, ‘v’ is its velocity, and ‘B’ is the magnetic field strength. The key characteristic of this force is that it is always perpendicular to both the velocity of the particle and the magnetic field. This perpendicularity means the force acts as a centripetal force, causing the particle to move in a circular or helical path, but it does not change the particle’s speed. Because the force is perpendicular, the work done by the magnetic field on the particle is zero, as work is only done when a force acts along the direction of motion. Since no work is done, the kinetic energy of the particle remains constant, and therefore its speed remains constant.
It is not necessary to know any specific quantity to determine the speed of the particle within the magnetic field. The speed remains constant because the Lorentz force does no work on the particle, as it acts perpendicularly to the particle’s velocity.
The charge of the particle is not required to calculate the speed. The Lorentz force, while dependent on the charge, does not alter the speed of the particle because it acts perpendicularly to the velocity vector, resulting in zero work done.
The magnitude of the magnetic field itself is not necessary to calculate the speed. While the magnetic field is essential for the Lorentz force to exist, it does not change the speed of the particle due to the perpendicular nature of the force.
The angle between the particle’s velocity and the magnetic field lines does not affect the speed calculation. The angle determines the direction and magnitude of the Lorentz force, but because this force is always perpendicular to the velocity, it does not change the particle’s speed.
