In the supplied beta-cyclodextrin host conformer, oxygen O3 bridges carbon C4 and carbon C37, forming a C4-O3-C37 linkage. What is the C-O-C bond angle at this bridging oxygen? Report the angle in degrees to two decimal places.
原参考答案（未改写）
116.0251189 degree
原评分
Compute the vector angle at the middle atom. Absolute tolerance 0.1 degree. Equivalent rounding within tolerance is accepted. No thermodynamic conclusion is requested.
In the indicated phenylene ring of the supplied WP6 conformer, what is the internal C3-C1-C4 bond angle at ring carbon C1? Report the angle in degrees to two decimal places.
原参考答案（未改写）
119.9446787 degree
原评分
Angle at carbon row 1, from vectors to carbons rows 3 and 4. Accept within 0.1 degree of the computed value; report two decimals.
For the indicated carboxylate arm in the supplied WP6 charge-state model, what is the C20-O22 bond length? Report the length in angstrom to three decimal places.
原参考答案（未改写）
1.272432316 angstrom
原评分
Euclidean distance between rows 20 and 22; tolerance 0.001 angstrom. Report three decimals.
For the supplied paired molecular geometries A and B, does the total force at B oppose or reinforce continuing the specific A-to-B deformation? Compute S = sum_i F_i(B) dot [r_i(B)-r_i(A)] in eV to four decimal places. For the path r_i(lambda) = r_i(A) + lambda [r_i(B)-r_i(A)], state the sign of dE/dlambda at lambda=1. Explain whether this establishes that A is a stable minimum of the PBE0+MBD energy.
原参考答案（未改写）
S = -1.8479097251 eV (rounded: -1.8479 eV), below the negative threshold: the instantaneous force projection opposes the specified A-to-B deformation at B. Since forces are minus the Cartesian energy gradient, dE/dlambda at lambda=1 = -S = 1.8479097251 eV, positive. This is a local directional derivative, not E(B)-E(A). It does not establish a stable PBE0+MBD minimum at A: A was optimized at a different level, and neither a stationary-point check nor curvature in all relevant directions is provided.
原评分
5 points: S within 0.0001 eV of independently calculated reference (1); oppose classification using stated threshold (1); positive dE/dlambda with value -S (1); distinguish local derivative from finite energy change (1); no claim of stable minimum, with differing optimization level or missing stationarity/curvature as a reason (1). Interpretive points require semantic review.
What is the donor-acceptor O6...O14 separation in the indicated intramolecular hydroxyl/carbonyl contact? Calculate it from the supplied coordinates and report angstrom to three decimals.
原参考答案（未改写）
2.538116731 angstrom
原评分
Euclidean separation of atom rows 6 and 14. Accept within 0.001 angstrom; use coordinates, not a published rounded annotation.
Which conformer satisfies both the stated short-contact and directional criteria for the indicated O-H...O arrangement? Report H...O distance and O-H...O angle for A and B, then classify each. Explain why donor-acceptor proximity alone is not sufficient.
原参考答案（未改写）
A: H...O = 1.611262 angstrom, angle = 151.490090 degrees; passes both. B: H...O = 3.733213 angstrom, angle = 8.966231 degrees; fails both. Close O...O centers do not establish a correctly oriented donor-H group.
原评分
1 point per distance within 0.002 angstrom and per angle within 0.2 degree (4); 1 for both classifications; 1 for distinguishing orientation from donor-acceptor distance.
Within the conventional red-shifting OH hydrogen-bond regime studied for this conformer pair, which structure is expected to have the lower OH stretching frequency? Support the qualitative assignment with the O-H bond lengths and the orientation/contact of the donor toward its carbonyl acceptor. Can these coordinates determine an exact frequency?
原参考答案（未改写）
A has the lower expected OH stretching frequency under the stated interpretation. Its O-H bond is 1.002922 angstrom versus 0.962927 angstrom for B. A also points toward the acceptor (H...O 1.611262 angstrom, angle 151.490090 degrees); B points away (H...O 3.733213 angstrom, angle 8.966231 degrees). This supports the qualitative red-shift assignment for this pair, not a universal law. Coordinates alone do not yield an exact frequency without a force-constant/Hessian calculation.
原评分
1 point for each O-H length within 0.002 angstrom (2); 1 for geometry-supported contact/orientation comparison; 1 for A as lower frequency under stated scope; 1 for refusing an exact frequency and naming missing curvature information.
Identify the closer donor-H...acceptor direction in each supplied conformer, assign its local interaction force constant using the pair-specific lookup, and state which local interaction is stiffer and by what ratio. Does this establish which entire conformer is thermodynamically more stable?
原参考答案（未改写）
A selects N3-H5...O13: H...O 2.008861 angstrom, angle 128.561836 degrees, hence 0.150 millidynes/angstrom. B selects O5-H4...N11: H...N 1.926478 angstrom, angle 138.283726 degrees, hence 0.201 millidynes/angstrom. B has the stiffer local interaction by 0.201/0.150 = 1.340. This does not establish total conformer thermodynamic stability: local curvature is not total energy or free energy.
原评分
1 point for each direction with geometry evidence (2); 1 for correct pair-specific assignments; 1 for ratio within 0.002; 1 for separating local stiffness from overall thermodynamic stability. Distances within 0.002 angstrom and angles within 0.2 degree.
For the crambin CA model, calculate (a) the N-to-C terminal separation and (b) the maximum distance over all pairs of CA sites. Identify the residues attaining the maximum and quantify how much the terminal distance underestimates this maximum. Would a terminal-distance measurement alone describe this model's full spatial extent?
原参考答案（未改写）
dNC=1.180957 nm; Dmax=2.902186 nm, attained by residues A:19 and A:42. Underestimate Dmax-dNC=1.721229 nm (dNC/Dmax=0.406920). The terminal separation samples just one pair and does not describe the full CA extent here.
原评分
6 points: dNC (1), Dmax (1), correct unordered maximizing residue pair (1), difference (1), explanation that terminal pair need not be maximally separated (1), restriction to CA model instead of all-atom/solvent envelope (1). Numeric nm tolerances 0.002.
Calculate the equal-weight CA radius of gyration of each supplied protein. Which has the larger spatial spread, and what is Rg(1UBQ)/Rg(1CRN)? Does that ranking alone establish which protein is less stably folded?
原参考答案（未改写）
Rg(1UBQ)=1.149289 nm and Rg(1CRN)=0.967550 nm; ratio=1.187835. 1UBQ has larger equal-CA spatial spread. Different chain lengths/compositions and lack of free-energy or ensemble information prevent inferring folding stability from this ranking. These are unweighted geometric values, not measured SAXS radii.
原评分
6 points: each Rg (2 total; tolerance 0.002 nm), ratio (1; tolerance 0.003), spatial-spread ranking (1), no folding-stability inference (1), equal-site rather than mass/electron-density interpretation (1).
For the supplied DNA paired-sugar marker sequence, calculate its polygonal path length L, its endpoint chord r, and the ratio r/L. Is this marker path a straight segment? Explain why this ratio should not be reported as a standard DNA helical-axis bend angle.
原参考答案（未改写）
L=4.101719 nm; r=3.686551 nm; r/L=0.898782. Since r<L the chosen marker path is not a straight segment. This ratio is a path straightness descriptor, not an angle. Sugar midpoints retain effects of helical sugar arrangement and are not a fitted helical axis; they cannot establish the paper's helical-axis bend angle.
原评分
6 points: L and r (2; tolerance 0.002 nm), ratio (1; tolerance 0.002), not straight based on r<L (1), dimensionless ratio is not angle (1), task-specific markers differ from helical-axis/base-pair frames (1).
Under the supplied equal-scatterer model, which protein has lower normalized scattering I(q)/I(0) at q=0.5 nm^-1? Calculate both values from the coordinates. At q=1.0 nm^-1, for which protein(s) does the stated Guinier-validity rule permit a prediction? Give qRg for both and predict intensity only where the rule permits.
原参考答案（未改写）
Rg: 1UBQ 1.149289 nm, 1CRN 0.967550 nm. At q=0.5, qRg=0.574645 and 0.483775; both satisfy the rule. Normalized intensities are 0.895769 and 0.924953, respectively; 1UBQ is lower. At q=1.0 the products are 1.149289 and 0.967550. Only 1CRN passes, giving 0.731944. The stipulated approximation supplies no prediction for 1UBQ at that q; this does not mean its scattering vanishes.
原评分
8 points: both geometric Rg values (1 combined; 0.002 nm), both q=0.5 intensities (2; 0.002), correct ranking (1), q=1 products (1 combined; 0.002), validity classifications (1), accepted 1CRN intensity (1; 0.002), explicit refusal for 1UBQ without zero/experimental-intensity claim (1).
In the specified virtual DNA-end FRET experiment, calculate the transfer efficiency. An analyst instead uses the length along the full sugar-marker path as the transfer distance: calculate that alternative efficiency and its signed error E_path-E_correct. Explain which distance the physical model requires and why this is not a predicted experimental efficiency for actual 1BNA.
原参考答案（未改写）
Interprobe straight-line r=3.686551 nm gives E_correct=0.422731. Marker path L=4.101719 nm gives E_path=0.278511; signed error=-0.144220. FRET uses spatial donor-acceptor separation, not distance along the molecular marker path. Actual dye locations, orientation factors, linker motion and photophysics are not established by these sugar coordinates or the assumed effective R0; this is a conditional virtual-probe calculation only.
原评分
7 points: r and L (2; 0.002 nm), correct and alternative E (2; 0.002), signed error (1; 0.003), spatial vs path explanation (1), distinction between assumed virtual probes and real measured FRET (1).
For the silicon atom at fractional (0,0,0), find every atom in its shortest-distance nonzero coordination shell. Give the coordination number, shell distance, each neighbor as basis-row plus integer-cell translation, and all distinct neighbor-center-neighbor angles. What coordination polyhedron does this geometry form?
原参考答案（未改写）
Coordination 4; distance 0.2351562 nm. Neighbors: row 3, translation [0, 0, 0]; row 4, translation [0, -1, -1]; row 7, translation [-1, 0, -1]; row 8, translation [-1, -1, 0]. All six pair angles are 109.471221 degrees, so the shell is tetrahedral. Periodic translations are essential to recover all neighbors.
原评分
7 points: count (1), distance within0.0001 nm (1), complete four row/translation pairs locating distinct correct images (2), all six angles / single unique value within0.1 degree (1), tetrahedral interpretation (1), periodic rather than isolated-cell treatment (1). Equivalent Cartesian neighbor vectors accepted.
Around the Na at fractional (0,0,0), determine separately (a) the closest Cl shell and (b) the closest noncentral Na shell. Report each distance and population, identify the first-shell coordination polyhedron, and explain why counting only basis atoms in the displayed cell can give a wrong population.
原参考答案（未改写）
Closest Cl: 0.2820280 nm, 6 neighbors along ±x, ±y, ±z at a/2, giving an octahedron. Closest other Na: 0.3988478 nm, 12 neighbors at a/sqrt(2), with two components ±a/2 and the third0. Some neighbors belong to translated cells; an eight-atom basis is not an isolated coordination cluster.
原评分
7 points: two distances (2;0.0001 nm), two populations (2), first-shell directions (1), octahedron (1), explanation of missing periodic images (1).
Determine the perpendicular spacings d111 and d220 for each supplied crystal, and the acute angle between the (111) and (100) plane normals. Are identical Miller indices and the same cubic structure type sufficient to make the two materials' plane spacings equal?
原参考答案（未改写）
Silicon: d111=0.3135416 nm; d220=0.1920042 nm. Diamond: d111=0.2059287 nm; d220=0.1261051 nm. The (111)/(100) normal angle is 54.735610 degrees for both. Spacing is a/sqrt(h^2+k^2+l^2); matching hkl/cubic type fixes geometry ratios and orientations but not the material-specific lattice parameter.
原评分
7 points: four spacings (4;0.0001 nm), normal angle (1;0.1 degree), correct perpendicular-plane derivation (1), explanation of lattice-parameter dependence without strain claim (1).
For first-order (111) diffraction at the stated wavelength, calculate the scattering angle 2theta for each supplied material. Which has the higher-angle reflection and what is the angular difference? Explain the direction of the shift using the structural spacing.
原参考答案（未改写）
Silicon d111=0.3135416 nm gives 2theta=28.443706 degrees. Diamond d111=0.2059287 nm gives 2theta=43.932637 degrees. Diamond is higher by 15.488931 degrees because its smaller d requires larger sin(theta) at fixed lambda. These are ideal calculated positions, not reported measured peaks or a strain measurement between these two materials.
From the supplied eight-atom silicon basis, calculate F and I* for (111), (200) and (220). Identify any cancelled reflection and demonstrate its cancellation by grouping the atomic phase contributions. Does cancellation mean the corresponding geometric lattice-plane family does not exist?
原参考答案（未改写）
111: F=4-4i, |F|²=32; 200: F=0+0i, |F|²=0; 220: F=8+0i, |F|²=64. For (200), four atomic factors are +1 and four are -1, hence cancellation. For (111), four are +1 and four are -i in this supplied origin; for (220), all8 are +1. The (200) plane family still exists geometrically; the ideal identical-atom basis contributions cancel there.
原评分
8 points: three F values (3; absolute tolerance0.000001 per real/imag part), three I* values (1 combined;0.000001), (200) cancellation and four-plus/four-minus groups (2), demonstration of the other phase groups (1), distinction between plane geometry and reflection amplitude (1).
Using the supplied NaCl basis and model amplitudes, calculate F111, F200 and I*111/I*200. Repeat the ratio after setting both species' amplitudes to1 without moving any atom. Explain what this comparison demonstrates about geometry and chemical scattering contrast.
原参考答案（未改写）
Case A: F111=-24, F200=112; I*111=576 and I*200=12544, giving ratio=0.045918367. Case B: F111=0, F200=8; ratio=0. At111 the Na and Cl sublattices contribute opposite phases, while at200 they add. Equal amplitudes cancel the odd reflection in this ideal model; unequal amplitudes leave a residual despite identical atom positions. Geometry fixes phases, but intensity also depends on scattering contrast. The ratio is not a corrected experimental powder intensity ratio.
原评分
8 points: case A F values (2;0.000001), case A ratio (1;0.000001), case B F values (2;0.000001), case B zero ratio (1), opposite/additive phase explanation (1), contrast dependence and nonexperimental scope (1).
Around the Mg at fractional (0,0,0), determine separately (a) the closest O shell and (b) the closest noncentral Mg shell. Report each distance and population, identify the first-shell coordination polyhedron, and explain why counting only basis atoms in the displayed cell can give a wrong population.
原参考答案（未改写）
Closest O: 0.2108500 nm, 6 neighbors along +-x, +-y, +-z at a/2, giving an octahedron. Closest other Mg: 0.2981869 nm, 12 neighbors at a/sqrt(2), with two Cartesian components +-a/2 and the third 0. Some neighbors belong to translated cells; the eight-atom basis stored in one cell is not itself an isolated coordination cluster, so counting only its 8 rows without periodic images would miss most of the 6 O and all but a few of the 12 Mg neighbors.
原评分
7 points: two distances (2; 0.0001 nm), two populations (2), first-shell directions (1), octahedron (1), explanation of missing periodic images (1).
Around the Cs at fractional (0,0,0), determine separately (a) the closest Cl shell and (b) the closest noncentral Cs shell. Report each distance and population, identify the first-shell coordination polyhedron and its full set of distinct neighbor-neighbor angles about the central Cs, and explain why counting only the two atoms in the displayed primitive cell cannot give either population.
原参考答案（未改写）
Closest Cl: 0.3570623 nm (=a*sqrt(3)/2), 8 neighbors at the corners of a cube centered on the Cs atom, at Cartesian displacements (+-a/2,+-a/2,+-a/2) for all 8 sign combinations. Closest other Cs: 0.4123000 nm (=a), 6 neighbors along +-x, +-y, +-z (the simple-cubic Cs sublattice). Among the 28 pairs of the 8 Cl neighbors, pairs differing in exactly one sign (12 pairs) subtend 70.53 degrees, pairs differing in exactly two signs (12 pairs) subtend 109.47 degrees, and the 4 antipodal (through-center) pairs subtend 180.00 degrees; these are the standard cube vertex-center-vertex angles. The stored primitive cell has only 2 atoms, so neither the 8-fold Cl shell nor the 6-fold Cs shell can be read off it directly; both require the translated images r_j+a*(n_x,n_y,n_z).
原评分
8 points: two distances (2; 0.0001 nm), two populations (2), cube (1), all three distinct angle values within 0.1 degree (2), explanation of missing periodic images (1).
判断：答案用“fewer atoms fit between planes”解释较小晶面间距，输入与公式没有支持该原子计数解释。
下一步：删除原子计数解释，改为固定波长下 d 与 sin(theta) 的关系。
审读的题干、参考答案与评分
题干
For first-order (111) and (200) diffraction at the stated wavelength, calculate the scattering angle 2theta for the supplied MgO structure. Which reflection is at the higher angle and what is the angular difference? Explain the direction of the shift using the structural spacing.
原参考答案（未改写）
d111=0.2434686 nm gives 2theta=36.888982 degrees. d200=0.2108500 nm gives 2theta=42.855885 degrees. (200) is at the higher angle, by 5.966903 degrees, because its smaller d (fewer atoms fit between planes normal to (200) than (111) for this rock-salt cell geometry) requires a larger sin(theta) at the same fixed wavelength. These are ideal calculated positions, not measured peak intensities, and do not by themselves establish whether the (200) reflection is experimentally strong or weak.
原评分
7 points: two spacings (2; 0.0001 nm), each 2theta (2; 0.02 degree), difference (1; 0.03 degree), correct ordering with fixed-wavelength geometric explanation (1), theta/2theta distinction (1).
Using the supplied CsCl basis and model amplitudes, calculate F and I* at (100), (110) and (111) for case A (real atomic-number weights) and case B (equal weights, both unchanged coordinates). Identify which reflections vanish in case B, name the resulting selection rule, and explain what the comparison shows about the difference between a plane family's geometric existence and its experimental diffraction visibility.
原参考答案（未改写）
Case A (f_Cs=55, f_Cl=17): F100=38, I*100=1444; F110=72, I*110=5184; F111=38, I*111=1444 (all real, zero imaginary part in this origin choice). Case B (f_Cs=f_Cl=1): F100=0, I*100=0; F110=2, I*110=4; F111=0, I*111=0. In case B, (100) and (111) vanish because h+k+l is odd there, which is exactly the body-centered-cubic selection rule F=0 for h+k+l odd (this primitive 2-atom basis with equal weights is geometrically equivalent to a BCC lattice); (110) survives because h+k+l=2 is even. The (100) and (111) plane families exist geometrically in either case (d_hkl=a/sqrt(h^2+k^2+l^2) is unaffected by the weights), but their case-A reflections are nonzero only because Cs and Cl scatter unequally: their presence in a real diffraction pattern is direct evidence of chemical contrast between the two sites, not merely of the lattice geometry.
原评分
9 points: case A three F values (3; exact integers), case A three I* values (1 combined), case B three F values (2; exact integers, includes the two zeros), naming the body-centered-cubic h+k+l-odd selection rule (1), geometric-existence-vs-diffraction-visibility explanation (2).
判断：只提供 CA 位点，却要求由 CA 间距论证侧链对底物的催化可达性与机制，且混合局部和整体任务。
下一步：保留整体尺寸题；移除机制评分，或另取侧链/底物及机制证据构建独立题。
审读的题干、参考答案与评分
题干
For the supplied hen egg-white lysozyme CA model, calculate (a) the N-to-C terminal separation, (b) the maximum distance over all pairs of CA sites, and (c) the CA-CA distance between the catalytic residues Glu35 and Asp52. Identify the residues attaining the maximum pairwise distance. Why must (c) be small enough for a two-carboxylate acid/base mechanism to work, yet nonzero?
原参考答案（未改写）
dNC=2.092 nm (residues Lys A:1 to Leu A:129). Dmax=4.384 nm, attained by residues A:70 and A:126. d_cat=0.973 nm between Glu35 and Asp52. A gap of about 1 nm across the substrate-binding cleft is small enough that both carboxylate side chains can simultaneously reach the sugar ring/glycosidic-bond region of a bound oligosaccharide substrate spanning the cleft, but the two residues remain on opposite faces of that substrate, letting one act as a general acid (protonating the leaving-group oxygen) while the other stabilizes the developing positive charge on the other face; if the gap were much larger neither residue could reach the reacting bond, and if it were near zero the two carboxylates could not independently address two different chemical roles on either side of the substrate.
原评分
8 points: dNC (1; 0.001 nm), Dmax (1; 0.001 nm), maximum-pair residue numbers (1), d_cat (2; 0.001 nm), explanation that a catalytic gap must be small enough for both carboxylates to contact the substrate's glycosidic bond region yet nonzero so each can independently protonate/stabilize different atoms (3).
For the supplied yeast phenylalanine tRNA C1' backbone model, calculate (a) the polygonal backbone path length from residue 1 to residue 76, (b) the endpoint chord (residue 1 to residue 76 straight-line distance), and (c) the path/chord ratio. Is this molecule's backbone close to a straight or gently curved chain, or is it folded back on itself? What does a large path/chord ratio show about the tertiary structure that the path length or chord alone would not?
原参考答案（未改写）
L=45.376 nm; c=1.900 nm; ratio c/L=0.042 (equivalently L/c=23.888). The chain is folded back on itself, not straight or gently curved: the backbone travels roughly 24 times farther than the direct distance between its two ends. Path length alone would only say the chain is long; the chord alone would only say the two ends happen to be close together. Their large ratio together is direct geometric evidence that the intervening backbone loops back through space, consistent with tRNA's known folded L-shaped tertiary structure rather than an extended or lightly bent single strand.
原评分
7 points: L (2; 0.001 nm), c (1; 0.001 nm), ratio (1; 0.001), correct qualitative statement that the chain is folded back on itself rather than straight or gently curved (2), explanation that a large ratio shows extensive compact folding that neither the path length nor the chord alone would reveal (1).
For the anticodon triplet (residues 34, 35, 36) of the supplied tRNA, calculate the three pairwise C1'-C1' distances (34-35, 35-36, 34-36) and the angle at residue 35 (vertex 35, using the 34-35 and 35-36 vectors). Is this triplet closer to a straight line, a right angle, or a folded-back hairpin? What does that geometry indicate about the availability of these three bases for pairing with an mRNA codon?
原参考答案（未改写）
d(34,35)=0.526 nm; d(35,36)=0.500 nm; d(34,36)=0.946 nm; angle at residue 35 = 134.53 degrees. This is neither a straight line (180 degrees) nor a folded-back hairpin (a small angle with d(34,36) much less than d(34,35)+d(35,36)); it is a gently bent, largely extended single-stranded segment (d(34,36)=0.946 nm is close to but slightly less than the 1.025 nm sum of the two adjacent distances). This open, non-folded-back conformation leaves the anticodon triplet's base-pairing edges exposed and roughly aligned along the loop, consistent with its role of presenting three consecutive bases for antiparallel Watson-Crick pairing with an mRNA codon rather than burying them in internal contacts.
原评分
7 points: three distances (3; 0.001 nm), angle (2; 0.02 degree), correct qualitative call of a gently bent, largely extended single-stranded segment rather than a straight line or a folded-back hairpin (1), correct explanation that this open, accessible conformation is what lets the three bases stack in a way that presents their Watson-Crick edges for antiparallel codon pairing (1).
Calculate the equal-weight CA radius of gyration of the supplied lysozyme structure. Using the already-established values Rg(1UBQ)=1.149289 nm and Rg(1CRN)=0.967550 nm, rank all three proteins by Rg and by chain length (129, 76, 46 residues respectively). Do the two rankings agree? Does that agreement alone establish that longer chains always fold less compactly, or that compactness is fully determined by length?
原参考答案（未改写）
Rg(lysozyme)=1.375 nm. Ranking by Rg: lysozyme (1.375 nm) > 1UBQ (1.149289 nm) > 1CRN (0.967550 nm). Ranking by chain length: lysozyme (129) > 1UBQ (76) > 1CRN (46). The two rankings agree exactly for these three proteins. This agreement does not establish a general rule that longer chains always fold less compactly: with only three data points, different folds (an alpha/beta hydrolase, a beta-grasp fold, and a small disulfide-stabilized fold), and no free-energy or ensemble information, this ranking cannot separate a genuine length-compactness trend from coincidence among these particular three structures, and radius of gyration by itself says nothing about folding stability.
原评分
7 points: Rg(lysozyme) (2; 0.001 nm), correct ranking by Rg and by length (2), correct statement that the rankings agree here (1), correct caveat that three data points with different folds cannot establish a general length-compactness law and that Rg alone does not measure folding stability or free energy (2).
In a virtual FRET experiment placing a point donor at the tRNA's residue-76 C1' site (the 3' acceptor terminus) and a point acceptor at the residue-35 C1' site (anticodon center), calculate the transfer efficiency assuming an effective R0=3.5 nm. Given that documented R0 values for real dye pairs are typically in the 1 to 7 nm range, is this R0/dye-pair choice well matched to the measured donor-acceptor distance? What would an experimentalist need to change to get a more informative efficiency reading for this particular distance?
原参考答案（未改写）
r=7.234 nm; E=0.012663. This R0/dye-pair choice is poorly matched to the measured distance: FRET efficiency falls off as the sixth power of r/R0, and at r/R0 approximately 2.07 the predicted efficiency is already down near the assay's practical noise floor, well past the steep, informative part of the E(r) curve that lies close to R0. Since documented R0 values for real dye pairs are typically in the 1 to 7 nm range, an experimentalist studying a ~7.2 nm separation should choose a donor-acceptor pair with a larger R0 (closer to this distance) so that E responds sensitively to conformational changes, or instead measure the distance with multiple shorter-range FRET pairs placed along intermediate points of the fold, or use a different distance-sensitive method (e.g. paramagnetic relaxation, cross-linking) better suited to this length scale.
原评分
8 points: r (2; 0.001 nm), E (2; 1e-6), correct statement that E is very small/near floor and that r is well beyond the R0-comparable range where FRET is most sensitive (2), concrete recommendation such as choosing a longer-R0 dye pair, using multiple shorter-range FRET pairs along the path, or an entirely different distance-sensitive method for r on this scale (2).
In the supplied H26DM-B-CD (heptakis-2,6-di-O-methyl-beta-cyclodextrin) host conformer, oxygen O3 bridges carbon C4 and carbon C37, forming a C4-O3-C37 linkage. What is the C-O-C bond angle at this bridging oxygen? Report the angle in degrees to two decimal places.
原参考答案（未改写）
116.0251189 degree.
原评分
Angle at O3 (row 46), from vectors to C4 (row 4) and C37 (row 37). Accept within 0.1 degree of the computed value; report two decimals.
In the supplied H26DM-B-CD host conformer, one of the exocyclic methyl-ether oxygens, O2, bridges ring carbon C3 and a methyl carbon (row 45). What is (a) the C3-O2-C(45) bond angle at O2, (b) the O2-C(45) bond length, and (c) the C3-O2 bond length? Report the angle in degrees and both lengths in angstrom, each to two decimal places.
原参考答案（未改写）
C3-O2-C(45) angle = 107.62 degree. O2-C(45) bond length = 1.43 angstrom. C3-O2 bond length = 1.43 angstrom. Both C-O bond lengths are typical sp3 ether single-bond distances (roughly 1.41-1.44 angstrom), and the O2 angle (107.62 degree) is close to but somewhat compressed from the ideal tetrahedral 109.47 degree, as is common for ether oxygens with two lone pairs.
原评分
Angle at O2 (row 44) from vectors to C3 (row 3) and methyl-C (row 45): tolerance 0.1 degree. O2-C(45) length: tolerance 0.01 angstrom. C3-O2 length: tolerance 0.01 angstrom. All three values required; report two decimals each.
CNP001 (already accepted in this project) reports the same-position C4-O3-C37 bridging angle for the unmethylated B-CD host conformer as 116.0251189 degree. Using your C4-O3-C37 measurement on the supplied H26DM-B-CD (2,6-di-O-methylated) conformer, calculate the signed difference (methylated minus unmethylated). Is this difference large, small, or essentially zero relative to the two-decimal precision these questions are reported at? What does that indicate, and what would it NOT be valid to conclude from a single conformer pair like this?
原参考答案（未改写）
This conformer's C4-O3-C37 angle is 116.0251189 degree (matching CNP005). Signed difference (methylated minus unmethylated) = 116.0251189 - 116.0251189 = 0.000000 degree (more precisely, about 4e-08 degree, numerical noise far below any meaningful precision). This is essentially zero: far smaller than the 0.1 degree tolerance these bond-angle questions are scored at. This indicates that, in these two independently generated MOE conformers, the 1,4-glycosidic ring-junction backbone is unchanged by the distal 2,6-di-O-methylation; the two structures appear to share the same macrocycle backbone coordinates and differ only in their exocyclic substituents (free OH replaced by O-CH3 at the 2,6 positions). It would NOT be valid to conclude from this single conformer pair that 2,6-di-O-methylation never perturbs glycosidic geometry in general: these are single static model conformers from the same generation method, not a conformational ensemble, not an experimental structure, and not evidence about behavior in solution or upon guest binding.
原评分
9 points: this conformer's angle restated or reused from CNP005 (1), signed difference computed correctly to 6 decimals (2; tolerance 1e-5 degree), correct characterization that the difference is essentially zero at reported precision (well under 0.001 degree, i.e. far below the 0.1 degree tolerance these questions are scored at) (2), correct qualitative explanation (this reflects the two MOE-generated conformers sharing an unchanged macrocycle ring backbone with only exocyclic substituents differing) (2), correct caveat that this single conformer pair does not establish a general rule that 2,6-di-O-methylation never affects glycosidic geometry in any conformer or in solution (2).
The supplied lysozyme structure's own SSBOND record pairs Cys6 and Cys127 as a disulfide bridge. Using the supplied SG (sulfur) coordinates, calculate the Cys6-Cys127 SG-SG distance. Report the distance in angstrom to three decimal places. Why must this distance be short enough to represent a real covalent S-S bond rather than two sulfurs that merely happen to be nearby?
原参考答案（未改写）
2.031 angstrom. This value is consistent with a real covalent S-S single bond, which has a well-defined length of about 2.0-2.1 angstrom; if the two sulfurs were merely nearby in space without a covalent bond, their separation would typically be several angstrom larger (van der Waals contact, not bonded contact), and this entry's own SSBOND record (2.03 angstrom) independently corroborates a bonded distance.
原评分
5 points: distance (3; 0.005 angstrom), correct explanation that a covalent single S-S bond has a well-defined short length (about 2.0-2.1 angstrom) so a much larger separation would indicate no real bond (2).
The supplied lysozyme structure's own SSBOND record pairs Cys30 and Cys115 as a second disulfide bridge. Using the supplied SG (sulfur) coordinates, calculate the Cys30-Cys115 SG-SG distance. Report the distance in angstrom to three decimal places.
原参考答案（未改写）
2.042 angstrom, matching this entry's own SSBOND record (2.04 angstrom).
Within the supplied lysozyme alpha helix (residues 25-35, per this entry's own HELIX record), calculate (a) the adjacent-residue Ca27-Ca28 distance and (b) the one-helical-turn Ca27-Ca31 (i, i+4) distance. Report both distances in angstrom to three decimal places. Which one is closer to the well-known typical peptide-backbone adjacent-Ca spacing of about 3.8 angstrom?
原参考答案（未改写）
Ca27-Ca28 = 3.798 angstrom; Ca27-Ca31 = 6.248 angstrom. The adjacent-residue distance, Ca27-Ca28 (3.798 angstrom), is the one close to the typical 3.8 angstrom peptide-backbone Ca-Ca spacing. The i,i+4 distance (Ca27-Ca31) is instead close to one helical turn's rise, which for a regular alpha helix is expected to be several angstrom larger than a single peptide unit, consistent with the measured 6.248 angstrom.
原评分
6 points: Ca27-Ca28 distance (2; 0.01 angstrom), Ca27-Ca31 distance (2; 0.01 angstrom), correct identification that the i,i+1 distance (Ca27-Ca28) is the one close to 3.8 angstrom, not the i,i+4 distance (2).
Two more disulfide bridges in the supplied lysozyme structure, Cys64-Cys80 and Cys76-Cys94, lie close enough together to compare directly. Calculate both SG-SG distances. Is the difference between them large or small relative to the typical covalent S-S single-bond length (about 2.0-2.1 angstrom)? What would a much larger difference (say, one bond at 2.0 angstrom and the other at 2.8 angstrom) have suggested instead?
原参考答案（未改写）
Cys64-Cys80 = 2.005 angstrom; Cys76-Cys94 = 2.110 angstrom. The difference (about 0.105 angstrom) is small relative to the typical covalent S-S bond length itself (2.0-2.1 angstrom): both values fall within the normal range expected for real disulfide bonds, whose exact length varies slightly with local strain and dihedral geometry even though each is a genuine covalent bond. A much larger difference, such as 2.0 versus 2.8 angstrom, would instead suggest that the longer 'bond' is not a real covalent disulfide at all (2.8 angstrom is well beyond a normal S-S single bond and closer to a non-bonded sulfur-sulfur contact), for example a broken, reduced, or merely spatially proximal cysteine pair rather than an intact disulfide bridge.
原评分
7 points: both distances (4; 0.005 angstrom each), correct statement that the ~0.1 angstrom difference is small and consistent with normal geometric variation among genuine covalent disulfide bonds, not a chemically different bond type (2), correct statement that a difference as large as 0.8 angstrom (2.0 vs 2.8) would instead suggest one of the two pairs is not a real covalent bond (a broken/reduced or merely proximal sulfur pair, not a disulfide) (1).
Using two independent i,i+4 Ca-Ca measurements within the supplied helix, Ca27-Ca31 and Ca28-Ca32, calculate both distances and their difference. Does this pair of measurements support treating residues 25-35 as a fairly regular alpha helix, or does it show a large, irregular jump between consecutive turns? What would a much larger difference between these two i,i+4 distances have indicated instead?
原参考答案（未改写）
Ca27-Ca31 = 6.248 angstrom; Ca28-Ca32 = 6.187 angstrom; signed difference (27-31 minus 28-32) = 0.061 angstrom. This is a small difference relative to the roughly 6.2 angstrom distances themselves (about 1 percent), and it supports treating residues 25-35 as a fairly regular alpha helix over this span, since consecutive helical turns advance by a consistent amount rather than jumping unpredictably. A much larger difference between these two i,i+4 measurements (for example, several angstrom) would instead indicate an irregular, kinked, or locally distorted segment rather than a uniform helical repeat, since a true regular helix produces nearly constant i,i+4 spacing along its length.
原评分
7 points: both distances (4; 0.01 angstrom each), signed difference (1; 0.01 angstrom), correct statement that a difference of about 0.06 angstrom out of roughly 6.2 angstrom is small and supports a fairly regular repeating helical geometry over this span (1), correct statement that a much larger difference would instead indicate an irregular, distorted, or kinked segment rather than a uniform helix turn (1).
The supplied crambin structure's own SSBOND record pairs Cys16 and Cys26 as a disulfide bridge. Using the supplied SG (sulfur) coordinates, calculate the Cys16-Cys26 SG-SG distance. Report the distance in angstrom to three decimal places.
原参考答案（未改写）
2.051 angstrom, matching this entry's own SSBOND record (2.05 angstrom).
Two disulfide bridges in the supplied crambin structure, Cys3-Cys40 and Cys4-Cys32, lie close enough together to compare directly. Calculate both SG-SG distances. Is the difference between them large or small relative to the typical covalent S-S single-bond length (about 2.0-2.1 angstrom)? Crambin packs three disulfide bonds into only 46 residues; does a small, chemically ordinary difference between these two bond lengths support or undermine describing all three as genuine stabilizing covalent crosslinks?
原参考答案（未改写）
Cys3-Cys40 = 2.004 angstrom; Cys4-Cys32 = 2.035 angstrom. The difference (about 0.030 angstrom) is very small relative to the typical covalent S-S bond length itself (2.0-2.1 angstrom), well within normal geometric variation among genuine disulfide bonds. This small, chemically ordinary difference supports, rather than undermines, describing both as real stabilizing covalent crosslinks; it is consistent with (though it alone does not prove) the third disulfide, Cys16-Cys26, also being genuine, since crambin's unusually high crosslink density for its size (three disulfides in 46 residues) is a well-documented feature of this reference structure, not an artifact of any single bond's exact length.
原评分
7 points: both distances (4; 0.005 angstrom each), correct statement that the ~0.04 angstrom difference is very small relative to typical covalent S-S bond length (2), correct statement that this small, chemically ordinary difference supports (does not undermine) treating both as genuine disulfide crosslinks, and by extension is consistent with (though does not by itself prove) the third bond also being a real crosslink (1).
Within the supplied crambin alpha helix (residues 7-19, per this entry's own HELIX record), calculate the one-helical-turn Ca9-Ca13 (i, i+4) distance. Report the distance in angstrom to three decimal places.
The already-accepted bio-nm-v04-003 candidate BNP009 reports an i,i+4 Ca-Ca distance of 6.248 angstrom within a real alpha helix of hen egg-white lysozyme (PDB 6LYZ). Using your Ca9-Ca13 measurement on the supplied crambin (PDB 1CRN) helix, calculate the difference between these two independently sourced, unrelated proteins' i,i+4 spacings. Is this difference small enough to treat both as examples of the same underlying regular alpha-helix geometry? What would a much larger difference (for example, several angstrom) have suggested instead?
原参考答案（未改写）
Crambin Ca9-Ca13 = 6.065 angstrom. Signed difference (crambin minus lysozyme's already-published BNP009 value, 6.248 angstrom) = -0.183 angstrom. This is small relative to the roughly 6.2 angstrom distances themselves (under 3 percent), and it supports treating both as real examples of the same underlying regular alpha-helix i,i+4 geometry, even though crambin and lysozyme are otherwise unrelated proteins with very different sequences, sizes and functions. A difference of several angstrom would instead have suggested that at least one of the two measured segments is not a regular alpha helix (for example, if it were actually a distorted, kinked, or different secondary-structure type), or that the two i,i+4 measurements were not really comparable structural features.
原评分
8 points: crambin distance restated or reused from BNP011 (1), signed difference computed correctly (2; tolerance 0.02 angstrom), correct statement that a difference of about 0.18 angstrom out of roughly 6.2 angstrom (under 3 percent) is small and supports treating both as instances of the same regular alpha-helix i,i+4 geometry despite being unrelated proteins (2), correct statement that a difference of several angstrom would instead suggest at least one of the two segments is not a regular alpha helix, or that the two structural features being compared are not actually equivalent (3).
The supplied atoms form the three Watson-Crick hydrogen bonds of a real G-C base pair (guanine, chain A residue 2; cytosine, chain B residue 23) in a B-DNA dodecamer. Calculate all three donor-acceptor distances: N1(G)-N3(C), N2(G)-O2(C), and O6(G)-N4(C). Report each distance in angstrom to three decimal places. A Watson-Crick A-T pair instead forms only two hydrogen bonds (N1-N3 and N6-O4). Which base pair type would you expect to be more thermally stable, and why?
原参考答案（未改写）
N1(G)-N3(C) = 2.827 angstrom; N2(G)-O2(C) = 2.817 angstrom; O6(G)-N4(C) = 2.709 angstrom. All three fall in the typical hydrogen-bond distance range (about 2.7-2.9 angstrom), consistent with a genuine Watson-Crick G-C pair. G-C base pairs are expected to be more thermally stable than A-T base pairs at the same position, because a G-C pair forms three hydrogen bonds versus an A-T pair's two: more hydrogen bonds means more favorable interactions must be broken to separate the two strands there, which is part of why GC-rich DNA regions generally have higher melting temperatures than AT-rich regions (base stacking also contributes and is not addressed by this hydrogen-bond count alone).
原评分
8 points: three distances (6; 0.01 angstrom each), correct identification that G-C pairs (3 hydrogen bonds) are expected to be more thermally stable than A-T pairs (2 hydrogen bonds), with the reasoning that more hydrogen bonds per base pair means more energy is required to separate the strands at that position (2).
In the attached optimized QM7-X geometry for molecule 7002 (11 atoms), row 5 and row 6 are the molecule's only two nitrogen atoms. What is the distance between row 5 and row 6? Report the value in angstrom to four decimal places.
原参考答案（未改写）
1.187247758 angstrom
原评分
Distance between row 5 and row 6 within 0.0001 angstrom of the computed reference value; report to four decimal places.
For the supplied paired molecular geometries A and B of QM7-X molecule 7002 (A is the attached optimized reference, B is its attached paired displaced structure), does the total force at B oppose or reinforce continuing the specific A-to-B deformation? Compute S = sum_i F_i(B) dot [r_i(B)-r_i(A)] in eV to four decimal places. For the path r_i(lambda) = r_i(A) + lambda [r_i(B)-r_i(A)], state the sign of dE/dlambda at lambda=1. Explain whether this establishes that A is a stable minimum of the PBE0+MBD energy.
原参考答案（未改写）
S = -0.4352795772 eV (rounded: -0.4353 eV), below the negative threshold: the instantaneous force projection opposes the specified A-to-B deformation at B. Since forces are minus the Cartesian energy gradient, dE/dlambda at lambda=1 = -S = 0.4352795772 eV, positive. This is a local directional derivative, not E(B)-E(A). It does not establish a stable PBE0+MBD minimum at A: A was optimized at a different level (DFTB3+MBD), and neither a stationary-point check nor curvature in all relevant directions is provided.
原评分
5 points: S within 0.0001 eV of the independently calculated reference (1); correct oppose/reinforce classification using the stated threshold (1); correct sign of dE/dlambda with value -S (1); distinguishes local directional derivative from a finite energy change (1); no claim of a stable minimum, citing the differing optimization level or missing stationarity/curvature check as the reason (1). Interpretive points require semantic review.
In the attached optimized QM7-X geometry for molecule 7035 (17 atoms), row 3 is the molecule's only sulfur atom, and row 4 is one of the two oxygen atoms with the shortest measured distance to row 3. What is the distance between row 3 and row 4? Report the value in angstrom to four decimal places.
原参考答案（未改写）
1.461565269 angstrom
原评分
Distance between row 3 and row 4 within 0.0001 angstrom of the computed reference value; report to four decimal places.
For the supplied paired molecular geometries A and B of QM7-X molecule 7035 (A is the attached optimized reference, B is its attached paired displaced structure), does the total force at B oppose or reinforce continuing the specific A-to-B deformation? Compute S = sum_i F_i(B) dot [r_i(B)-r_i(A)] in eV to four decimal places. For the path r_i(lambda) = r_i(A) + lambda [r_i(B)-r_i(A)], state the sign of dE/dlambda at lambda=1. Explain whether this establishes that A is a stable minimum of the PBE0+MBD energy.
原参考答案（未改写）
S = -5.5319173445 eV (rounded: -5.5319 eV), below the negative threshold: the instantaneous force projection opposes the specified A-to-B deformation at B. Since forces are minus the Cartesian energy gradient, dE/dlambda at lambda=1 = -S = 5.5319173445 eV, positive. This is a local directional derivative, not E(B)-E(A). It does not establish a stable PBE0+MBD minimum at A: A was optimized at a different level (DFTB3+MBD), and neither a stationary-point check nor curvature in all relevant directions is provided.
原评分
5 points: S within 0.0001 eV of the independently calculated reference (1); correct oppose/reinforce classification using the stated threshold (1); correct sign of dE/dlambda with value -S (1); distinguishes local directional derivative from a finite energy change (1); no claim of a stable minimum, citing the differing optimization level or missing stationarity/curvature check as the reason (1). Interpretive points require semantic review.
In the attached optimized QM7-X geometry for molecule 7040 (16 atoms), row 3 is the molecule's only sulfur atom, and rows 4 and 5 are the two oxygen atoms with the shortest measured distance to row 3. What is the angle at row 3, measured from row 4 to row 3 to row 5? Report the value in degrees to two decimal places.
原参考答案（未改写）
121.9601735 degree
原评分
Angle at row 3 from row 4 to row 5 within 0.1 degree of the computed reference value; report to two decimal places.
QM7-X molecule 7040 has one optimized reference A (attached) and two independently generated displaced structures, d1 and d2 (both attached), each paired with A in the same atom order. For each displacement, compute S = sum_i F_i(displaced) dot [r_i(displaced)-r_i(A)] in eV using the supplied total force table at that displaced geometry. Report S_d1 and S_d2 to four decimal places, state whether the total force opposes or reinforces continuing each specified deformation, and identify which of the two displacements the total force opposes more strongly (larger magnitude of a negative S).
原参考答案（未改写）
S_d1 = -11.8341124224 eV (rounded: -11.8341 eV): opposes, below the negative threshold, opposes the A-to-d1 deformation. S_d2 = -5.0423222646 eV (rounded: -5.0423 eV): opposes, below the negative threshold, opposes the A-to-d2 deformation. |S_d1| = 11.8341 eV > |S_d2| = 5.0423 eV, so the total force opposes continuing the d1 deformation more strongly than the d2 deformation. This compares only the instantaneous directional force response along these two specific supplied displacements; it does not indicate which displacement is dynamically more likely or which geometry is closer to a true minimum.
原评分
6 points: S_d1 within 0.0001 eV of reference (1); S_d2 within 0.0001 eV of reference (1); correct oppose/reinforce classification for d1 (1); correct oppose/reinforce classification for d2 (1); correct identification of which displacement the force opposes more strongly, i.e. larger |S| among the two opposing values (1); no claim about which displacement is more physically likely or which reference geometry is more stable (1). Interpretive points require semantic review.
In the attached optimized QM7-X geometry for molecule 7070 (10 atoms), row 1 is a chlorine atom and row 2 is the carbon atom with the shortest measured distance to row 1. What is the distance between row 1 and row 2? Report the value in angstrom to four decimal places.
原参考答案（未改写）
1.737175201 angstrom
原评分
Distance between row 1 and row 2 within 0.0001 angstrom of the computed reference value; report to four decimal places.
QM7-X molecule 7070 has an optimized reference geometry (opt) and a paired displaced geometry (d1). Using the attached official 3-component dipole vector (vDIP) at each geometry, compute the dipole moment magnitude at opt and at d1, then report the signed change (d1 minus opt) in e.Ang to four decimal places. Does the total dipole moment magnitude increase or decrease under this displacement?
原参考答案（未改写）
|vDIP(opt)| = 0.9173500511 e.Ang (rounded: 0.9174 e.Ang), consistent with the officially reported scalar DIP = 0.91735 e.Ang. |vDIP(d1)| = 0.8124573237 e.Ang (rounded: 0.8125 e.Ang), consistent with the officially reported scalar DIP = 0.812457 e.Ang. Signed change (d1 minus opt) = -0.1048927274 e.Ang (rounded: -0.1049 e.Ang): the total dipole moment magnitude decreases under this displacement.
原评分
5 points: magnitude at opt within 0.0001 e.Ang of reference (1); magnitude at d1 within 0.0001 e.Ang of reference (1); signed change (d1 minus opt) within 0.0001 e.Ang and correct sign (1); correctly states the magnitude decreases (1); does not overclaim a general trend beyond this one supplied displacement (1).
In the attached optimized QM7-X geometry for molecule 7122 (11 atoms), row 7 is the molecule's only nitrogen atom, and row 5 is the carbon atom with the shortest measured distance to row 7. What is the distance between row 5 and row 7? Report the value in angstrom to four decimal places.
原参考答案（未改写）
1.40133892 angstrom
原评分
Distance between row 5 and row 7 within 0.0001 angstrom of the computed reference value; report to four decimal places.
In the attached PBE+vdW-relaxed geometry of OE62 molecule MAJHUB (41 atoms), what is the maximum pairwise Cartesian distance between any two atoms in the molecule? Identify the two atom rows that attain this maximum. Report the distance in angstrom to four decimal places.
原参考答案（未改写）
Dmax = 10.0003 angstrom, attained by row 10 and row 30 (exhaustive scan over all 820 atom pairs).
原评分
Distance within 0.0001 angstrom of the computed reference value (1); correct unordered pair of attaining rows (1). Any pair equally attaining the maximum within numerical precision is accepted.
For OE62 molecule MAJHUB, using the attached geometry and the official PBE0(vacuum) and G0W0@PBE0(vacuum, complete-basis-set) orbital-energy lists, identify the HOMO and LUMO at each level and compute both HOMO-LUMO gaps. Report both gaps in eV to four decimal places, state the signed difference (GW gap minus PBE0 gap), and explain whether this single molecule alone establishes a general relationship between the two levels of theory.
原参考答案（未改写）
PBE0 gap = 3.8834700000 eV (rounded: 3.8835 eV). G0W0@PBE0 (CBS) gap = 6.8329100000 eV (rounded: 6.8329 eV). Signed difference (GW minus PBE0) = 2.9494400000 eV (rounded: 2.9494 eV): the GW gap is larger. This is a single-molecule, single-instance comparison; it does not by itself establish a general relationship between PBE0 and G0W0@PBE0 gaps across other molecules.
原评分
5 points: PBE0 gap within 0.0001 eV of reference (1); GW gap within 0.0001 eV of reference (1); correct signed difference within 0.0001 eV (1); correctly states the GW gap is larger (1); explicitly states this single molecule does not establish a general PBE0-vs-GW trend (1).
In the attached PBE+vdW-relaxed geometry of OE62 molecule JIMHOC (57 atoms), what is the maximum pairwise Cartesian distance between any two atoms in the molecule? Identify the two atom rows that attain this maximum. Report the distance in angstrom to four decimal places.
原参考答案（未改写）
Dmax = 11.8784 angstrom, attained by row 15 and row 40 (exhaustive scan over all 1596 atom pairs).
原评分
Distance within 0.0001 angstrom of the computed reference value (1); correct unordered pair of attaining rows (1).
In the attached PBE+vdW-relaxed geometry of OE62 molecule KEGPOC (64 atoms), what is the maximum pairwise Cartesian distance between any two atoms in the molecule? Identify the two atom rows that attain this maximum. Report the distance in angstrom to four decimal places.
原参考答案（未改写）
Dmax = 14.0870 angstrom, attained by row 23 and row 62 (exhaustive scan over all 2016 atom pairs).
原评分
Distance within 0.0001 angstrom of the computed reference value (1); correct unordered pair of attaining rows (1).
In the attached PBE+vdW-relaxed geometry of OE62 molecule KEWGID (48 atoms), compute the equal-weight radius of gyration Rg = sqrt[(1/N) sum_i ||r_i - r_centroid||^2] over all N=48 atoms, where r_centroid is the mean of all atom positions. Report Rg in angstrom to four decimal places.
In the attached PBE+vdW-relaxed geometry of OE62 molecule PEKZAG (70 atoms), compute the equal-weight radius of gyration Rg = sqrt[(1/N) sum_i ||r_i - r_centroid||^2] over all N=70 atoms. Report Rg in angstrom to four decimal places.
For OE62 molecule GALPAL, using the attached geometry and the official PBE+vdW(vacuum) occupied/unoccupied orbital-energy lists, identify the HOMO and LUMO and compute the HOMO-LUMO gap. Report all three values in eV to four decimal places.
原参考答案（未改写）
HOMO = -5.6107500000 eV, LUMO = -1.6380900000 eV, gap = 3.9726600000 eV (rounded: HOMO -5.6108 eV, LUMO -1.6381 eV, gap 3.9727 eV).
原评分
3 points: HOMO within 0.0001 eV (1); LUMO within 0.0001 eV (1); gap (LUMO-HOMO) within 0.0001 eV (1).
For OE62 molecule MEHLOA, using the attached geometry and the official PBE+vdW(vacuum) and PBE0(vacuum) orbital-energy lists, identify the HOMO and LUMO at each level and compute both HOMO-LUMO gaps. Report both gaps in eV to four decimal places, state the signed difference (PBE0 gap minus PBE gap), and explain whether this single molecule alone establishes a general relationship between the two functionals.
原参考答案（未改写）
PBE gap = 3.1507500000 eV (rounded: 3.1507 eV). PBE0 gap = 5.1545200000 eV (rounded: 5.1545 eV). Signed difference (PBE0 minus PBE) = 2.0037700000 eV (rounded: 2.0038 eV): the PBE0 gap is larger. This is a single-molecule, single-instance comparison; it does not by itself establish a general relationship between PBE and PBE0 gaps across other molecules.
原评分
5 points: PBE gap within 0.0001 eV (1); PBE0 gap within 0.0001 eV (1); correct signed difference within 0.0001 eV (1); correctly states the PBE0 gap is larger (1); explicitly states this single molecule does not establish a general PBE-vs-PBE0 trend (1).
OE62 molecules EZUTAU (66 atoms) and ZOGVEW (79 atoms) are two independent real molecules, both with their own maximum pairwise atom distance in [1,10) nm. Using each molecule's attached geometry and official PBE+vdW(vacuum) orbital-energy lists, compute the HOMO-LUMO gap of each. Report both gaps in eV to four decimal places, state which is larger, and explain whether comparing just these two molecules establishes a general relationship between molecular size and gap.
原参考答案（未改写）
EZUTAU gap = 4.2659300000 eV (rounded: 4.2659 eV, Dmax=1.4960 nm). ZOGVEW gap = 2.3149200000 eV (rounded: 2.3149 eV, Dmax=1.6472 nm). EZUTAU's gap is larger by 1.9510100000 eV (rounded: 1.9510 eV), even though ZOGVEW is the larger molecule by both atom count and global extent. This ordering by itself does not establish a general relationship between molecular size and HOMO-LUMO gap: the two molecules differ in composition (P/O/N/C/H vs P/O/C/H), connectivity and functional groups, any of which can dominate the gap independent of overall size; two data points cannot separate a size effect from these confounds.
原评分
5 points: EZUTAU gap within 0.0001 eV (1); ZOGVEW gap within 0.0001 eV (1); correctly identifies EZUTAU as larger (1); correct signed difference within 0.0001 eV (1); explicitly states that a two-molecule comparison cannot establish a general size-gap trend, citing at least one confounding difference (composition, connectivity, etc.) (1).
For OE62 molecule JIMHOC (the same geometry already used in QNP008), using the attached geometry and the official PBE+vdW(vacuum) and PBE0(vacuum) orbital-energy lists, identify the HOMO and LUMO at each level and compute both HOMO-LUMO gaps. Report both gaps in eV to four decimal places, state the signed difference (PBE0 gap minus PBE gap), and explain whether this single molecule alone establishes a general relationship between the two functionals.
原参考答案（未改写）
PBE gap = 3.5756300000 eV (rounded: 3.5756 eV). PBE0 gap = 5.2964300000 eV (rounded: 5.2964 eV). Signed difference (PBE0 minus PBE) = 1.7208000000 eV (rounded: 1.7208 eV): the PBE0 gap is larger. This is a single-molecule, single-instance comparison; it does not by itself establish a general relationship between PBE and PBE0 gaps across other molecules.
原评分
5 points: PBE gap within 0.0001 eV (1); PBE0 gap within 0.0001 eV (1); correct signed difference within 0.0001 eV (1); correctly states the PBE0 gap is larger (1); explicitly states this single molecule does not establish a general PBE-vs-PBE0 trend (1).
The attached structure is the Zn sublattice of ZIF-8 (COD 4118891), 12 zinc atoms forming one symmetry-expanded conventional cubic cell. What is the maximum pairwise Cartesian distance between any two Zn atoms in this cell? Identify the two atom rows that attain this maximum. Report the distance in angstrom to four decimal places.
原参考答案（未改写）
Dmax = 17.8512 angstrom, attained by row 3 and row 12 (exhaustive scan over all 66 atom pairs in the 12-atom set).
原评分
Distance within 0.0001 angstrom of the computed reference value (1); correct unordered pair of attaining rows (1).
The attached structure is the Zn sublattice of MOF-5 (COD 1516287), 32 zinc atoms forming one symmetry-expanded conventional cubic cell. What is the maximum pairwise Cartesian distance between any two Zn atoms in this cell? Identify the two atom rows that attain this maximum. Report the distance in angstrom to four decimal places.
原参考答案（未改写）
Dmax = 23.9459 angstrom, attained by row 1 and row 32 (exhaustive scan over all 496 atom pairs in the 32-atom set).
原评分
Distance within 0.0001 angstrom of the computed reference value (1); correct unordered pair of attaining rows (1).
ZIF-8 (COD 4118891) has a cubic conventional cell with a=16.8303 angstrom. For a cubic cell, the interplanar spacing of the (h k l) lattice planes is d=a/sqrt(h^2+k^2+l^2). What is d for the (1 1 0) planes? Report the value in nanometers to six decimal places.
原参考答案（未改写）
d(110) = 1.1900819259 nm (rounded: 1.190082 nm).
原评分
d(110) within 0.000001 nm of the computed reference value; report six decimals.
MOF-5 (COD 1516287) has a cubic conventional cell with a=25.8247 angstrom. For a cubic cell, the interplanar spacing of the (h k l) lattice planes is d=a/sqrt(h^2+k^2+l^2). What is d for the (1 1 1) planes? Report the value in nanometers to six decimal places.
原参考答案（未改写）
d(111) = 1.4909897497 nm (rounded: 1.490990 nm).
原评分
d(111) within 0.000001 nm of the computed reference value; report six decimals.
The attached structure is the full ordered framework of MOF-5 (COD 1516287), all 424 atoms (Zn, O, C, H) in one symmetry-expanded conventional cubic cell. What is the maximum pairwise Cartesian distance between any two atoms of any element in this cell? Identify the two atom rows that attain this maximum. Report the distance in angstrom to four decimal places.
原参考答案（未改写）
Dmax = 30.6102 angstrom, attained by row 233 and row 328 (exhaustive scan over all 89676 atom pairs in the 424-atom set).
原评分
Distance within 0.0001 angstrom of the computed reference value (1); correct unordered pair of attaining rows (1).
判断：比较对象写成含糊的 typical salt / a1-2 angstrom scale reflection，不能得到确定比较条件。
下一步：给明确的参考 d 与相同波长/级次，限制为条件衍射几何。
审读的题干、参考答案与评分
题干
Using d(110)=1.1900819259 nm for ZIF-8 (from the preceding plane-spacing calculation) and a CuKa X-ray wavelength of 0.15406 nm, apply Bragg's law 2*d*sin(theta)=n*lambda (n=1) to find the first-order diffraction angle 2*theta. Report the value in degrees to two decimal places. Is this angle larger or smaller than you would get from a typical small-unit-cell inorganic salt (a a1-2 angstrom scale reflection)? Explain why in one sentence referencing the cell size.
原参考答案（未改写）
2*theta = 7.4223150084 degrees (rounded: 7.42 degrees). This is smaller than the diffraction angle a similarly low-index reflection would give for a typical small-unit-cell inorganic salt (a~4-6 angstrom, sub-nm d-spacing): at fixed wavelength, Bragg's law gives sin(theta) inversely proportional to d, so ZIF-8's much larger ~1.19 nm d(110) spacing produces a correspondingly smaller diffraction angle.
原评分
4 points: 2*theta within 0.01 degree of the computed reference value (1); correct application of Bragg's law with n=1 shown (1); correctly states the angle is smaller than for a typical small-cell salt (1); correct reasoning that a larger d-spacing (from the larger cell) gives a smaller diffraction angle at fixed wavelength (1).
Using d(111)=1.4909897497 nm for MOF-5 (from the preceding plane-spacing calculation) and a CuKa X-ray wavelength of 0.15406 nm, apply Bragg's law 2*d*sin(theta)=n*lambda (n=1) to find the first-order diffraction angle 2*theta. Report the value in degrees to two decimal places. Is this angle larger or smaller than the ZIF-8 (110) angle computed previously, and why?
原参考答案（未改写）
2*theta = 5.9228571183 degrees (rounded: 5.92 degrees). This is smaller than ZIF-8's (110) angle (7.4223 degrees): MOF-5's d(111)=1.490990 nm is larger than ZIF-8's d(110)=1.190082 nm, and at fixed wavelength a larger d-spacing gives a smaller Bragg angle.
原评分
4 points: 2*theta within 0.01 degree of the computed reference value (1); correct application of Bragg's law with n=1 shown (1); correctly states the angle is smaller than ZIF-8's (1); correct reasoning that MOF-5's larger d(111) spacing (from its larger cell) gives a smaller angle (1).
Using the attached full ZIF-8 structure (204 atoms) and the fractional coordinates you obtain by dividing each Cartesian coordinate by a=16.8303 angstrom, compute the structure factor F(hkl)=sum_j Z_j*exp(2*pi*i*(h*x_j+k*y_j+l*z_j)) for (h k l) = (1,0,0), (1,1,0), (2,0,0) and (1,1,1), using atomic number as the weight Z_j (Zn=30, N=7, C=6, H=1). Report the amplitude |F(hkl)| for each of the four reflections to one decimal place. Which reflections are systematically absent (amplitude effectively zero), and what single rule (in terms of h+k+l) predicts this pattern for a body-centered (I-centered) lattice?
原参考答案（未改写）
|F(100)| = 0.000000 (~0, absent). |F(110)| = 331.5155 (nonzero, present). |F(200)| = 196.9412 (nonzero, present). |F(111)| = 0.000000 (~0, absent). (100) has h+k+l=1 (odd) and is absent; (111) has h+k+l=3 (odd) and is absent; (110) and (200) both have h+k+l even and are present. This matches the standard body-centered (I-centering) selection rule: only reflections with h+k+l=even survive.
原评分
6 points: |F(100)| within 1.0 of reference and correctly identified as ~0 (1); |F(110)| within 1.0 of reference and correctly identified as nonzero (1); |F(200)| within 1.0 of reference and correctly identified as nonzero (1); |F(111)| within 1.0 of reference and correctly identified as ~0 (1); correctly states the h+k+l=odd reflections are the absent ones (1); correctly states the h+k+l=even rule for body-centered lattices, or an operationally equivalent statement of the I-centering selection rule (1).
Using the attached full MOF-5 structure (424 atoms) and the fractional coordinates you obtain by dividing each Cartesian coordinate by a=25.8247 angstrom, compute the structure factor F(hkl)=sum_j Z_j*exp(2*pi*i*(h*x_j+k*y_j+l*z_j)) for (h k l) = (1,0,0), (1,1,0), (2,0,0) and (1,1,1), using atomic number as the weight Z_j (Zn=30, O=8, C=6, H=1). Report the amplitude |F(hkl)| for each of the four reflections to one decimal place. Which reflections are systematically absent, and what single rule (in terms of the parities of h, k and l) predicts this pattern for a face-centered (F-centered) lattice?
原参考答案（未改写）
|F(100)| = 0.000000 (~0, absent). |F(110)| = 0.000000 (~0, absent). |F(200)| = 1874.1986 (nonzero, present). |F(111)| = 70.6477 (nonzero, present). (100) is (odd,even,even) mixed parity and absent; (110) is (odd,odd,even) mixed parity and absent; (200) is (even,even,even) and present; (111) is (odd,odd,odd) and present. This matches the standard face-centered (F-centering) selection rule: only reflections with h, k, l all even or all odd survive.
原评分
6 points: |F(100)| within 1.0 of reference and correctly identified as ~0 (1); |F(110)| within 1.0 of reference and correctly identified as ~0 (1); |F(200)| within 1.0 of reference and correctly identified as nonzero (1); |F(111)| within 1.0 of reference and correctly identified as nonzero (1); correctly states the mixed-parity reflections are the absent ones (1); correctly states the all-even-or-all-odd rule for face-centered lattices (1).
ZIF-8 (body-centered, I-43m) obeys the selection rule "h+k+l=even survives" (from the preceding calculation); MOF-5 (face-centered, Fm-3m) obeys "h,k,l all even or all odd survives" (from the preceding calculation). Both are real nm-scale MOF frameworks with cubic cells an order of magnitude larger than simple inorganic salts. Does the fact that these two structures obey different systematic-absence rules mean that larger unit cells generally have different (or more restrictive) selection rules than small ones? Explain what actually determines each rule, using both structures' own computed results as evidence, and state what would and would not be established by comparing only these two examples.
原参考答案（未改写）
No: the difference in selection rule between ZIF-8 and MOF-5 is not caused by their cell sizes (1.68 nm vs 2.58 nm). It is caused by their different lattice centering types: ZIF-8's space group I-43m has a body-centering translation (+1/2,+1/2,+1/2), which forces reflections with h+k+l=odd to cancel (MNI008: |F(100)|~0, |F(111)|~0, while |F(110)| and |F(200)| are nonzero). MOF-5's space group Fm-3m has face-centering translations, which force mixed-parity reflections to cancel (MNI009: |F(100)|~0, |F(110)|~0, while |F(200)| and |F(111)| are nonzero). The centering type, a property of the space group's translational symmetry, determines the rule regardless of how large a is. Comparing only these two structures establishes that centering type (not size) explains this pair's difference; it does not by itself establish that cell size never matters for any structure or that every I-centered or F-centered nm-scale material behaves identically, since only two specific examples were examined.
原评分
4 points: correctly states that unit cell size is not what determines the selection rule (1); correctly states that lattice centering type (I vs F) is what determines the rule, citing both computed patterns as evidence (1); explicitly states that two examples cannot establish a general size-independence claim beyond these two specific structures (1); does not claim the rules generalize to all body-centered or face-centered structures without qualification, though citing the well-established crystallographic convention that identical centering types share the same rule is acceptable background context (1).
The attached structure is one vancomycin monomer, PDB 1AA5 chain A (132 atoms, major altLoc/occupancy only). What is the maximum pairwise Cartesian distance between any two atoms in this molecule? Identify the two atom rows that attain this maximum. Report the distance in angstrom to four decimal places.
原参考答案（未改写）
Dmax = 19.5698 angstrom, attained by row 11 and row 132 (exhaustive scan over all 8646 atom pairs in the 132-atom set).
原评分
Distance within 0.0001 angstrom of the computed reference value (1); correct unordered pair of attaining rows (1).
The attached structure is the second vancomycin monomer of the same asymmetric dimer, PDB 1AA5 chain B (132 atoms, major altLoc/occupancy only). What is the maximum pairwise Cartesian distance between any two atoms in this molecule? Identify the two atom rows that attain this maximum. Report the distance in angstrom to four decimal places.
原参考答案（未改写）
Dmax = 19.2646 angstrom, attained by row 11 and row 132 (exhaustive scan over all 8646 atom pairs in the 132-atom set).
原评分
Distance within 0.0001 angstrom of the computed reference value (1); correct unordered pair of attaining rows (1).
The attached structure is the full vancomycin asymmetric dimer, PDB 1AA5 chains A+B combined (264 atoms: chain A as rows 1-132, chain B as rows 133-264; major altLoc/occupancy only). What is the maximum pairwise Cartesian distance between any two atoms in this assembly, considering atoms from both chains together? Identify the two atom rows that attain this maximum. Report the distance in angstrom to four decimal places.
原参考答案（未改写）
Dmax = 28.7192 angstrom, attained by row 11 and row 143 (exhaustive scan over all 34716 atom pairs in the 264-atom set).
原评分
Distance within 0.0001 angstrom of the computed reference value (1); correct unordered pair of attaining rows (1).
Using the attached chain-A (132 atoms) and chain-B (132 atoms) structures of the vancomycin asymmetric dimer (PDB 1AA5), find the minimum Cartesian distance between any atom in chain A and any atom in chain B. Identify the chain-A row and the chain-B row that attain this minimum, and state the element symbol of each. Report the distance in angstrom to four decimal places.
原参考答案（未改写）
Dmin = 2.0332 angstrom, attained by chain-A row 78 (element O) and chain-B row 86 (element H) (exhaustive scan over all 17424 cross-chain atom pairs).
原评分
3 points: distance within 0.0001 angstrom of the computed reference value (1); correct chain-A row and chain-B row (1); correctly reports the element symbols (O and H) of the attaining atoms (1).
Using the attached full vancomycin dimer structure (PDB 1AA5 chains A+B combined, 264 atoms), compute the equal-weight radius of gyration Rg = sqrt[(1/N) sum_i ||r_i - r_centroid||^2] over all N=264 atoms, where r_centroid is the mean of all atom positions. Report Rg in angstrom to four decimal places.
In the attached vancomycin chain-A structure (PDB 1AA5, 132 atoms), row 3 is the CA atom of residue 1 (MLU, the sequence N-terminus of this 7-residue backbone) and row 122 is the CA atom of residue 7 (3FG, the sequence C-terminus). Compute the distance between these two rows, and compare it to the whole-molecule maximum pairwise distance already established for this same chain (Dmax = 19.5698 angstrom, from the preceding question). Does this terminal-to-terminal distance alone describe the molecule's full spatial extent? Report the terminal distance in angstrom to four decimal places.
原参考答案（未改写）
Terminal (residue 1 to residue 7 CA-CA) distance = 13.8392621913 angstrom (rounded: 13.8393 angstrom). This underestimates the molecule's full extent: Dmax = 19.5698 angstrom (established previously), a difference of 5.7305 angstrom. The terminal pair samples only one of 8646 possible atom pairs and is not guaranteed to be the maximally separated one; here it clearly is not.
原评分
3 points: terminal distance within 0.0001 angstrom of the computed reference value (1); correctly states the terminal distance underestimates Dmax (1); correctly explains that the terminal pair samples only one of many atom pairs and need not be the maximally separated one (1).
In the attached vancomycin chain-B structure (PDB 1AA5, 132 atoms), row 3 is the CA atom of residue 1 (MLU) and row 122 is the CA atom of residue 7 (3FG). Compute the distance between these two rows, and compare it to the whole-molecule maximum pairwise distance already established for chain B (Dmax = 19.2646 angstrom, from the preceding question). Report the terminal distance in angstrom to four decimal places.
原参考答案（未改写）
Terminal (residue 1 to residue 7 CA-CA) distance = 13.8114848224 angstrom (rounded: 13.8115 angstrom). This underestimates the molecule's full extent: Dmax = 19.2646 angstrom, a difference of 5.4531 angstrom.
原评分
3 points: terminal distance within 0.0001 angstrom of the computed reference value (1); correctly states the terminal distance underestimates Dmax (1); correctly explains the terminal pair need not be the maximally separated one (1).
Using the attached chain-A and chain-B structures (PDB 1AA5, 132 atoms each, identical 7-residue covalent sequence), compute the equal-weight radius of gyration Rg for each chain independently (same formula as the dimer Rg computed previously). Report both Rg values in angstrom to four decimal places and their signed difference (chain A minus chain B). The original publication describes this structure as an "asymmetric" vancomycin dimer. Does the Rg comparison alone establish that the two chains adopt different conformations, and if so, how strong is that evidence?
原参考答案（未改写）
Rg(chain A) = 6.4796736273 angstrom (rounded: 6.4797 angstrom). Rg(chain B) = 6.5288208793 angstrom (rounded: 6.5288 angstrom). Signed difference (A minus B) = -0.0491472520 angstrom (rounded: -0.0491 angstrom). The two Rg values are very close (differing by less than 0.05 angstrom out of ~6.5 angstrom), so this comparison alone is weak evidence either way about conformational asymmetry: Rg is a single whole-molecule size/compactness scalar, and two chains could have nearly identical Rg while still differing substantially in local conformation (e.g. side-chain rotamers, ring puckering). The paper's 'asymmetric 'dimer' description is not confirmed or refuted by Rg alone; a full atom-by-atom structural comparison would be needed.
原评分
4 points: Rg(A) within 0.0001 angstrom (1); Rg(B) within 0.0001 angstrom (1); correct signed difference within 0.0001 angstrom (1); correctly characterizes the difference as small/consistent with the two chains being similar overall in size while still allowing for local conformational asymmetry, without overclaiming that Rg alone proves or disproves the paper's "asymmetric" description (1).
Using the already-established results Dmax(chain A) = 19.5698 angstrom, Dmax(chain B) = 19.2646 angstrom, and Dmax(dimer) = 28.7192 angstrom, compute the sum of the two individual monomer extents and compare it to the dimer's own extent. Report the sum and the signed difference (sum minus dimer extent) in angstrom to four decimal places. Is the dimer's extent close to, less than, or greater than the sum of the two monomer extents, and what does this imply about how the two monomers are arranged relative to each other (e.g. extending end-to-end versus overlapping side-by-side)?
原参考答案（未改写）
Sum of monomer extents = 38.8343580910 angstrom (rounded: 38.8344 angstrom). Signed difference (sum minus dimer extent) = 10.1151097373 angstrom (rounded: 10.1151 angstrom). The dimer's own extent (28.7192 angstrom) is noticeably less than the sum of the two monomer extents (38.8344 angstrom). If the two monomers were arranged fully end-to-end along a common axis, the dimer extent would approach the sum of their individual extents; the fact that it is substantially smaller is consistent with the two monomers associating side-by-side or back-to-back (overlapping in extent) rather than extending collinearly away from each other.
原评分
3 points: correct sum within 0.0001 angstrom (1); correct signed difference within 0.0001 angstrom, correctly identifying the dimer extent as less than the sum (1); reasonable geometric interpretation that the two monomers overlap/associate side-by-side or back-to-back rather than extending fully end-to-end, since a purely end-to-end (collinear) arrangement would make the dimer extent approach the sum (1).
The closest inter-chain contact in the vancomycin dimer (established previously) is between an O atom (chain A, residue GHP5) and an H atom (chain B, residue GHP5), at 2.0332 angstrom. Does this single distance measurement by itself confirm that this is a hydrogen bond, and does it by itself confirm that this is THE literature-documented dimerization interface reported in the original 1997 publication? Explain what additional evidence each of these two stronger claims would require.
原参考答案（未改写）
No to both. The O...H distance (2.0332 angstrom) and element combination are consistent with a hydrogen bond, but distance alone does not confirm one: a proper hydrogen-bond assessment also needs the donor-hydrogen-acceptor angle (typically expected to be close to linear, roughly 120-180 degrees) and the identity of the covalently bonded donor heavy atom, none of which was computed here. Separately, identifying this contact as THE dimerization interface documented in Loll et al. (1997) would require comparing it against the full set of interactions those authors reported and characterized (they describe multiple interactions, not a single contact), not merely finding the closest pair by distance in an independent re-derivation from the deposited coordinates. This batch establishes only the geometric fact of a short O...H contact at a specific, real location in the structure.
原评分
4 points: correctly states the O...H distance and element combination are consistent with (but do not alone confirm) a hydrogen bond (1); correctly notes that confirming a functional hydrogen bond would additionally require donor-hydrogen-acceptor angle information, not distance alone (1); correctly states that identifying this as THE literature-documented dimerization interface would require comparing against the full interaction pattern reported in the original paper, not a single distance from this batch's own re-derivation (1); does not overclaim beyond these stated limits (1).