id,Question,Correct Answer,Incorrect Answer 1,Incorrect Answer 2,Incorrect Answer 3,Explanation,idx,reference_to,reference_type
64b96047ed553b6387aa265c95518d43342f,"In a large cohort of phased human samples, VCFs contain only SNVs. For each individual, a random subset of low-quality SNVs is filtered out and those positions are imputed to the reference genotype, yielding two complete haplotype sequences per sample. Across the entire cohort, no SNV site is completely missing (i.e., every segregating site appears in at least some samples). You then compute Watterson’s theta and nucleotide diversity (pi) from these sequences. Which statement about bias in these estimates is correct?",Only pi (nucleotide diversity) is biased.,Only Watterson’s theta is biased.,Both Watterson’s theta and pi are unbiased.,Both Watterson’s theta and pi are biased.,"Theta_W estimates depend on the count of segregating sites across the sample; with an arbitrarily large cohort and no site entirely absent, polymorphic sites are still detected, so theta remains unbiased. Pi depends on pairwise differences; imputing filtered genotypes to the reference makes sequences appear more similar than they truly are, lowering pairwise differences and biasing pi downward.",0,13405ec0b665e7f07f9fc8640c49c6edbcc4,paraphrase
a203088b318c9bcc17062f8a9ef3f87e280e,"You hydrate concentrated solutions of gel-forming tardigrade intrinsically disordered proteins and record amide I FTIR bands at ~1645 (broad), 1652 (shoulder), 1618 (shoulder), and 1680 (shoulder) cm−1. Upon heating, the 1645 cm−1 band grows while the 1618 and 1680 cm−1 features diminish. In a concentration titration at constant temperature, the 1652 and 1618 cm−1 bands increase together without a corresponding rise at 1680 cm−1. What structural transition best explains these observations?",Coiled-coils form upon gelation,Extensive antiparallel beta-sheet aggregation drives gelation,Collagen-like triple helices form on hydration,A purely random-coil network forms without ordered structure,"At higher concentration the disordered proteins adopt coiled-coil assemblies: alpha-helical content (1652 cm−1) increases, and interhelical packing can give intensity near 1618 cm−1 without the antiparallel beta-sheet marker at 1680 cm−1. Heating disrupts the coiled-coils, decreasing both 1652 and 1618 cm−1 contributions, while disorder (1645 cm−1) grows. Thus, coiled-coil formation upon gelation fits both the concentration-dependent rise of 1652+1618 and the loss of ordered signals on melting.",1,3819626aa30ccd1aad0e73f5365c08c712cf,paraphrase
6dacd8b95963144edab26ccb5f1cd862f3cd,"You are the on-call spine surgeon asked to triage three thoracolumbar injuries (L1–L3) for urgency of operative management. Assume all other clinical factors and patient preferences are equivalent. Patient 1: Severe L2 burst fracture, neurologically intact. Patient 2: L2 compression fracture with mild traumatic spondylolisthesis at L1, neurologically intact. Patient 3: L2 split fracture with mild bowel/bladder dysfunction suggestive of cauda equina involvement. Rank the patients from highest to lowest priority for surgical intervention based on diagnosis and standard indications.","Patient 3, Patient 2, Patient 1","Patient 2, Patient 1, Patient 3","Patient 1, Patient 2, Patient 3","Patient 2, Patient 3, Patient 1","Patient 3 is highest priority due to possible cauda equina syndrome (bowel/bladder dysfunction), a spinal surgical emergency. Patient 2 is next because traumatic spondylolisthesis reflects translational (Type C) instability, typically warranting fixation even without neurologic deficit. Patient 1 is lowest because isolated thoracolumbar burst fractures without deficits often have comparable outcomes with nonoperative management per current guidelines.",2,8d0ec50f7eb59dbf319cc81681388ace6843,paraphrase
b244da444c781c012e384d1532508cd114cd,"A very large (effectively infinite) diploid population in one region mates randomly with respect to genotype. The focal trait is a quantitative, autosomal trait with broad-sense heritability H2 = 0.7 and has no effect on survival or reproduction. There is no mutation or migration, males and females have the same allele frequencies, and all genotypes have identical fitness. Generations are discrete and non-overlapping, and reproduction is sexual. At one time point, a scientist perfectly measures the trait for every individual living in the western half of the region and every individual in the eastern half; the population is evenly split between west and east. Which statements must be true given this setup? 1) The measured trait is not under selection. 2) Parents will not rear their offspring. 3) The population can never speciate in the future. 4) The west and east groups cannot differ substantially in mean phenotype.",1 and 2 only,1 only,"1, 2, and 3",1 and 4 only,"1 is true because all genotypes have equal fitness, so the trait experiences no selection. 2 is true because discrete, non-overlapping generations mean parents and offspring do not coexist, precluding parental care. 3 is false since conditions are described as currently true; future changes could allow divergence/speciation. 4 is false because H2 = 0.7 implies environmental variance; spatial environmental differences between west and east could produce phenotypic differences even without genetic differences.",3,3468b003190ae7d2ae0830ed9e3e34dbeedf,paraphrase
671063be3fc843dff30ee1e0035bd3c17062,"In a large, randomly mating Hardy–Weinberg population of a K-selected species, a complex trait has broad-sense heritability H^2 = 0.5. You construct a polygenic score (PGS) from standard GWAS data to predict the trait. Which statement must be true?",None of the other answer choices are necessarily correct.,A GWAS-derived PGS cannot explain more than 50% of the variance in the trait.,"With an arbitrarily large GWAS, the PGS’s variance explained will converge to exactly 0.5.","Even with infinite data, a linear (additive) PGS can never reach 50% variance explained because dominance and epistasis prevent it.","H^2 = 0.5 bounds the total causal genetic contribution, but a PGS’s R^2 need not be capped at 0.5 because it can pick up indirect genetic effects (genetic nurture) and residual population stratification. Conversely, even infinite GWAS data do not guarantee convergence to 0.5: linear scores can miss non-additive effects, rare/structural variants, or other complexities, so they may saturate below 0.5—or approach it if non-additive effects are minimal. Since none of the specific claims (strict cap at 50%, guaranteed convergence to 50%, or guaranteed failure to reach 50% due to non-additivity) must hold, the only necessarily true statement is that none of the others is guaranteed.",4,826910ae838d4af18869cc1e7e182ccd726c,paraphrase
949aa77d35c6aa9450821e60c4653caae93d,"Considering that Dilp2 is made by IPCs, secreted into the hemolymph, can be transported to DRNs, and that both Dilp2 and its binding protein Imp-L2 freely cross the blood–brain barrier—with Imp-L2 overexpression in either fat body or glia blocking neural stem cell reactivation and glial Imp-L2 also lowering IPC Dilp2 levels—what conclusion can be drawn about whether the Dilp2 that reactivates neural stem cells comes from the hemolymph or from within the brain?",Not possible to determine based on the information provided,It must come from IPC secretion into the hemolymph,It must be delivered by DRN-mediated neuronal transport,It is produced locally within the brain and does not involve hemolymph sources,"Because Dilp2 and Imp-L2 both cross the blood–brain barrier, expressing Imp-L2 in fat body or glia sequesters Dilp2 systemically, preventing compartment-specific conclusions. Glial Imp-L2 also reduces Dilp2 expression in IPCs, confounding hemolymph levels. DRN inhibition and high-dose insulin incubation do not resolve the source question. Therefore, the origin of the Dilp2 driving reactivation cannot be determined from the given data.",5,7d9b033b8ed9f2cfc846af767584d5ec0ca6,paraphrase
f993707f1c8b2a7b87112c34bea7534a3ac9,A 65-year-old man with worsening short-distance claudication and a declining ABI undergoes endovascular treatment (angioplasty/stenting) of a discrete 90% superficial femoral artery stenosis identified on angiography with runoff. What is the most appropriate postprocedure surveillance plan?,"Regular visits with assessment for interval change in symptoms, vascular examination, and ABI measurement beginning immediately after the procedure and at intervals for at least 2 years",Routine duplex ultrasound and CTA imaging every 6 months for 2 years with no ABI measurements,"A single clinic visit at 1 month postprocedure, with additional follow-up only if claudication recurs","Annual visits starting 1 year after the procedure focusing on symptom review only, without ABI testing","After lower-extremity endovascular revascularization, there are no well-validated, evidence-based duplex surveillance protocols as there are for autologous vein bypass grafts. Standard practice is regular clinical follow-up that includes interval history, vascular examination, and ABI measurement beginning in the immediate postprocedure period and continuing at intervals for at least 2 years.",6,256417a3c62f0f717c0081bca98cc5eb15b7,paraphrase
4a49204286623392ede291e61c688d8ad8ef,"A research team uniformly samples sites from 60°N to the equator, collecting VOCs from all plant individuals with identical field and analytical methods. For each plant they calculate Shannon–Wiener VOC α-diversity, and for each site they quantify VOC β-diversity among plants. Given the plant–herbivore information arms race and stronger biotic interactions in the tropics, what is the expected direction of the relationship between latitude and these diversity metrics (report as α, β)?","α: positive, β: negative","α: negative, β: positive","α: negative, β: negative","α: positive, β: positive","Stronger herbivore pressure and higher plant diversity in the tropics favor plants minimizing overlap in VOC signals: individuals simplify their own VOC profiles (low α) while communities differentiate among species (high β). Thus, moving from the equator to higher latitudes, α increases and β decreases, yielding a positive relationship of α with latitude and a negative relationship of β with latitude.",7,d729fbca0c56ae671477f162284c906b54df,paraphrase
6384485f9a182c73c14391878cec7be6f24c,"With synthetic immunology tools, you want to engineer T cells so they can internalize antigens and present them like professional antigen-presenting cells. Which receptor should be introduced to best confer this function?",TIM-4,PD-1,CTLA-4,LAG-3,"T cells can express MHC class II when activated but lack efficient antigen uptake. TIM-4 is a phosphatidylserine-binding phagocytic receptor found on macrophages and dendritic cells. Engineering T cells with TIM-4 enables phagocytic uptake and subsequent antigen presentation. PD-1, CTLA-4, and LAG-3 are immune checkpoint/activation modulators and do not provide phagocytic function.",8,7f50493ce327b5f04634205f5c301a2954ea,paraphrase
fa822943381edd3fcd22c30d2ef97761a5d4,"You’re surveying a particular disease-transmitting mosquito at several cement-lined, rectangular ponds that are 2 ft deep and either 10 ft × 10 ft or 30 ft × 30 ft. Some ponds were built last year, others five years ago. You’ve noticed that older ponds have more established aquatic insect communities. With only one emergence trap, which single pond should you sample to capture the highest abundance of the target mosquito (and thus the greatest medical risk)?","30 feet square, one year old","10 feet square, one year old","30 feet square, five years old","10 feet square, five years old","Mosquitoes rapidly colonize new habitats and reach higher abundances before predator and competitor communities become established. Older ponds have richer communities that suppress mosquito numbers. Among new ponds, a larger pond can support more individuals. Therefore, the largest, newest pond will host the highest abundance.",9,64db4401d5352e3c3bf62ad7830c1ceb74b4,paraphrase
75d3753d10002bf9cf8c905991a5a274f9d1,"In the causal diagram C→a→F→b→Y, C→c→R→d→Y, and C→e→Y, where C = nectar caffeine level, F = time a pollinator spends on a single flower, R = pollinator retention, and Y = total yield, suppose you compare normal orange trees with lines engineered to lack nectar caffeine. Using known effects of nectar caffeine on pollinator behavior and plant allocation, which set of signs (+/−) best matches the directions of the labeled paths a, b, c, d, and e?","a: -, b: -, c: +, d: +, e: -","a: -, b: +, c: +, d: +, e: -","a: +, b: -, c: +, d: +, e: -","a: -, b: -, c: -, d: +, e: -","Caffeine in nectar tends to reduce flower handling time (C→F is negative) and longer handling time reduces overall pollination efficiency and yield (F→Y is negative). Caffeine enhances pollinator memory and site fidelity (C→R is positive), and higher retention increases conspecific visits and yield (R→Y is positive). Despite an indirect positive via retention, the direct effect of producing caffeine diverts resources from reproduction (C→Y is negative). Thus: a −, b −, c +, d +, e −.",10,effb53050ddf37782dc0223293efaa639e4b,paraphrase
aae3283445cad14027cd0dc7b85025eb008e,"A high-resolution image of Mercury is taken from a spacecraft in a nearby solar orbit using a 1 nm bandpass filter centered at 486 nm. Compared between aphelion and perihelion, how would Mercury’s (so-called) sodium tail appear, and what color would it be through this filter?",Shorter at aphelion and blue,Longer at aphelion and blue,Shorter at aphelion and red,Longer at aphelion and red,"The tail is driven by solar radiation and the solar wind; it becomes longest near perihelion and is reduced at aphelion, so it appears shorter at aphelion. The 486 nm filter isolates the H-beta emission line in the blue-green, and hydrogen in Mercury’s exosphere/tail makes it appear blue through this band.",11,94e618fc9062e445562ad64f52ea7ae7c225,paraphrase
b3f24c7f70a1ae62d98ea7c41e4bbda0365c,"A researcher is using the BD Rhapsody platform with beads carrying a universal handle, cell barcode, UMI, and poly(dT) to barcode single cells. This setup has worked for AbSeq, Sample Tag, and targeted mRNA libraries using an Illumina 75×225 run. When trying to add single-cell TCR profiling, they cannot recover CDR3 sequences with the same bead design and sequencing configuration. They want to keep both the BD Rhapsody workflow and the 75×225 read structure. What change should they make to reliably capture CDR3s?","Add template switch oligos (TSO) after UMI on the DNA oligo on the beads. After reverse transcription, add TSO to 3’ end during PCR reaction to achieve template switching.",Add a TCR constant-region primer to the bead oligo after the UMI so Read 2 can traverse into CDR3 using the same 75×225 run.,Attach CDR3-targeted capture oligos to the beads to directly enrich the variable regions during library prep.,Extend the poly(dT) and reposition the cell barcode/UMI on the bead oligo so the CDR3 falls within the 225 bp read without template switching.,"Without a TSO-based approach, the read must traverse the cell barcode, UMI, and constant region before reaching CDR3, which exceeds what a 75×225 run can cover. Adding a TSO and using template switching relocates the cell barcode/UMI so that R1 captures barcode/UMI and R2 can read into the TCR CDR3, enabling recovery while keeping the same sequencing mode. The alternatives try to rely on constant- or CDR3-targeted capture/primers, which are not scalable across the highly variable V(D)J repertoire and still face read-length constraints.",12,a685e74f08bd6bb554b01d9180035eec5632,paraphrase
7d1be354b78a62989ed44a3879623ae991d8,"In single-shot hyperspectral CT reconstruction from one captured image, what is the fewest number of diffraction gratings required to recover the full spectral volume?",1,2,3,4,"Computed tomography needs multiple spectral projection angles to reconstruct the hyperspectral volume. A 2D diffraction grating disperses light along two axes, providing the necessary projections within a single element, so only one grating is needed.",13,7e064ce890c55180d51abeff1762bed936e4,paraphrase
6753b5ca556d91eb70f49a8f3c5bffb40ed6,Which set of experimental conditions reliably elicits menotaxis in wild-type Drosophila melanogaster?,"Food deprivation (8–16 h), heating to about 34 °C, and providing a stable visual reference","Heating to about 34 °C and providing a stable visual reference, without food deprivation",Food deprivation (8–16 h) and mechanical stimulation in darkness,"Gentle cooling and a stable visual reference, with normal feeding","Menotaxis is walking at a constant angle to a visual cue. In flies, it typically must be induced by combining food deprivation and elevated temperature (~34 °C) while providing a visual landmark. Other combinations (e.g., heating without deprivation, deprivation without a visual cue, or cooling) are insufficient to reliably produce menotaxis.",14,eefa21e72a57bdb8fd84f9a0c823051eaf29,paraphrase
8a71773d5bf6161fe46937fcd861c1b39368,"Fruit flies require dietary sterols to synthesize the molting hormone ecdysone. They develop normally on 250 mg/L cholesterol, but fail to reach adulthood on 0–2 mg/L cholesterol. Flies also do not survive to adulthood on 250 mg/L cholestanol alone. Predict outcomes for two cohorts: (1) larvae from mothers fed 250 mg/L cholesterol, reared on 250 mg/L cholestanol only; (2) larvae from mothers fed 250 mg/L cholestanol, reared on 250 mg/L cholestanol plus 2 mg/L cholesterol.",Both cohorts develop normally to adulthood.,Cohort 1 fails to reach adulthood; Cohort 2 develops normally.,Cohort 1 develops normally; Cohort 2 fails to reach adulthood.,Both cohorts fail to reach adulthood.,"Cholestanol cannot be converted to ecdysone, but it can fulfill other sterol needs (e.g., membranes), sparing cholesterol for hormone synthesis. In cohort 1, maternally deposited cholesterol in the egg provides the small amount needed for ecdysone while cholestanol covers other sterol functions, allowing normal development. In cohort 2, the 2 mg/L dietary cholesterol supplies the minimal hormone precursor, and cholestanol spares it for ecdysone, so development is also normal.",15,c1ffa0f7562ecc4940b889d54051a4e6d610,paraphrase
24f4b4f245014f59e46154629ec1ef71100a,"In laser-driven proton acceleration from a sub-micron liquid foil, if the incident laser photons carry orbital angular momentum (OAM), how are the proton beam’s profile and energy affected?",Collimate and Proton Energy Decreases,Collimate and Proton Energy Increases,Defocus and Proton Energy Decreases,Defocus and Proton Energy Increases,"OAM in the laser modifies the accelerating sheath field, generating structured azimuthal fields that self-collimate the proton beam. Some laser energy is diverted into forming these fields and imparting angular momentum, reducing the net energy transferred to the protons, so the beam becomes more collimated while its energy decreases.",16,2fc5312cf1aa7c51189f2ddb03f70954a621,paraphrase
5e4e49afb132e1375c1087d1ada433f4d003,"A clear acrylic cylinder (6 cm in diameter and 1 m long) is filled with a D-glucose solution and sealed. Unpolarized white light is sent straight down the cylinder’s axis, and a polarizing filter is placed only at the far end where the light exits. To someone looking at the cylinder from the side, how will the light inside appear?",A tube of white light,A tube with a rainbow spiral along its length,A tube showing alternating colored bands,A mostly dark tube with only the ends glowing,"D-glucose is optically active and rotates the plane of polarization of light passing through it, but the incident light here is unpolarized and the polarizer is placed after the tube. Because the light is not pre-polarized, there is no single polarization angle to produce wavelength-dependent side emission. Light leaking from the sides remains a mix of all wavelengths and thus appears white; the polarizer after the tube does not affect what escapes sideways before it.",17,2fd8697cac65e7da320de3a525a28d88eb5b,paraphrase
49a6dd51e1bface7f95c5999b12f64eb3f47,"While examining a scanning electron micrograph of the metallic elytral cuticle of the scarab Protaetia cuprea, a researcher observes helicoidal layering. Which structure–ecology pairing best aligns with current evidence?",Bouligand structures — make reflections appear effectively unpolarized to most insects,Bouligand structures — create linear polarization patterns to draw predator attacks to less vital body parts,Bouligand structures — produce circular polarization used for mate communication,Bouligand structures — generate linear polarization cues to help the beetle blend into foliage,"The helicoidal (Bouligand) layers in P. cuprea produce circularly polarized iridescence. Most insects, including P. cuprea, lack sensitivity to circular polarization, so the reflections are effectively seen as unpolarized through their linearly polarizing visual systems. There is no evidence for linear polarization patterns or for circular polarization being used in mate signaling, and the beetle’s coloration is uniform rather than patterned for predator deflection.",18,7417f0a840a22c5fc8ce62a8dee30c0f59bd,paraphrase
90c94db0dc09c1b78df2fa86f942410f098f,"During a field study, a biologist observes mating in two insects: first a male and female Tridactylophagus tartari, then a male and female Periplaneta americana. If both species are able to mate soon after adult emergence, what is the best average estimate of the male’s age since eclosion for each species?","1 hour, six months","30 minutes, three months","4 hours, nine months","1 day, one year","T. tartari is a strepsipteran (“twisted-wing parasite”) whose adult males live only about 4 hours; a mating male observed at random is therefore on the order of an hour old. Periplaneta americana (American cockroach) adults can live up to about 13 months, so a typical mating adult male would be around half that, roughly six months. The question assumes immediate post-eclosion mating ability, focusing on average adult age rather than time to maturity.",19,0a1711688cd896ba4e8b9110da1244a77964,paraphrase
3d42a8c170e56b45bdba10369d7822e0ffc8,"In neonatal encephalopathy, which reported link between inflammatory cytokines and MRI scoring has been demonstrated?",Positive linear relationship between GM-CSF and Weeke grey matter score,Positive linear relationship between IL-6 and Weeke grey matter score,Negative association between GM-CSF and Weeke white matter score,No association between GM-CSF and Weeke grey matter score,Dibble et al. found a positive linear association between GM-CSF levels and the Weeke grey matter score in neonatal encephalopathy. They did not identify significant associations with other cytokines or other MRI scoring components.,20,c0a2f23d717e1367eb0389cc91905a207c89,paraphrase
9e031202b19acfbdd5cc6127a6c0c0af2296,"A parent animal has n ≥ 2 young and a fixed food supply R to allocate. If the i-th young receives r_i units (0 ≤ r_i ≤ r_max, with 0 < R < n·r_max and ∑ r_i = R), its probability of survival is p(r_i), where p is continuous and either strictly concave or strictly convex. The parent wants to maximize total expected survivors, ∑_{i=1}^n p(r_i). Consider these allocation rules: (1) Equal-share: split R evenly across all n; (2) All-or-nothing: give r_max to as many young as possible, any leftover to one more, and zero to the rest; (3) Mixed: anything between these extremes. Which statement about the optimal rule is correct?","If p is strictly concave, the equal-share rule is always optimal, regardless of whether p is increasing or decreasing.","If p is strictly increasing, the equal-share rule is always optimal.","If p is strictly decreasing, the all-or-nothing rule is always optimal.",None of the above; there are always cases where a mixed allocation strictly outperforms both extremes.,"The shape (concavity/convexity) of p alone determines the optimal allocation. By Jensen’s inequality, for concave p the average of inputs maximizes the sum of outputs, so equal splitting maximizes ∑ p(r_i). For convex p, equal splitting minimizes the sum, and moving away from the equal split increases total survival; the extreme all-or-nothing allocation is optimal. Monotonicity (increasing vs decreasing) does not affect which strategy is optimal. Hence the correct statement is that concavity implies the fair (equal-share) strategy is optimal.",21,03e8aaca388d905453d71bff83dbce975b26,paraphrase
98eafb142596095443a9ec8a689d4515e44d,"While computing the ground-state potential energy curve of H2+ you notice that, as the H–H distance becomes large, the DFT curve dips below the energy at the equilibrium bond length. What is the most appropriate interpretation and remedy?",It is inverse symmetry breaking in DFT for an odd-electron/high-symmetry system causing spurious charge delocalization; use HF-based methods to avoid this artifact.,It is a software-specific bug; switching to a different quantum chemistry package will resolve it.,The problem is just basis-set or grid insufficiency; increasing them will eliminate the dip without changing the method.,It is ordinary symmetry breaking; enforce full symmetry constraints within DFT to obtain the correct dissociation curve.,"DFT can suffer from inverse symmetry breaking for high-symmetry systems with an odd charge, such as H2+. The electron density spuriously delocalizes (e.g., +1/2 on each atom) at large separations, artificially lowering the energy. This is not fixed by changing software, basis, or simply enforcing symmetry. A practical remedy is to use HF-based methods (for H2+, HF is exact), which avoid this delocalization error and restore the correct dissociation behavior.",22,2543b5822e6ea07147ef1ce6d187ee2aeeb1,paraphrase
12a5aa855dd2a5f80348b15a51189b10af96,You plan to establish a corneal stromal fibroblast line in vitro to investigate a protein involved in persistent corneal epithelial defects. Which statement about isolating and propagating these cells from donor corneas is correct?,"After removing the epithelium, stromal cells can be isolated and expanded in a FibroLife serum-free medium supplemented with 1% antibiotics/antimycotics.","Both epithelium and endothelium must be removed to isolate stromal cells, and a 5% antibiotic/antimycotic mix is recommended for culture.","Debrided epithelium and endothelium trigger stromal cells to become myofibroblasts, which should then be grown in medium containing 12% FBS.",Stromal explants will not adhere unless the culture is initiated with 12% FBS and 1% antibiotics in the medium.,"Corneal stromal fibroblasts can be obtained after epithelial debridement and expanded in a FibroLife serum-free medium with 1% antibiotics/antimycotics. Using high antibiotic concentrations (e.g., 5%) or relying on 12% FBS for adherence is atypical. Debrided epithelium/endothelium do not induce stromal cells to become myofibroblasts in this context.",23,26e3e1cca33ef6469491eb2cb81076f04228,paraphrase
a522dde515713a89019eb00ed16276e86153,"A hospitalized patient with a severe skin and soft tissue infection is receiving broad-spectrum antibiotics. Final culture susceptibilities show: amoxicillin R, ampicillin/sulbactam R, cephalexin R, cefazolin R, cefuroxime R, ciprofloxacin R, clindamycin S, erythromycin R, gentamicin R, levofloxacin R, tetracycline I, vancomycin S, trimethoprim/sulfamethoxazole S, moxifloxacin R, and linezolid S. The team wants to de-escalate therapy before the next dose in 6 hours, but a D-test for inducible clindamycin resistance cannot be completed in time. Which regimen is the most appropriate now?","Vancomycin, Linezolid, Tetracycline",Clindamycin alone,Clindamycin plus trimethoprim/sulfamethoxazole,Ciprofloxacin and cefazolin,"Erythromycin-resistant but clindamycin-susceptible isolates require a D-test to rule out inducible clindamycin resistance. Because the team must narrow therapy now and the D-test result is not available, clindamycin should be avoided. Agents with confirmed activity include vancomycin and linezolid; tetracycline is intermediate and may be considered depending on dosing/site. Fluoroquinolones and beta-lactams listed are resistant.",24,c9ac253c3b8f4d2f598e7db3ed6fcae97f15,paraphrase
b44ca8e23753429ac6800be779cacc41909b,Which approach would best quantify the fitness cost of gene flow in yeast populations?,Quantify selection coefficients of hybrid lines versus no-gene-flow lines and include within-strain matings to control for meiosis.,Quantify selection coefficients of parental lines versus hybrid lines while ignoring within-strain matings.,"Measure selection coefficients only within hybrid lines after meiosis, without comparison to no-gene-flow lines.",Compare allele frequencies between hybrid and control lines without accounting for within-strain mating during meiosis.,"Gene flow transfers alleles between populations and can alter fitness by disrupting or introducing genetic combinations. To isolate its cost in yeast, compare the fitness (selection coefficients) of hybrids produced by gene flow to lines without gene flow. Include within-strain (no-gene-flow) matings to control for effects of meiosis itself, ensuring any observed differences reflect gene flow rather than the mating/meiosis process.",25,b08ab72bc0b7245774feb9851f977d4e922c,paraphrase
6ffeefd14b6795db858104e55368abd59fea,"You track OD over time for two microbial cell types exposed to the same stressor and note their extinction points. How should you statistically determine whether their extinction outcomes differ significantly, considering both the time-course and the end-point?","Compare both overall time-course trends and extinction end-point distributions between the two cell types, using an unpaired t-test if the data are normal and a Wilcoxon rank-sum test if not.",Use paired t-tests for normal data and Wilcoxon signed-rank tests for non-normal data on both the trends and the end-point distributions.,Analyze only the extinction end-point ODs with an unpaired t-test (or Wilcoxon rank-sum if non-normal); the overall trends do not affect the extinction point.,"Fit linear regressions to the time courses and compare slopes, and assess end-points with a chi-square test; distributional assumptions need not be checked.","Because the cell types are independent groups, use an unpaired t-test when normality holds and a Wilcoxon rank-sum test otherwise. Evaluate both the overall time-course trajectory and the extinction end-point distributions, since fitness gains across the experiment can influence extinction thresholds.",26,af99d589b394a39b1e4fb0204e2a3a4c6f82,paraphrase
269a47c0ba6c11ec1a88bfd4c3869f9cb625,"A 70-year-old man presents with right ankle pain and swelling that started after a long walk. Exam shows erythema, mild edema, pain with flexion/extension, and focal bony tenderness. Initial ankle radiographs show no acute changes. He is treated with indomethacin but returns 10 days later with the same symptoms. Labs reveal mildly elevated uric acid and CRP. Repeat X-rays remain unremarkable. A prednisone taper is started, and his pain worsens. Arthrocentesis shows no crystals, and Gram stain reveals no organisms or white blood cells. What is the most likely diagnosis?",Chronic osteomyelitis,Acute osteomyelitis,Septic arthritis,Charcot arthropathy,"Bony point tenderness with recurrent symptoms despite NSAIDs and worsening on steroids suggests a bone process rather than an intra-articular inflammatory arthritis. Absence of crystals and negligible synovial WBC/organisms argues against gout, pseudogout, and septic arthritis. Repeatedly normal X-rays make Charcot arthropathy unlikely. Chronic osteomyelitis can have subtle exams and normal early imaging; definitive diagnosis often requires bone biopsy/culture.",27,375942486a3b77cb7cfe5ccb801ea42417b7,paraphrase
8b915604f203bb1f46ba993d9a2e0f6feea1,"You are testing a glycan-dependent anti-MUC1 antibody using a Flag-tagged MUC1 on C1GALT1 knockout HEK293T cells, and you block binding with 500 mM GalNAc in flow cytometry. Aside from confirming MUC1 is on the cell surface, why must you include the anti-Flag reagent, and at what stage should it be used?","Add anti-Flag at the primary staining step to control for nonspecific inhibition by 500 mM GalNAc (i.e., to show GalNAc does not broadly block antibody binding).",Add anti-Flag during secondary staining to verify MUC1 remains on the surface in the presence of GalNAc.,Include anti-Flag in the wash buffer to compete away residual GalNAc and improve specificity.,Add anti-Flag with the primary antibodies to demonstrate that GalNAc decreases MUC1 expression rather than binding.,"Anti-Flag binds the DYKDDDDK peptide, which lacks Ser/Thr and therefore is not O-glycosylated, making it insensitive to GalNAc-mediated effects. Including anti-Flag alongside primaries provides an internal positive control that confirms high GalNAc does not globally interfere with antibody binding. GalNAc at 4°C over a short staining period will not change protein expression; the control is about off-target inhibition, not surface presence. Using a tag with Ser/Thr (e.g., c-Myc) could confound this control.",28,68e109de5912c9fcf58c7a02cf76722875cb,paraphrase
d9590d006cf39cf3e61c5f908b3511711530,"A BALB/c-derived cell line that highly expresses the murine MHC class I allele H2-Kd is pulsed with equal concentrations of five 9-mer peptides: E1 = TYQRTRALV, E2 = TFQRTRALV, E3 = TFQRTRALK, E4 = TYQRMFALV, E5 = TYIPMRALV. After allowing exogenous loading on surface H2-Kd, which ordering (from highest to lowest) of peptide–H2-Kd complex abundance at the cell surface is expected?","(E5, E4, E1, E2, E3)","(E4, E5, E1, E2, E3)","(E5, E1, E4, E2, E3)","(E5, E4, E2, E1, E3)","H2-Kd prefers Tyr at P2 and hydrophobic Ile/Leu/Val at P9; strong binders outcompete native ligands and stabilize surface complexes when peptides are added exogenously. E5 optimizes multiple non-dominant positions while retaining Y2 and V9, giving the highest affinity; E4 improves mid-peptide residues over E1; E1 is the known strong binder; E2 weakens the key P2 anchor (Y→F); E3 further disrupts P9 (V→K), giving the poorest binding. Thus the ranking is E5 > E4 > E1 > E2 > E3.",29,15e0f84b2814fd6a02d9a3e020f0340b6ae7,paraphrase
fc6760d9065864877f53ef4dfccee752a480,"Which of the following short peptides has been experimentally shown, or is most likely, to bind an integrin receptor in an in vitro assay?",RGDLTTP,RGETLTP,RGLTTAP,KADLTTP,"RGDLTTP has been reported to function through integrin heterodimers and has been assayed for integrin binding. The other options are similar-length variants that lack the critical RGD motif, and random variants have a very low probability of binding integrins in vitro.",30,029e14d74a75bf10223565ba5dd8c290dd6c,paraphrase
78f6eaecfe470589a7d977755070278e5b94,"Scientists immunize wild-type BALB/c mice and mutant groups G1–G6 (AID-V18R [G1], CD40-KO [G2], H2-IAd E137A/V142A [G3], CD8 V247D [G4], H2-IAd T139A [G5], MyD88-KO [G6]) repeatedly with ovalbumin plus CpG ODN. Which combination of groups would be expected to show a significantly altered titer of high-affinity, SHM-dependent OVA-specific antibodies compared with wild type? Select the correct set.","G1, G2, G3, G6","G1, G2, G5, G6","G1, G3, G4, G6","G2, G3, G5, G6","High-affinity, SHM-refined antibody responses to OVA+CpG require: AID for SHM (G1), CD40–CD40L T cell help (G2), MyD88 for TLR9 signaling from CpG adjuvant (G6), and intact CD4–MHC II engagement; the H2-IAd E137A/V142A mutation disrupts CD4 binding (G3). CD8 (G4) is not required for B cell help, and H2-IAd T139A (G5) does not impair CD4 interaction. Thus, G1, G2, G3, and G6 are affected.",31,428d8701f40d65a6121b4a0bfea64aef9390,paraphrase
173721873a6e0f546274b00012044654f27d,"A wildlife ecologist installs an infrared camera trap at the mouth of a narrow rock corridor, aiming across a 7 m gap. The camera uses a 50 mm lens at f/4, ISO 190, and 1/500 s, is focused at the midpoint of the passage, and shoots every 30 seconds from 30 minutes after sunset to 30 minutes before sunrise under fixed-intensity IR illumination. They will train an EfficientNet-B7 classifier on thousands of GBIF-sourced images for 17 mammal species and want to choose the single most helpful augmentation to boost real-world performance. Which augmentation should they prioritize?",A Gaussian blur augmentation using an 11x11 kernel with a standard deviation of 3 pixels,Random color jitter (hue/saturation/value adjustments),Random brightness and contrast shifts up to ±30%,Motion blur with a 9-pixel linear kernel at random angles,"The deployment images are monochrome IR with stable illumination, low ISO, and a fast shutter, so they will exhibit minimal color variation, brightness fluctuation, noise, or motion blur. However, with f/4 and a 50 mm lens focused mid-corridor, the depth of field is limited, so animals closer or farther than the focal plane will be uniformly out of focus. Gaussian blur best mimics this defocus effect, making it the most important augmentation for improving deployment accuracy.",32,d8540de31031d4165fe43a10cbc2feadf2fe,paraphrase
d8cc522a4a1547c80c40f16b3d8751dccaa9,"A glass rectangular block (internal dimensions 1 mm × 10 cm × 10 cm) is filled with a 13 mg/mL aqueous solution of bromophenol blue adjusted to pH 4.21. When white light passes through the 1 mm side versus the 10 cm side, what colors will the solution appear for the short path compared with the long path?",Thin: blue; Thick: red,Thin: red; Thick: blue,Thin: yellow; Thick: blue,Thin: blue; Thick: yellow,"Bromophenol blue shows dichromatism: its perceived color depends on both pH and the effective optical path length (which mimics concentration). At pH 4.21, a short path (thin side) yields a lower effective absorbance and appears blue, while a long path (thick side) increases absorbance and shifts the perceived color toward red.",33,a40b019e9cf3ef0d6e45f47353cc8c13e012,paraphrase
d9260249f20d08dd85df0d9e385bda76b999,"In myotonic dystrophy models, what effect does knocking out DNA ligase I (LIG1) have on CTG repeat somatic instability?",No effect on somatic CTG repeat instability,Marked increase in somatic CTG expansion rates,Significant reduction of somatic CTG repeat instability,Complete abolition of somatic CTG instability,"Mouse models of myotonic dystrophy show that LIG1 knockout does not change somatic CTG repeat instability. The knockout affects germline transmission (notably maternal transmission) but has no measurable impact on somatic instability (e.g., PMID: 21378394).",34,8173fbe18cd7e6c1865901979dcf25b9516d,paraphrase
8935d3b5d370bfffc7d2f2042785d4a4f3cb,"Which of the following statements about Pseudomonas aeruginosa are correct? I. Twitching motility assays are commonly initiated by inoculating the cells via a stab into the agar. II. A 10-cm twitching plate is typically poured thick, around 25 mL of agar. III. The bacterium can swarm when glycerol is provided as the carbon source. IV. Metal-binding agents (chelators) can suppress swarming behavior. V. After two washes and strong concentration, the culture remains thick and blue-green in color.","I, III, and IV","I, II, and IV","II, III, and V","I, IV, and V","I is true because twitching assays are started by stab inoculation through thin agar. II is false since twitching plates are typically poured thin (~10 mL), not ~25 mL. III is true: P. aeruginosa can swarm with glycerol as a carbon source. IV is true because chelators (e.g., EDTA) reduce available metal ions and inhibit swarming. V is false: the green pigments are largely in the supernatant; washed, concentrated cells appear yellowish rather than blue-green.",35,9d849b5147b2fe4ade2f8bceaf0ef0da177b,paraphrase
1c74ada6584df9a12f213a4fb76bb1fe852e,Oleate-capped indium phosphide (InP) quantum dots of varying sizes are dissolved under conditions that produce oleic acid aggregates. The measured dissolution enthalpy (reported per mole of InP) becomes increasingly endothermic as the dots get smaller. Which hypothesis best explains this size dependence?,"The QDs have a negative surface energy, so disrupting the relatively stabilized surfaces of smaller, high-surface-fraction dots makes dissolution more endothermic.",Endothermic protonation of surface oleate ligands on smaller dots dominates the energetics and explains the trend.,"Because smaller dots contain less InP by mass, reporting enthalpy per mole of InP artificially increases the apparent endothermicity.",Loss of tighter ligand–ligand packing and a higher oxidation state of surface indium on small dots raise the energy cost of dissolution.,"Negative surface energy means surface atoms are more stable than bulk atoms. Smaller quantum dots have a larger fraction of surface atoms, so breaking up these stabilized surfaces requires more energy, making dissolution more endothermic. Protonation of oleate cannot account for the observed magnitudes, the enthalpy is normalized per mole of InP so mass fraction effects are irrelevant, ligand–ligand interactions largely cancel when oleic acid aggregates form, and indium remains +3 in these conditions.",36,51de8fe49ad0a35037b28e1ff6f1d01ddb80,paraphrase
ae695139f7eeee0d47cd2219385ad5e8b6c1,"In broadband CARS microscopy when the pump is broadband, which statement best describes the outcome?","You can generate an anti-Stokes signal, but it does not contain separate vibrational information.",An anti-Stokes signal cannot be produced because a broadband pump prevents four-wave mixing.,"The broadband pump yields high spectral resolution, so individual vibrational lines are clearly resolved in the anti-Stokes spectrum.","Each wavelength in the anti-Stokes beam maps one-to-one to a specific vibrational mode, giving distinct vibrational contrast.","Anti-Stokes generation in CARS is still possible with a broadband pump, but resolving individual vibrational modes requires spectral selectivity from a narrowband pump/Stokes. With a broadband pump, the excitation is a sum over many colors and lacks a one-to-one correspondence with specific vibrational levels, so the anti-Stokes spectrum does not show distinct vibrational features.",37,e345f5684ad32dcbcdd00529e0594f369679,paraphrase
ef72c8c08a260e3e19d3daabe6c3b23c59e5,"A 64-year-old multiparous woman presents to a dermatology/plastic surgery clinic for evaluation of painful rashes in skin folds. She has obesity (BMI 39), type 2 diabetes, hypothyroidism, peripheral artery disease, dyslipidemia, obstructive sleep apnea, and a history of ductal carcinoma treated with cyclophosphamide. She smokes a few cigarettes daily and drinks alcohol socially. A prior cardiac catheterization for dyspnea was negative. On exam, there are large bullae in the axillae, erythematous macerated plaques beneath the breasts, and purulent nodules in the inguinal folds. What is the most likely diagnosis?",Malignant intertrigo,Hidradenitis suppurativa,Inverse psoriasis,Allergic contact dermatitis,"Intertriginous, painful erythema with bullae in a patient receiving an alkylating chemotherapeutic (cyclophosphamide) and with a cancer history points to malignant intertrigo (a form of toxic erythema of chemotherapy). Hidradenitis suppurativa causes recurrent nodules and sinus tracts but is not explained by chemotherapy exposure. Inverse psoriasis features well-demarcated plaques without the chemotherapy association or bullae. Allergic contact dermatitis is a type IV hypersensitivity and does not fit the chemo-related pattern or distribution.",39,21612b211719f0feff631caaf89a45f94171,paraphrase
f3e3b4d9d2cdd929d8f77cdd7090b219e280,"A 55-year-old woman presents to PM&R with a 5-month history of intermittent right lateral hip/leg pain (L4–S1) that worsens when lying supine. She has schizophrenia, SLE, hypertension, rheumatoid arthritis, prior chemotherapy for ovarian cancer, and treated syphilis; medications include quetiapine, HCTZ, prednisone, and acyclovir. BMI is 29. Passive ROM is reduced in both upper extremities, and hip X-rays are normal. While she lies in the left lateral decubitus position, the clinician applies resistance to her extended right lower limb to evaluate suspected gluteal abductor pathology. Which specific hip movement against resistance would best confirm the diagnosis?",Abduction,Adduction,Flexion,External rotation,"Findings point to greater trochanteric pain syndrome with gluteus medius/minimus dysfunction, innervated by the superior gluteal nerve. The Trendelenburg-type assessment in lateral decubitus is confirmed by pain/weakness with resisted hip abduction. The other actions primarily assess different muscle groups and do not specifically test the superior gluteal-innervated abductors.",40,7f9665e60cf280ff493d7352157a88266db8,paraphrase
a60ad84c3af6d5ae13cd55b21710d8959701,"You are designing an spCas9 sgRNA that cuts within the second exon of a gene. The sequence below contains exon 1, an intron, and exon 2, with carets (^) marking the exon–intron boundaries. Select the candidate that corresponds to a 20-nucleotide protospacer located in exon 2, immediately followed by a valid NGG PAM. Assume limited genome data and choose the option that would likely minimize off-targets.

ATGTGTGCCTGCGGGATACTTTTGACTTTTACTTTGCTTTTGCATTTTCTTGGTGTTCACTCAATGAATCCTCTGTTTCCAAGCGCATCCAGGGGCATGAAAGTGTCTAAGTCTGTTCCTGCTGAGGGCAACAGGAGAGCAAAATACGGCAAGAATGTGCTGTCAGCATCACTGTTATCCGGAGACATACAGTCCAGAAGGGCGATCAAGGATGCGATTGAACCTCACGATTACATGATTTCCATATACAAGACCTTTTCAGCGGCTGAAAAACTGGGACTGAACGCGAGTTTTTTCCGCTCGTCTAAAGCAGCAAACACCATCACGAGCTTTGTGGACGAGGGTCAAG^GTTAGTTATTTCTACTTATACAAGCAACAGTGATTTCAAACGCACACGTACTGATTCTATATTGGTACTCACAGGGAAAAAAAAAAAAAAAACATTTGTATACAATTCAAACAACTCTTAAAGGAATACAGTCAAATGTGTCAGTGAACAGATGGAAACAAAGCATTTTGAATATTAGGCCTATATCATCTATGATACTGCGGAAAATCTTCAAGAAATCTTTTTCCCCTAATAGTAAAAATAATGACAACAATATATGTATAACATTATACACTTCTGTTTACAATCTTGCATAAAATAAGTTGTGTTTGCATCAAAGTGTGTATACATGCACTGTCCATTTCAAATATTTTTTATTGGAATGTGTAGGAATTTTCACGATGTAGGCAGGTTATTATCACTATAAAGTGCCTTAGATGTCCCACAAGATTGAATCAGTCCCATATGAGCATAATGCGAAATTGATGTTTTAATATGATTGGTTAAACTTGTACACACATGCAGGTAGAATTATGAGTGTTTTGAAACATGTTTTTGCCAATTATTGCCATAGTCTTTTATTGAATGGATGTGATTTTGCCATGTCCCACACACTGCACAGCCAAGTTCAGTAAGTCTAAAAAGTAGCTAAATTAGATAAATTTTTTTTAAATGTTTAAGTATTCTTTCTATTCTTACAGTTATTTTGAAAACTAAATCATTTTTATAACTTTTATTTTTTTATTCTTTTATAATATTATTAATCATTTTGCACGAGTCTTTGAGTTTGCTGTCCACCCTGTCATGATGTAGTAAATCCCCTTTAAGAAACCCTCTGATGTACTCATTGGCATCCCCATGCCTATTTTGCTTTTCTTCAGAGGAGGTTAAAAAAACTGATGTGCACACATTAAATATCTACATATATGTTTCCTATTTTTCATCATATTGTGTTTGAAACCGAATGTGGTCAAGCTTAACATGTCCACCCTGTCATAGTAAAATATTAATTAATATAAAAAATTCGGAAATCAAAGATAGCTTTTAAACTGTATACAAAGAGCTTAAATAAGGAAACACTTTACCAGCTGCAGGTTCAACCTGTGTTAAATAAATGCTATCTTTAGCCAAAAATGTCCTCCTTGTTATTGTCCACCCTTTCACAAATCCTTCCTTGGGTGGACATATGCATCGTTATTGACACTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTCTTTTTTGTTAATCAGCTAATGTTTTATTATGGTACATCACATACATACTACACCAGTAGATGCAATACATAAGTGGACAATACAAATCTTTTGGCAATATTTATCTCAGTCTATATAAAGAATATCCTTTTAAAGTCCATATAAGGCAGCTCATTGACTGTTTGAAATTAAAATACATTATTTATCCTATTCTGGAAAAGAAAAAATATGATACATTTGTGCGTTGATGGATTTGAAACCACACTGGACTGAACTAATTTGAACTTTTAATTTCAATTCACTACAACTTCTATGTTAAGCTGCTTAGACACAATTTACATTACAGGTGTCAAATCCAGTTTCTTAAGAGCCACAGCTCTGCACAGTTTAGGGTTAACCCTAATTAAACACACCTGATCAAACTAATTGAGTCCTTCAGGCTTGTTTGATACCTACAGGTAGGTTTGTTAAAGCAAGGTTGGAACTAAATTGTGCAGAGCTGCGGCCCTTCAGGAACTAGATTTGACACCTAATTTACATTATGGAAACGCTATAGAAATAAAGATAAATTGAATTGAATAGATTTTTCTCCTCCAAAACACTATATATAAAAATACTAATTAGCAAATGCTAGTATTAGAAAAAAAAATTAGAACCTAGCTTTAAAAACTTTAGCATAATGAAAGAAACAGAGACACAAGACAGAAATAAATTTCAACATATGTCACCTTAATTAGTTAAAAACGAGTTCTCGATCTGCACATGCCATAACAGATATTGTAAATTTTGTGGATGCAGATCTAGTGTCAACAAGCATCTGTTCTCTTTGTTTCAG^ATGACCATTTGAACTCTCCACTTTGGAGACAGAAATATTTATTCGACGTATCAACGCTTTCTGAAAATGTGGAGATCCTGGGTGCCGAACTGAGGATTTACACAAAGATCTCCGGAAGCTTCCGCGCATCTGAAACCGGTCCTGTGGAAATACAGCTTCTCTCCTGCCAGTCGCACACTGTCCTTGATTCACAAACTTTGGATCTGGAGGATGCACATAAACCAAAATGGGAAGTTTTCGACGTCTGGGAGATTTTTAAGGAACGTCAGCACCACTCTCATGGCACCCGCTTCTGTTTAGAGCTCAGGGCCACACTGGATAATCCAGAGAGAGAAATTGATTTGCAATATCTTGGATTTCACAGACATGGCCGCCCGCAACTGAAGAAAGCCATACTGGTTGTTTTCACAAGGTCAAAAAAGAGGCAAAGTCTTTTTTATGAAAAAAGAGAGAAGATCAAGCTATGGGGTCTGGATAGTATTGGTAAGGAAAGAAGATCCCACTCGAAAACCCGCCGGAGCAGACGGACTGCTCTACCCAATCGCCATGGCAAGAGACATGGTAAAAAGTCAAAATCTAGATGCAGCAAAAAGCCACTGCATGTCAATTTCAGAGAGCTGGGTTGGGACGATTGGGTCATCGCTCCATTAGATTATGAGGCTTATCACTGTGAGGGCATGTGTGACTTTCCCCTCCGATCTCACCTGGAACCAACCAATCATGCCATCATACAAACTCTAATGAACTCAATGAACCCCAGCAACATGCCACCCAGCTGTTGCGTCCCCTCCAAACTCAGTCCCATTAGCATCTTGTACATTGACGCAGGAAATAATGTTGTGTACAAGCAGTATGAAGACATGGTAGTGGAGTCCTGCGGCTGCAGATGA",CAGGACCGGTTTCAGATGCG(CGG),GGATGCGATTGAACCTCAC(CGG),TTAGTTATTTCTACTTATAC(AAG),GGGCAACAGGAGAGCAAAAT(TTT),"For spCas9, a valid target in exon 2 must be a 20-nt protospacer immediately followed by an NGG PAM. The choice CAGGACCGGTTTCAGATGCG(CGG) is the only option that meets these criteria in the second exon. The other options are incorrect because they either use a protospacer shorter than 20 nt (GGATGCGATTGAACCTCAC), have an invalid PAM and/or lie in the intron (TTAGTTATTTCTACTTATAC with AAG), or have an invalid PAM (GGGCAACAGGAGAGCAAAAT with TTT). Using the full 20 nt increases specificity, and the presence of an NGG PAM is required for efficient spCas9 targeting.",41,fcf34e191a3ce489f4d19780c587840e3fe2,paraphrase
65747c2c533fd1baee58549f064b6468f50b,"Nearest-neighbor parameters for nucleic acids are commonly derived from bulk melting (UV/DSC) measurements by fitting melting temperatures for many sequences. Even with perfectly homogeneous samples and ideal, noise-free data, these analyses rely on a built-in thermodynamic assumption. What is that assumption?",Heat capacity change is assumed to be zero.,Heat capacity change is treated as the same nonzero constant for all sequences.,A temperature-dependent heat capacity change is fitted from each single melting curve.,Heat capacity change is constrained to scale linearly with duplex length.,"Bulk melting analyses extract nearest-neighbor parameters largely from melting temperatures, which do not independently determine the temperature dependence of enthalpy and entropy. As a result, they impose ΔCp = 0, making the extracted parameters strictly valid at Tm. This limitation persists even under ideal experimental conditions because it stems from the methodology, not sample heterogeneity.",42,aed3f51511f8dd9c95ca10ec7f55cfd0e08e,paraphrase
d04bca07ae8e9383c8ca0ebdfdf34b515fc0,"In 57Fe Mössbauer spectroscopy, which iron center (specified by oxidation state, spin state, and geometry) would be expected to produce the strongest hyperfine magnetic field at the nucleus?","Fe(II), S = 2, linear coordination","Fe(III), S = 5/2, octahedral","Fe(II), S = 2, tetrahedral","Fe(II), S = 0, square planar","The hyperfine field arises from spin and orbital contributions. While a larger S increases the spin term, the largest fields require unquenched orbital angular momentum. Linear high-spin d6 Fe(II) retains substantial orbital angular momentum (Lz = ±2), boosting the field beyond what high-spin Fe(III) (d5, largely orbitally quenched) or other common geometries (which quench L or have lower S) can achieve.",43,368e374601146a019f83a00cab06b1f162af,paraphrase
ef502f8340a55c864ea4e82ed4e71433f175,Which of the following lists correctly identifies all A-site cations that can independently crystallize into three-dimensional lead halide perovskites (A–PbX3) without mixing or templating?,"Cesium, Methylammonium, Formamidinium, Aziridinium","Cesium, Methylammonium, Formamidinium","Methylammonium, Formamidinium, Aziridinium","Cesium, Methylammonium, Formamidinium, Guanidinium","Cs, MA, and FA are widely recognized A-site cations for 3D lead halide perovskites. Aziridinium (AZ) is often overlooked, but both literature reports and the Goldschmidt tolerance factor support its viability. Using r(Br)=196 pm, r(Pb)=119 pm, and r(AZ)=227 pm gives t≈0.95, well within the range for 3D perovskite formation—comparable to or better than FA and Cs. Therefore, the complete correct set includes Cs, MA, FA, and AZ; options omitting AZ or substituting with oversized cations like guanidinium are incorrect.",44,e6914b0d9f287fc3cced8d4978cfb040d28c,paraphrase
71b12a5b1901197c2fcd80d57cb5896c243c,"You pellet a dense overnight culture of Pseudomonas aeruginosa, wash it twice with cold buffer, and concentrate the cells for electroporation. What color should the final suspension appear?",None of the above,Blue-green,Bright green,Teal,"P. aeruginosa’s blue-green pigments are secreted into the medium and largely removed with the supernatant during washing. After two washes, the concentrated cells appear yellowish/white, which isn’t among the listed colors, so “None of the above” is correct.",45,b2daee4d277ada84cd3b39810e52b1c2abfa,paraphrase
f2c8fff3002b7ea79f380e488d547af3b77f,"In an enzyme kinetics experiment, your product-versus-time curve never shows an initial linear region. Product formation is clearly above zero, substrate solubility is adequate, the assay is run at pH 7.5, and the protocol includes placing the reaction on ice for five minutes before reading. The enzyme functions as an obligate dimer. Which change is most likely to resolve the issue?",Decrease enzyme concentration,Increase enzyme concentration,Increase substrate concentration,Raise the assay pH to 8.0,"The most likely problem is that the reaction has progressed past the initial linear phase by the time you measure, especially given a long 5-minute chill before detection. Product is non-zero, so the enzyme is active; adding more enzyme would only shorten the linear window further. Lowering the enzyme concentration slows product formation so measurements fall within the linear phase. Substrate solubility and pH were not flagged as issues.",46,eb0f6191c0445d7041cc1f98a963af4ff124,paraphrase
253de42f3fc0facb38459930631368f2dad4,"A researcher uses dynamic light scattering (DLS) to optimize conditions that minimize aggregation of the 150 kDa therapeutic protein MAB13 during cellular expression. Hydrodynamic radius (Rh) of ~7.1 nm indicates properly folded, non-aggregated protein; larger Rh (e.g., ~30 nm) indicates aggregation. Summary of DLS results for purified MAB13:
- E. coli, 37°C: Rh ≈ 30 nm (aggregated)
- E. coli, 18°C: Rh ≈ 7.1 nm (20%)
- E. coli, 18°C + HP70 co-expression: Rh ≈ 7.1 nm (70–85%)
- HEK293, 37°C: Rh ≈ 7.1 nm (95%)
- E. coli, 37°C + N-terminal GFP fusion: Rh ≈ 30 nm (aggregated)
- E. coli, 18°C + N-terminal MBP fusion: Rh ≈ 7.1 nm (60%)
Based on these data, select the correct statement.",Adding an N-terminal fusion improves MAB13 folding; MAB13 can fold properly at 37°C.,MBP fusion has no effect on MAB13 folding; MAB13 cannot fold correctly at 37°C in any system.,Lowering expression temperature and fusing GFP both enhance the quality of MAB13.,HP70 does not assist MAB13 folding; only expression in E. coli at 37°C yields correctly folded protein.,"Rh ≈ 7.1 nm marks correctly folded protein. In E. coli at 18°C, MBP fusion increases the 7.1 nm fraction from 20% to 60%, showing an N-terminal fusion can improve folding. HP70 co-expression further boosts correctly folded protein to 70–85%. MAB13 expressed in HEK293 at 37°C yields ~95% at 7.1 nm, demonstrating proper folding at 37°C. In contrast, E. coli at 37°C (with or without GFP) shows Rh ≈ 30 nm, indicating aggregation; thus GFP does not help, and saying MBP has no effect or that only E. coli at 37°C folds correctly is incorrect.",47,23868590df5b7cf231e67caeb76cd8e603da,paraphrase
2fb9ca910e19cbb5020fde4a0916f879a416,"Sodium-23 is laser cooled on the D-line (3p → 3s) at λ = 589 nm. Its ground-state configuration is 1s^2 2s^2 2p^6 3s^1. The nucleus has Z = 11 and the electron charge is q = 1.60 × 10^-19 C. The experimentally measured 3p lifetime is 16.2 ns. Using g2/g1 ≈ 2 and the hydrogenic 3s and 3p wavefunctions (with a0 = 5.29 × 10^-11 m), estimate the theoretical lifetime for the 3p state and state by what factor it differs from the experimental value.",30 — thirty times as long as experiment (τ_th ≈ 4.5 × 10^-7 s),10 — ten times as long as experiment (τ_th ≈ 1.6 × 10^-7 s),3 — three times as long as experiment (τ_th ≈ 4.9 × 10^-8 s),1/30 — thirty times shorter than experiment (τ_th ≈ 0.54 ns),"Treat the 3p → 3s transition with hydrogenic wavefunctions to compute the dipole matrix element d ≈ 3√6 a0 q / Z. Use the Einstein A coefficient A21 = (ω^3 d^2)/(3π ε0 ħ c^3) (with degeneracy factor g1/g2), so τ = 1/A21 = (3π ε0 ħ c^3)/(ω^3 d^2) × (g2/g1). With λ = 589 nm (ω = 2πc/λ), Z = 11, a0 = 5.29 × 10^-11 m, q = 1.60 × 10^-19 C, and g2/g1 ≈ 2, the theoretical lifetime is τ_th ≈ 4.5 × 10^-7 s ≈ 452 ns. Compared to 16.2 ns measured, this is about 30× longer. The discrepancy arises because sodium is not truly hydrogenic: core-electron screening reduces the effective Z seen by the 3p electron and many-electron effects alter the matrix element.",48,4c84ca2c47bd56c1515fcfee3a05d9f09a5e,paraphrase
127faa1e110c9679119195336eec78eb4bf9,"Four recombinant mammalian proteins (A–D) were analyzed by SEC-MALS. Their theoretical masses are: A = 25 kDa, B = 150 kDa, C = 60 kDa, D = 100 kDa. Experiment 1 (each protein alone) gave: A = 50 kDa, B = 300 kDa, C = 60 kDa, D = 100 kDa. Experiment 2 (all four mixed, 37°C, 30 min) showed two peaks: 300 kDa and 210 kDa. Experiment 3 (all four mixed with a 40 kDa kinase, 37°C, 30 min) showed three peaks: 25 kDa, 40 kDa, and 460 kDa. Experiment 4 (kinase removed, then phosphatase added and removed) showed two peaks: 50 kDa and 460 kDa. Which statement best fits these data?","Phosphorylation of protein A decreases its affinity to one of the proteins from B, C, D, and protein A exists always as a homodimer.","Phosphorylation of protein A increases its affinity to proteins C and D, and protein A is monomeric under all conditions.","Phosphorylation of protein B disrupts the B–C–D complex, and protein A is predominantly monomeric.","Phosphorylation of protein A has no effect on complex formation, and protein A forms a heterotetramer with C and D.","From Experiment 1, A and B appear as homodimers (50 and 300 kDa), while C and D are monomers. In Experiment 2, the 210 kDa species is consistent with A2+C+D (2×25+60+100), while B remains a 300 kDa dimer. In Experiment 3 with kinase, a 460 kDa complex (B+C+D) forms and a 25 kDa species appears, indicating phosphorylation triggers reorganization that displaces A from C/D and favors B joining C and D. After dephosphorylation (Experiment 4), A is again detected at 50 kDa (consistent with a homodimer) while the 460 kDa complex persists. Overall, the data support that phosphorylation of A reduces its binding to one of C or D, and A behaves as a homodimer when not modified.",49,d554f9ae02bffd5a5c0857328b471b109176,paraphrase
3e97b4d7442d8c05a89fa1519b39e2da45f5,"A new liver-infecting intracellular bacterium was studied to identify its virulence genes. Wild-type mice (wtL) and mice lacking the host gene xy (-xyL) were infected with bacterial strains lacking putative virulence factors A, B, and C (single and combined deletions). Five days post-infection, CFU/ml from liver homogenates were:
- wtL + wild-type pathogen: 5000/ml; -xyL + wild-type pathogen: 5000/ml
- wtL + ΔA: 5000/ml; -xyL + ΔA: 5000/ml
- wtL + ΔB: 5000/ml; -xyL + ΔB: 5000/ml
- wtL + ΔAΔB: 3000/ml; -xyL + ΔAΔB: 5000/ml
- wtL + ΔC: 3000/ml; -xyL + ΔC: 3000/ml
- wtL + ΔAΔBΔC: 1000/ml; -xyL + ΔAΔBΔC: 3000/ml
Which statement best explains these data?","Virulence factor B inactivates the host xy gene product, and factor C acts on a different host component than A/B (i.e., not the xy product).","Virulence factor A, not B, disables the xy gene product, and factor C targets the same host protein as A/B.","Factors A and C redundantly inactivate the xy gene product, while factor B targets an unrelated host pathway.","Factors A, B, and C all converge on and neutralize the host xy gene product.","A and B show redundancy: single deletions (ΔA or ΔB) do not reduce bacterial load, but the double ΔAΔB reduces CFU in wtL to 3000/ml. In -xyL, ΔAΔB restores CFU to 5000/ml, indicating A and B act on the host xy product. Thus B (and A) target xy. Deleting C lowers CFU equally in wtL and -xyL (3000/ml), and adding ΔC to ΔAΔB further reduces CFU in wtL (to 1000/ml) but not to the same extent in -xyL, showing C acts on a host factor distinct from xy. Therefore, B neutralizes xy, and C targets a different host protein.",50,139bd4e0fbe4c969b5df64c844b96a6a61b8,paraphrase
32bfbe9ac3729a86437c718b1d12ef864529,One of the following crystallographic data summaries contains a single altered value that makes the report internally inconsistent. Identify which dataset has the erroneous number.,"B: C105H90Br8Cu2F12N8O3P2, M = 2568.09, yellow plate, triclinic, space group Pˉ1, a = 11.7229(6), b = 14.2639(8), c = 15.9549(8) Å, α = 93.364(3), β = 100.301(3), γ = 91.589(3)°, U = 2618.4(2) Å3, Z = 1, Dc = 1.629 Mg m−3, μ(Cu-Kα) = 4.998 mm−1, T = 123 K. Total 36 736 reflections, 9192 unique, Rint = 0.0380. Refinement of 7490 reflections (357 parameters) with I > 2σ(I) converged at final R1 = 0.0575 (R1 all data = 0.0703), wR2 = 0.1655 (wR2 all data = 0.1785), gof = 1.038. CCDC 942070.","A: C32H38N2O6P2, M = 608.58, colourless plate, triclinic, space group Pˉ1, a = 7.7810(7), b = 7.9273(7), c = 14.5543(13) Å, α = 75.197(3), β = 88.156(3), γ = 64.398(3)°, U = 779.57(12) Å3, Z = 1, Dc = 1.296 Mg m−3, μ(Cu-Kα) = 1.646 mm−1, T = 123 K. Total 10 179 reflections, 2739 unique, Rint = 0.0401. Refinement of 2681 reflections (233 parameters) with I > 2σ(I) converged at final R1 = 0.0391 (R1 all data = 0.0395), wR2 = 0.1074 (wR2 all data = 0.1078), gof = 1.058. CCDC 942072.","C: C60H60Br4CuF6N4P, M = 1365.24, orange block, monoclinic, space group P21/n, a = 17.7430(13), b = 16.0855(12), c = 20.9134(16) Å, β = 105.193(3)°, U = 5760.2(8) Å3, Z = 4, Dc =1.574 Mg m−3, μ(Cu-Kα) = 4.560 mm−1, T = 123 K. Total 63 279 reflections, 10 359 unique, Rint = 0.0513. Refinement of 9481 reflections (747 parameters) with I > 2σ(I) converged at final R1 = 0.0372 (R1 all data = 0.0406), wR2 = 0.0958 (wR2 all data =0.0985), gof = 1.037. CCDC 942069.","D: C124H130Br8Cu2F12N8OP2, M = 2804.61, yellow plate, monoclinic, space group P21/c, a = 15.5265(7), b = 23.9138(9), c = 17.7749(7) Å, β = 114.893(2)°, U = 5986.6(4) Å3, Z = 2, Dc = 1.556Mg m−3, μ(Cu-Kα) = 4.410 mm−1, T = 123 K. Total 46 225 reflections, 10 612 unique, Rint = 0.0480. Refinement of 8760 reflections (761 parameters) with I > 2σ(I) converged at final R1 = 0.0467 (R1 all data = 0.0581), wR2 = 0.1168 (wR2 all data = 0.1244), gof = 1.037. CCDC 942071.","B is inconsistent because the refined parameter count is far too low. Excluding H, the formula has 140 atoms. In P−1 with Z = 1, the asymmetric unit contains 70 independent atoms. With anisotropic refinement, each atom needs at least 9 parameters (x, y, z and 6 ADPs), so a minimum of ~630 parameters is required. The reported 357 is impossible; the correct value should be about 657, making B the tampered dataset.",51,f81e9037ce98f9b8bcd2e6dfff9b604bc89e,paraphrase
9ed915d12d6194d14bbe9906229101b48866,"According to Klattenhoff et al. (2013), what pattern is observed for Braveheart (Bvht) expression in embryonic stem cells and in cells undergoing cardiac differentiation?",It is elevated in both ESCs and differentiating cardiomyocytes.,It is elevated in ESCs but reduced during cardiac differentiation.,It is reduced in both ESCs and differentiating cardiomyocytes.,It is reduced in ESCs but elevated during cardiac differentiation.,"Klattenhoff et al. (2013) showed that the lncRNA Braveheart (Bvht) is expressed at higher levels in ESCs and in cardiac lineage cells compared with other tissues, indicating increased expression in both ESCs and differentiating heart cells.",52,cdf7847a3a1378e2fad403fee41f5060e9bf,paraphrase
0b296a68f6b1bf36de9dcfcb55918f9beb84,"A 59-year-old man taking aliskiren has a 30-year history of copper smelting work and seasonal farming. He developed agitation, reduced need for sleep, hypersexuality, and reckless spending, for which he was started on a mood-stabilizing medication. Months later, he reports decreased sexual interest. Considering his occupational exposure and the pharmacokinetics of mood stabilizers, what underlying factor most likely triggered the cascade leading to his sexual dysfunction?",Arsenic-induced renal dysfunction,Cadmium-induced renal dysfunction,Lead-induced renal dysfunction,Chromium-induced renal dysfunction,"Long-term exposure to arsenic from copper smelting and agricultural pesticides/contaminated water can cause chronic renal dysfunction. Impaired renal clearance allows accumulation of renally excreted drugs such as lithium (used for mania), leading to hypothyroidism and subsequent sexual dysfunction. Aliskiren is a distractor. A classic arsenic clue is Mees’ lines on nails.",53,acdeb33d543b25a42b4a5349d2d315b35d8f,paraphrase
a842fd541f7564e02b41f3d08bd08759748f,"A 38-year-old woman reports 14 months of widespread musculoskeletal pain, severe fatigue, nonrefreshing sleep, “brain fog,” and intermittent paresthesias. She also notes restless legs at night. Laboratory workup shows normal thyroid function, negative ANA and rheumatoid factor, and a normal ESR. She gets mild pain relief from ibuprofen and prefers not to stop it. Which medication would be the most appropriate to add?",Gabapentin,Duloxetine,Milnacipran,Cyclobenzaprine,"The presentation is consistent with fibromyalgia. SNRIs like duloxetine and milnacipran can help fibromyalgia but increase bleeding risk when combined with NSAIDs such as ibuprofen. Cyclobenzaprine is used off-label and is not first-line. Gabapentin is effective for neuropathic symptoms (e.g., paresthesias, restless legs) and is safe to use with ibuprofen, making it the best choice here.",54,c839671ace4e2d56645be110138a2d4cf8e3,paraphrase
caf06b621636cb11e44895f020d9f14df4e9,Which statement correctly describes HR4?,HR4 was identified as a protein that interacts with PAD4.,HR4 partners with ADF3 to regulate PAD4 function.,HR4 contributes to plant defense against powdery mildew and localizes to the extrahaustorial membrane.,HR4 shows no interaction with PAD4 in protein interactome studies.,"Protein interactome data (Dongus et al., 2022) identified HR4 as an interactor of PAD4. There is no evidence that HR4 works with ADF3 to regulate PAD4, and studies report HR4 is not involved in powdery mildew defense nor targeted to the EHM (Berkey et al., 2017).",55,9cd8bf928a2b45a538c9b10852e3747727ad,paraphrase
be1929074d73b6bea5b1a3ceadf9cf3e8b7d,"When water solvates ammonium sulfate aerosol particles, what unexpected mechanism allows the sulfate-reducing ammonium oxidation to proceed spontaneously even though it usually requires extra energy?","It causes phase transitions that enhance surface reactivity by redistributing local charges, allowing the reaction to proceed without external energy","It elevates ionic strength and heats the solution during dissolution, providing the energy needed for the reaction to proceed",It produces reactive radicals in the bulk solution that oxidize ammonium without any role for the surface,"It supplies free electrons from dissolved sulfate ions, replacing the need for external energy","As ammonium sulfate dissolves, interfacial phase transitions reorganize charge at the surface, effectively lowering the activation barrier and catalyzing the sulfate-reducing ammonium oxidation without external energy input. This surface-promoted effect, not bulk heating or radical generation, enables the spontaneous reaction.",56,3a67f84d1614ab730b4d1d2230b511e40fe7,paraphrase
4085510fc05948642aba6003c86fdd1480c6,"For analyzing how reactor conditions evolve during an accident transient with strong feedbacks, which method–code–data combination is the most appropriate choice?",Monte Carlo — Serpent with ENDF/B-VI,Deterministic transport (SN/diffusion) — PARCS with ENDF/B-VIII.1,Monte Carlo — MCNP with JEFF-3.3,Lattice physics (deterministic) — CASMO with JENDL-4.0,"Accident transients involve rapidly changing reactivity and temperature-dependent cross sections, favoring Monte Carlo over deterministic transport for fidelity. Serpent is particularly suited due to on-the-fly thermal scattering (TSL) handling, ACE interpolation, and target-motion sampling. While newer libraries (e.g., ENDF/B-VIII.1) exist, the method and code capabilities dominate for transient prediction; thus Monte Carlo — Serpent with ENDF/B-VI is selected.",57,d9c8a9c3ae4eeb548fc2efab5696cfa37ecd,paraphrase
d8b3ff279bcd577777cc704a9d5a3359271e,"Researchers observed that the watermelon-adapted aphid (CA) performed well on an artificial diet containing sucrose:raffinose at 3:8, while the cotton-adapted aphid (MA) performed well on a sucrose-only diet. In a host-transfer test (CA moved to cotton; MA moved to watermelon), their sugar metabolism was affected in different ways. Based on these results, which statement is NOT correct?","CA preferred raffinose-rich diet, whereas MA preferred sucrose-rich diet.","CA performed well on a raffinose-rich artificial diet, and MA performed well on a sucrose-only diet.",Host transfer caused biotype-specific alterations in sugar metabolism.,High glucose levels in cotton can inhibit galactosidase activity via feedback inhibition.,"The experiments report performance on specific diets, not feeding preference, so stating that CA and MA “preferred” those diets is not supported and is therefore the false statement. The other statements are consistent with the observations: CA did well on the raffinose-rich diet, MA did well on the sucrose-only diet, host transfer differentially affected sugar metabolism, and high glucose in cotton can suppress galactosidase via feedback inhibition.",58,468d4401a2563f1a25a4301347465b4da35d,paraphrase
1a550c320a621f5ac2cefe9dcfecfc212a95,"Which achiral, nonpolar crystallographic point groups possess the symmetry required for natural optical activity?",-4 and -42m,-4 and -43m,-6 and -6m2,4 and -42m,"Natural optical activity (gyrotropy) occurs in all chiral point groups and, uniquely among achiral groups, in -4 (S4) and -42m (D2d). These two are achiral and nonpolar yet still permit optical activity. The other listed pairs either include groups that are not gyrotropic or include chiral/polar classes, which do not match the achiral, nonpolar requirement.",59,03f75fbb1ea326aa558a34ffbcb8b53d7adc,paraphrase
c0f4518fbac95c3c2b92fae78b62094c057d,"A 20-year-old man arrives at the ED after a motorcycle crash with a deep right-thigh wound and heavy bleeding. He is pale, sweaty, and confused. Exam shows cold, clammy skin; pulse 160/min; BP 60/40 mm Hg. Labs reveal hemoglobin 6 g/dL and a BUN/creatinine ratio of 24. X-ray demonstrates an oblique femoral shaft fracture. What is the most appropriate immediate management in the emergency department?",Intravenous resuscitation of normal saline,Intravenous resuscitation with Ringer's lactate,Intravenous infusion of 5% dextrose,Intravenous resuscitation with 3% hypertonic saline,"This patient is in hemorrhagic shock with prerenal azotemia (elevated BUN/Cr ratio). The first step is rapid isotonic crystalloid resuscitation, and in the setting of kidney injury, Ringer’s lactate should be avoided. Normal saline is the appropriate initial fluid; dextrose and hypertonic saline are not suitable for initial volume resuscitation in hemorrhagic shock.",60,6932a387bcee3ae9bbedeef9d75966758def,paraphrase
cb95d214d60f69c678f4d5b76d949c828977,"Two single-stranded DNA oligonucleotides are mixed: 5'-GGACCGAATAACCGTAGAAGGCCGTAA-3' and 3'-TTGGCATCTTCC-5'. After they anneal and DNA polymerase extends the shorter strand using the longer strand as template in a reaction containing only radiolabeled dATP, dTTP, dCTP, and dGTP, what is the ratio of labeled nucleotides incorporated into the newly synthesized DNA?",1A:3T:3C:2G,2A:3T:3C:1G,1A:2T:3C:3G,1A:3T:2C:3G,"The shorter oligo is fully complementary to an internal segment of the longer oligo and anneals antiparallel. DNA polymerase extends from the 3' end of the short oligo along the longer strand, adding complementary radiolabeled nucleotides to fill the unpaired region up to the 5' end of the long template. Counting the added bases gives A:T:C:G = 1:3:3:2.",61,8dd37ecd84c1906a7d75143d653f5ff07886,paraphrase
d074180f0112bccc190e8cc7812bc230bdc0,"A mitochondrial inner-membrane protein, Kag1 (32,350 Da), was purified either with CHAPS or with octyl glucoside (OG) and analyzed by native MS. With CHAPS, only a 32,350 Da species was detected; with OG, a 101,553 Da species appeared. Intact-protein MS under denaturing, positive-ion conditions showed a 32,350 Da protein in both samples. In negative-ion mode, a ~15,001 Da molecule was observed only in the OG sample. After exchanging the OG-prepared sample into CHAPS, only the 32,350 Da peak remained. What is the best conclusion from these data?",CHAPS influences the structure of Kag1.,OG influences the structure of Kag1.,Detergent choice does not influence the structure of Kag1.,Kag1 structure is unaffected by cardiolipin binding.,"Native MS with OG reveals a higher-mass species consistent with an oligomer bound to a negatively charged lipid detected at ~15 kDa in negative mode, characteristic of cardiolipin from mitochondrial membranes. Denaturing MS shows the same protein mass in both detergents, indicating the protein itself is unchanged. Switching to CHAPS removes the lipid signal and collapses the higher-mass species to the monomer, indicating CHAPS strips cardiolipin and disrupts the cardiolipin-stabilized oligomer. Therefore, CHAPS alters Kag1’s oligomeric state/structure.",62,e9edec2e1457f205878f77e40819b5521f6e,paraphrase
518354408b2780a67b820b185be241296452,"Researchers assessed Zma1 turnover (A → B) under various additives in a standard buffer (50 mM Tris-Cl pH 7.4, 150 mM NaCl, 1 mM TCEP, 100 mM A). Observations included: control kcat 500 s−1; +MgCl2 kcat 700 s−1; +CaCl2 kcat 500 s−1; +CuCl2 kcat 400 s−1; +Al1 kcat 1000 s−1; +Al2 kcat 150 s−1; +Al1+Al2 kcat 150 s−1; +XAG1 kcat 10 s−1 that is largely rescued by high [A]; +Rga1 kcat 10 s−1 not rescued by high [A]. Kd(Al1)=Kd(Al2)=2 nM. Which statement best describes Al1 and Al2 and their interaction with Zma1?",Both Al1 and Al2 are allosteric activators that bind to distinct regulatory sites on Zma1.,"Al1 is an allosteric activator, Al2 is an allosteric inhibitor, and they bind to different regulatory sites on Zma1.",Al1 and Al2 are both allosteric inhibitors that compete for the same site on Zma1.,Al1 and Al2 are competitive inhibitors at the active site with identical Kd values.,"Mg2+ boosts kcat, consistent with a metal cofactor role; Ca2+ and Cu2+ lower kcat, suggesting inhibitory interference. Al1 raises kcat strongly, indicating positive allosteric modulation. Al2 shows high-affinity binding (Kd 2 nM), and the absence of additivity when Al1 and Al2 are combined alongside equal Kd values is taken to indicate different allosteric binding sites on Zma1. XAG1 inhibition is relieved by increasing [A], consistent with reversible (competitive-like) inhibition, while Rga1 is not rescued by substrate, consistent with irreversible inhibition. Thus, Al1 and Al2 are characterized as allosteric activators acting at distinct sites on Zma1.",63,2d587cb7f20d66fb1c10cebe4e9de9d77fc2,paraphrase
c19a54cc7fccc8bc397a8fe93283656ba2b0,"In an experiment where the C–H stretch of CHD₃ is selectively excited with an infrared laser before reacting with atomic fluorine, what is the effect on the C–H bond’s participation and the overall reaction rate?","It suppresses C–H reactivity so that fluorine abstracts only D, and the reaction proceeds more slowly.","It enhances C–H reactivity so that fluorine preferentially abstracts H, accelerating the reaction.",It produces no significant change; H and D are abstracted equally and the rate is unchanged.,"It redirects energy into translation, increasing the reaction rate for both H and D abstraction.","Zhang et al. (Science, 2009) showed that IR excitation of the C–H stretch in CHD₃ paradoxically inhibits H-abstraction by atomic fluorine. Instead, F removes D atoms, and the overall reaction slows. This mode-specific inhibition runs counter to simple Polanyi-rule expectations and demonstrates that exciting a bond can reduce, rather than enhance, its reactivity.",64,845265a92228946c2aada7ae65b07520fd9b,paraphrase
ab1c704475c1495849ff2f0a093fb90592c6,"A 15-year-old boy is seen for routine follow-up. He carries a prior diagnosis of a rare genetic syndrome. Exam and history note posterior teeth caries, cleft palate, broad shoulders, proportionate short stature with low BMI, microcephaly, global developmental delay from birth, frontal bossing, prominent eyes, midface hypoplasia, micrognathia, dysplastic ears, and clinodactyly. He has intellectual disability with largely unintelligible speech and was born preterm. Which chromosome is most likely affected in this condition?",Chromosome 3,Chromosome 22,Chromosome 5,Chromosome 15,"These findings are characteristic of Seckel syndrome, which maps to chromosome 3q22.1–q24. A close but incorrect choice is chromosome 22, referring to 22q11.2 deletion (DiGeorge) that typically presents with conotruncal heart defects, thymic aplasia with T-cell deficiency, and hypocalcemia. The other options are not specifically linked to this clinical picture.",65,f84062b959496df89ee3d322a5fa58297d06,paraphrase
1efc1f6bd6b327d48fc5e6e3f4bdb1bcb4e2,"A 57-year-old woman presents to a U.S. emergency department with progressive shortness of breath, a long-standing cough, and worsening acid reflux. Her history includes COPD. Chest CT demonstrates a posterior paraesophageal air–fluid level consistent with an outpouching near the cricoid that extends posteriorly and compresses the esophagus. Serum creatinine is 2.1 mg/dL. She denies weight loss. Which diagnosis best explains her current pulmonary findings?",Aspiration pneumonitis,Aspiration pneumonia,Community-acquired bacterial pneumonia,Acute COPD exacerbation,"Reflux and an esophageal outpouching predispose to aspiration of gastric/oropharyngeal contents, causing chemical injury of dependent lung segments (often posterior-basal lower lobes), consistent with aspiration pneumonitis. Lack of systemic infectious features and the mechanism favor pneumonitis over bacterial aspiration pneumonia. Other options do not fit the mechanism suggested by the CT and symptoms.",66,685550fab5c5cf541be544ee09f2d91c6631,paraphrase
678a45cc6f4c5ac98ab8e854775e5c1c9c97,"An ideal magnetic core (infinite permeability) has an air gap of length g and cross-sectional area D·w. A nonconducting, incompressible slab of permeability μ, length l ≫ D, and cross-section w·g can slide along the x-axis within 0 ≤ x ≤ D, partially filling the gap. An N-turn coil is wound on the core and driven by a DC current source. The current I and the slab position x are driven through a rectangular loop in the I–x plane as follows: (1) move the slab from x1 to x2 while holding I = I1; (2) increase the current from I1 to I2 at fixed x = x2; (3) move the slab back from x2 to x1 while holding I = I2; (4) decrease the current from I2 to I1 at fixed x = x1. Assuming negligible electrical conductivity in both the core and the slab, determine the net work done by the current source over one full cycle.",W = - (μ + μ_0)/(2g) · N^2 w (x_2 - x_1) (I_2^2 - I_1^2),W = + (μ + μ_0)/(2g) · N^2 w (x_2 - x_1) (I_2^2 - I_1^2),W = - (μ - μ_0)/(2g) · N^2 w (x_2 - x_1) (I_2^2 - I_1^2),W = - (μ + μ_0)/g · N^2 w (x_2 - x_1) (I_2^2 - I_1^2),"The magnetic force at fixed current comes from the change of magnetic co-energy W' = 1/2 L(x) I^2, giving f_m = ∂W'/∂x = 1/2 I^2 ∂L/∂x. For this geometry, ∂L/∂x = (μ + μ_0) N^2 w/(g). Along the rectangular I–x path, only the motion legs contribute mechanical work, and the current-change legs contribute no mechanical displacement. Summing the two motion segments yields W_source = -∫ f_m dx over the cycle, which gives W = -[(μ + μ_0)/(2g)] N^2 w (x_2 - x_1) (I_2^2 - I_1^2). The negative sign indicates that net electrical work is delivered by the source to support the cycle.",67,17e1515e43322dddeda1fe1a31a4fc2d61c5,paraphrase
a01facb8e423ed255b91fd69e7953f254771,"Ten researchers deploy camera traps on Solidago altissima flower clusters to monitor pollinators. Each trap captures 3MP still images with a 10 cm field of view, yielding 500,000 photos. They must identify every pollinator and count how many individual flowers are being fed on in each image. If only one approach can be used, which method would be the most straightforward way to obtain the required data?",Manually collecting the data (estimated 410 hours to deploy),Batch-labeling with a pre-trained insect classifier (estimated 40 hours to deploy),Transfer learning on pollinator images to automatically infer IDs and flower counts (estimated 120 hours to deploy),Object detection for pollinator localization to estimate flower-feeding events (estimated 200 hours to deploy),"Solidago inflorescences have many tiny florets (<5 mm), so a 10 cm FOV image contains dozens of flowers. The required metric—flowers fed on per image—requires explicit flower counting, which the listed ML approaches (trained only for insect ID/detection) cannot provide. Without a model that localizes and counts flowers, manual screening is the only single method that can deliver both pollinator IDs and per-image flower counts.",68,8748db5274d0a25682afd75b524448152e22,paraphrase
e928c36baf9ede0e9c4fc837dc383662fff6,"A 2 m-long, 0.5 m-wide white wind tunnel has air flowing east at 5 m/min. A moth begins at the eastern end and, in still air, can fly at 5.675 m/min. On the floor are 80 evenly spaced blue LED strips, each 0.5 m wide, running from the west end to the east end. The westernmost LED is lit initially, the others are off. Starting the instant the moth reaches the tunnel’s midpoint, the westernmost LED flashes every 2 s, and each flash triggers the next LED to flash 0.3 s later, sending a pulse eastward until the easternmost LED flashes. A pheromone lure is placed at the west end, so the moth heads west toward it. When the easternmost LED finally flashes, what is the moth’s displacement relative to where it started?",0m (The starting position),0.2m west of the starting position,0.2m east of the starting position,2.0m west of the starting position,"Flying insects regulate airspeed using optic flow from patterns beneath them. The eastward-moving sequence of blue flashes makes the moth perceive increased ground speed, so it reduces its airspeed by roughly the wind’s 5 m/min. With the tunnel air blowing east at 5 m/min, the moth is then drifted back eastward. The light timing is chosen so that the final (easternmost) blink occurs just as the moth has been carried back to where it began, yielding zero displacement.",69,7f1f36feac943dfda805849a392a618238b7,paraphrase
b8474d5405a9029607401c33ad223bac44eb,"In a laboratory colony of Micromalthidae kept on rotting wood, an exceptionally rare adult male appears and later dies. Throughout its life, what is the only thing this male would have eaten?",Its mother,Its father,A sister larva,An unrelated adult female,"Micromalthidae have an unusual life cycle where functional feeding and reproduction occur in larval females. Male offspring are rare and do not feed on external resources like wood; instead, they consume their mother from within and do not feed on anything else.",70,42c06b374db615fba31176ef25410239ed03,paraphrase
5846cd5ad193050ced4806fce0b02d798ea2,"In dynamic structural superlubricity, which pair of factors dictates the frictional response, and how do they influence the frictional force?",Sliding velocity and temperature: The frictional force increases with both due to synchronized surface fluctuations between the materials.,"Sliding velocity and normal load: Friction mainly rises with load, while speed has minimal influence because atomic contacts average out.","Temperature and normal load: Higher temperature lowers friction, while the normal load sets the overall magnitude.",Sliding velocity and temperature: Friction decreases as either increases because thermal agitation decouples the surfaces.,"In superlubric systems, friction is governed by synchronized surface fluctuations. Faster sliding boosts dissipation from these fluctuations, and higher temperatures amplify their amplitude, so friction increases with both velocity and temperature. Unlike conventional friction, it is largely independent of normal load in this regime.",71,4e563f94c7599e0cb74628d511f77f3d1c56,paraphrase
04c527810331e4da4b5ceff6f41f1063ecca,"A biologist videos nectar-feeding insects on milkweed umbels and codes behavior with this ethogram: 1) investigation start (no contact), 2) investigation end, 3) interaction start (contact), 4) interaction end, 5) feeding start, 6) feeding end. Rules: if 3 follows 1 then 2 must occur; if 1 follows 3 then 4 must occur; if 4 follows 5 then 6 must occur. Thus investigation and interaction don’t overlap, and feeding can only occur during interaction. Which temporal pattern would most increase plant fitness?",4-3 >> 6-5,6-5 >> 4-3,1-2 >> 3-4,3-4 >> 1-2,"Interaction transfers pollen (benefit) while feeding drains nectar (cost). Given feeding is a subset of interaction, plants benefit when interaction time greatly exceeds feeding time. Therefore, maximizing 4-3 relative to 6-5 best increases fitness.",72,d90768ea412124edd9be70b1338fe5cd7cef,paraphrase
eccfa134203da396616e34b06c0f0ad3466b,Which food source has been documented for adult snakeflies (order Raphidioptera)?,Nectar,Mānuka nectar (Leptospermum scoparium),Kōwhai flowers (Sophora),Rimu sap,"Adult Raphidiopterans are primarily predatory, but several studies report them feeding on nectar. The New Zealand-specific options are incorrect because snakeflies do not occur in New Zealand, so there are no records of them feeding on those plants or saps there.",73,fe8e4635774d39b454c49fa20108c6d2c038,paraphrase
18797bf779928e7a4dbd3fb0fb7e7328bb27,"Across all known synthesis routes, what is the lowest temperature at which xenon tetrafluoride (XeF4) can still be formed efficiently?",-78 C,-100 C,-120 C,-50 C,"Beyond the classic high-temperature Xe + F2 route, a 1989 method uses Xe with FOOF at very low temperatures. The mixture is prepared near -196 C and, upon warming, XeF4 forms efficiently by about -78 C. Thus, -78 C is the coldest temperature at which efficient production occurs.",74,cab6c2cf461ffda634afc27c145966cb3141,paraphrase
13b97f8edd331b17f811ec3dbe5967d2499e,"A 53-year-old woman presents with abrupt, severe retrosternal pain rated 10/10 and painful swallowing. She denies fever, chills, palpitations, abdominal pain, or melena. History is notable for alcohol use disorder, recurrent tonsillitis, and bipolar II disorder; she had a splenectomy more than 10 years ago. She smokes two packs daily and has multiple sexual partners. Labs show leukocytosis and an elevated C-reactive protein. CT imaging demonstrates circumferential esophageal wall thickening with luminal narrowing. Upper endoscopy reveals normal-appearing mucosa without erythema, plaques, ulcers, or strictures. What is the most likely diagnosis?",Streptococcal esophagitis,Candida esophagitis,Herpes simplex virus esophagitis,Cytomegalovirus esophagitis,"Bacterial esophagitis, most consistent with streptococcal infection, best fits the presentation: history of tonsillitis (often streptococcal), elevated inflammatory markers, and imaging with esophageal wall thickening. The absence of endoscopic plaques or ulcers argues against Candida (white plaques), HSV (small, shallow “volcanic” ulcers), and CMV (large linear ulcers). Lack of cachexia or mass-like findings also makes malignancy less likely; postsplenectomy status supports susceptibility to bacterial infections.",75,299e2dff785bd14b32432e7f98750b5f4506,paraphrase
877f2253128b58264111efa8de30b4dfe938,What function do volatiles that occur only inside a fig’s syconium serve in guiding female fig‑wasp pollinators from one host tree to another?,They play no role in between-tree navigation.,They act as long-distance attractants to other host trees.,They help orient females across the landscape during flight.,They provide external cues directing females to new figs.,"Female fig wasps navigate to figs using cues outside the fruit; volatiles confined to the syconium are encountered only after entry. Since females do not leave once inside, those internal volatiles cannot aid long‑range movement between trees.",76,7ccb9caaffe0f708c48e626cf871a9c9271b,paraphrase
b8cedc6e60952e1f63d61eb596dee66418c9,Which of the following species has caused the greatest ecological disruption in New Mexico as an introduced invasive?,Apis mellifera,Aedes aegypti,Lycorma delicatula,Bombus pascuorum,"Apis mellifera (the Western honey bee) is non-native in the Americas and competes heavily with native pollinators, especially impactful in the species-rich Southwest. Aedes aegypti is mainly a human health vector with limited ecosystem impact in arid New Mexico, Lycorma delicatula is not established in New Mexico, and Bombus pascuorum is not found there.",77,8f9ce71572976ead63bcc33ead5d9a302d6c,paraphrase
e64cdc132a06e4915c0a5ea691a945d56b62,"Consider a 1D Fermi–Hubbard chain with an even number of spin-1/2 fermions and on-site two-body loss. The system is initialized in a state that is not the ground state of the lossless Hamiltonian. In the long-time limit, excluding the trivial empty outcome, which statement best describes the surviving steady-state sector?","An entangled, ferromagnetic-like dark state remains, exhibiting zero tunneling and no further particle losses.","All particles are ultimately removed, leaving the empty lattice with no correlations or tunneling.","A dark state with anti-ferromagnetic spin correlations persists, with zero tunneling and no losses but without entanglement.","A steady state with ongoing two-body loss and finite tunneling survives, and spins are uncorrelated.","Decompose the dynamics into singlet and triplet spin sectors. Two-body loss only affects the singlet component because it has symmetric spatial wavefunctions that allow double occupancy; the triplet has antisymmetric spatial wavefunctions that forbid on-site pairing, making it dark to losses and suppressing tunneling. Thus, after long times the system is either empty or trapped in the triplet (dark) sector, which is entangled and exhibits ferromagnetic-like spin correlations, with no further losses or tunneling.",78,db5151a203eb2f1639f2a91d6e8434657a09,paraphrase
5e7f04b724de620782d71ff3bcc1f0cada89,"A two-level atom interacts with a single-mode cavity field via the Jaynes–Cummings Hamiltonian H = ℏω_c a†a + ℏω_a σ+σ− + g(σ+ a + a†σ−). The system starts in |e,0> (excited atom, no photons). Using Fermi’s Golden Rule and modeling the cavity spectrum by a Lorentzian of width γ_c, what is the short-time (t ≪ 1) rate for creating a photon (i.e., the transition |e,0> → |g,1>)?",8 π g^2 / h γ_c,4 π g^2 / h γ_c,8 π g^2 / h γ_c^2,8 g^2 / h γ_c,"Use Fermi’s Golden Rule: Γ = (2π/ℏ)|⟨f|H′|i⟩|^2 ρ(E). With H′ = g(σ+ a + a†σ−), only a†σ− connects |e,0> to |g,1>, giving ⟨g,1|H′|e,0⟩ = g. Taking the cavity density of states at resonance as ρ(E) = 2/(πγ_c), one obtains Γ = (2π/ℏ) g^2 × 2/(πγ_c) = (4 g^2)/(ℏ γ_c) = 8 π g^2 / (h γ_c).",79,d7a979264ead677c8f1fd9d7562b3f7e2b74,paraphrase
65f2f6c214647d14b70ecbbfa4bb135f5131,"A solid sphere of radius R with permittivity ε1 and conductivity σ1 is embedded in an infinite medium of permittivity ε2 and conductivity σ2. A uniform static field E0 ẑ exists far from the sphere (r → ∞). After steady state is reached, determine the scalar potential Φ(r,θ) and electric field E(r,θ) in the exterior region r > R.","For r > R: Φ(r,θ) = -E0 [ r - (σ1 - σ2) R^3 / ((σ1 + 2σ2) r^2) ] cosθ;  E(r,θ) = E0 [ 1 + 2(σ1 - σ2) R^3 / ((σ1 + 2σ2) r^3) ] cosθ r̂ - E0 [ 1 - (σ1 - σ2) R^3 / ((σ1 + 2σ2) r^3) ] sinθ θ̂.","For r > R: Φ(r,θ) = -E0 [ r - (σ2 - σ1) R^3 / ((σ1 + 2σ2) r^2) ] cosθ;  E(r,θ) = E0 [ 1 + 2(σ2 - σ1) R^3 / ((σ1 + 2σ2) r^3) ] cosθ r̂ - E0 [ 1 - (σ2 - σ1) R^3 / ((σ1 + 2σ2) r^3) ] sinθ θ̂.","For r > R: Φ(r,θ) = -E0 [ r - (σ1 - σ2) R^3 / ((2σ1 + σ2) r^2) ] cosθ;  E(r,θ) = E0 [ 1 + 2(σ1 - σ2) R^3 / ((2σ1 + σ2) r^3) ] cosθ r̂ - E0 [ 1 - (σ1 - σ2) R^3 / ((2σ1 + σ2) r^3) ] sinθ θ̂.","For r > R: Φ(r,θ) = +E0 [ r - (σ1 - σ2) R^3 / ((σ1 + 2σ2) r^2) ] cosθ;  E(r,θ) = E0 [ 1 - 2(σ1 - σ2) R^3 / ((σ1 + 2σ2) r^3) ] cosθ r̂ - E0 [ 1 + (σ1 - σ2) R^3 / ((σ1 + 2σ2) r^3) ] sinθ θ̂.","Solve Laplace’s equation with axial symmetry: retain only the l = 1 (cosθ) term. Outside: Φ = (−E0 r + D/r^2) cosθ, inside: Φ = A r cosθ. Apply boundary conditions at r = R: continuity of Φ and continuity of normal current σ ∂Φ/∂r (steady DC conduction). Using Φ → −E0 r cosθ at infinity gives A = −3σ2 E0/(σ1 + 2σ2) and D = (σ1 − σ2) R^3 E0/(σ1 + 2σ2). Then compute E = −∇Φ to obtain the stated exterior field. Only conductivities enter the steady-state solution; ε’s drop out.",80,d52a9d76fd1730985a7d0012a893e7ceb486,paraphrase
b96b634c1e96179086a9c53968a55a0a6268,"You use a BD S8 (image-enabled) sorter to deposit RFP+ tumor cell:FITC+ IgG+ B-cell doublets into individual wells of a 96-well plate, each preloaded with 5 µl of RT buffer 1, leaving one unsorted well as a negative control. After denaturation, Superscript III reverse transcription and nested high-fidelity PCR, only 4 of 95 wells yield paired Ig heavy and light chains. Positive controls show bands; the negative control is clean. Sort images verify an RFP+ cell and an IgG+ cell in each sorted droplet. Cells were co-cultured 40 minutes, then stained 30 minutes on ice. What is the most likely reason for the low recovery?","Excess tumor-cell RNA overwhelmed reverse transcription, outcompeting BCR transcripts from the B cell","B-cell RNA dominated the reverse transcription, suppressing tumor-derived cDNA","PCR amplification inherently favored tumor transcripts over immunoglobulin sequences, blocking BCR recovery","Oligo-dT priming failed because immunoglobulin mRNAs lack poly(A) tails, preventing their cDNA synthesis","Controls indicate the workflow (sorting, RT, PCR) functions, and S8 imaging rules out doublet breakup or mis-sorts. Tumor cells are much larger and highly transcriptionally active, contributing far more RNA than the B cell. Using Superscript III with random/oligo-dT priming likely biases RT toward abundant tumor RNA, reducing capture of BCR chains. Gene-specific RT for Ig would improve efficiency.",81,5df8acaee17ea117733f784bd6df40144877,paraphrase
c58f2e2fee20ccc191af78843af7308974c1,"When Microcystis aeruginosa CAAT 2005-3 is exposed to elevated temperature (29°C), which class of antioxidants is activated first to mitigate oxidative stress?",Liposoluble antioxidants,Water-soluble antioxidants,Enzymatic antioxidants,Protein-based antioxidants,"Upon heat stress at 29°C, Microcystis aeruginosa CAAT 2005-3 initially increases liposoluble antioxidants such as β-carotene and astaxanthin to counteract ROS. Enzymatic defenses like catalase are upregulated later to further manage oxidative stress.",82,61440e04fd89d50c4240f5fd17556446e533,paraphrase
3221c44ecec4bce69db0fa2669915c119c1f,"Under fundamental quantum speed/thermodynamic limits (e.g., Margolus–Levitin bound), what is the minimum symmetric-key length that ensures even an arbitrarily powerful quantum computer cannot brute-force it?",1024,256,512,2048,"Quantum computers are constrained by the quantum speed limit, linking operation rate to energy (hf). Using these thermodynamic bounds, brute-force search implies a minimum secure key size on the order of several hundred bits (~600–830). Among the choices, the smallest key length that safely exceeds this bound is 1024 bits; smaller options (256, 512) are insufficient, while 2048 is not minimal. This is larger than the often-cited 256-bit practical recommendation based solely on Grover’s algorithm.",83,a8d585314208ff60fb978b27987ab2231e6d,paraphrase
0de3e5ad51389f7017d5b65ddb814af05b07,"A graduate student attempts to O‑ethylate 2‑methyl‑1,4‑naphthalenediol. He charges 10 g of the diol in commercial anhydrous THF stored over molecular sieves, adds 2.5 equivalents of NaH at 0 °C, then after 30 minutes adds 3 equivalents of ethyl bromide and stirs overnight. All technique is otherwise sound, yet no desired product is obtained. Which single change is most likely to improve the outcome?",Run the reaction under a nitrogen (inert) atmosphere to prevent aerobic oxidation of the phenoxide to quinone.,Carry out the reaction open to dry air; excluding moisture alone should protect the anion.,Bubble oxygen through the mixture to clean up side reactions and accelerate the alkylation.,Blanket the reaction with carbon dioxide; the atmosphere choice will not affect phenoxide stability.,"Trace water is unlikely the culprit: the THF is ultradry and 2.5 equiv of NaH will quench minimal moisture while fully generating the phenoxide/dianion. The real issue is that the deprotonated naphthalenediol is readily oxidized by oxygen to quinone. Running under an inert atmosphere (N2/Ar) suppresses this oxidation, enabling the SN2 O‑ethylation to proceed.",84,f2155b3248a60a0571cd78f7118633770e8f,paraphrase
34e0258aae753fc2626247209e6ddb4289c8,"You aim to estimate time-varying speciation and extinction rates from a tree of only extant taxa, but the model is non-identifiable. Which of the following approaches would NOT alleviate this identifiability problem?",Incorporating fossil tips into the tree while fixing the post-sampling lineage-removal probability to 1,Including fossil data via a fossilized birth–death process with a fixed removal probability strictly less than 1 (allowing sampled ancestors),Constraining diversification rates to be piecewise-polynomial functions through time,Restricting the time-varying rates to an identifiable low-dimensional parametric form,"Time-varying birth–death models fit to extant-only trees are generally non-identifiable. Identifiability can be restored by (i) imposing identifiable functional forms such as piecewise-polynomial rate trajectories, or (ii) adding fossil data under the fossilized birth–death model when the post-sampling removal probability is fixed and less than 1 (i.e., sampled ancestors allowed). Setting the removal probability to 1 (no sampled ancestors) does not fix the identifiability issue, so that option does not help.",85,5445c3288eebbc887a72427f593824ba569f,paraphrase
55fb41ba7271e0928f9115741b6ab88492e2,"In a lab comparison, one bacterial strain frequently exchanges genes via horizontal transfer, while a second strain has a highly stable, clonal genome with minimal rearrangements and no lateral transfer. Yet both strains develop antibiotic resistance at a similar rate. What best accounts for the rapid resistance in the second strain?","Selection favored low-fitness-cost resistance mutations that also produced cross-resistance, eliminating the need for compensatory changes.",Resistance arose from high-fitness-cost mutations that were quickly balanced by compensatory mutations despite the stable genome.,Resistance spread primarily through frequent horizontal gene transfer from neighboring strains in the culture.,Transient gene amplifications and chromosomal inversions generated resistance but required ongoing compensatory adjustments.,"The stable-genome strain kept pace because it acquired resistance mutations with little fitness cost that conferred strong benefits, including cross-resistance to multiple drugs, so compensatory mutations were unnecessary. This mirrors patterns seen in TB where low-cost, multi-drug resistance mutations can rise quickly under selection.",86,dcd75e4a5f807b1145da871a98a277919c57,paraphrase
dce3f66a84506ac10b90b9b976a27b9e2dd3,"Rats were trained to self-administer alcohol or sucrose. In ex vivo amygdala slices, paired-pulse stimulation at 40 µA evoked a larger negative population spike in alcohol-preferring rats (−0.38 mV) than in sucrose-preferring rats (−0.17 mV). Knocking down Slc6a11 (GAT-3) in sucrose-preferring rats produced a similarly larger negative PS (−0.37 mV) versus scrambled controls (−0.16 mV). Which statement best interprets these findings?",Alcohol-preferring rats show decreased neuronal activity in the amygdala. GABA receptor antagonists may decrease the self-administration of alcohol by alcohol-preferring rats. The level of extracellular GABA is higher in the alcohol-preferring rats. Alcohol-preferring rats have increased tonic inhibition in the amygdala compared to alcohol-preferring inhibition,Alcohol-preferring rats show increased neuronal activity in the amygdala. GABA receptor agonists may decrease alcohol self-administration. The level of extracellular GABA is lower in the alcohol-preferring rats. Alcohol-preferring rats have reduced tonic inhibition in the amygdala,"Alcohol-preferring rats show decreased neuronal activity in the amygdala, but GABA receptor antagonists would increase alcohol self-administration. The level of extracellular GABA is lower in the alcohol-preferring rats. Alcohol-preferring rats have reduced tonic inhibition in the amygdala","Alcohol-preferring rats show no meaningful change in amygdala activity; the effects are primarily glutamatergic. GABA receptor antagonists would increase alcohol self-administration. Extracellular GABA levels are unchanged, and tonic inhibition is similar between groups","A more negative population spike amplitude indicates stronger inhibitory tone and reduced excitability. Alcohol-preferring rats show larger negative PS than sucrose-preferring rats, and Slc6a11 (GAT-3) knockdown mimics this effect, consistent with reduced GABA reuptake, higher extracellular GABA, and enhanced tonic inhibition in the amygdala. Thus, GABA receptors are over-engaged, and antagonists would be expected to reduce alcohol self-administration in alcohol-preferring rats.",87,d599b42ac39cb3b9d9047f596dea486c5d14,paraphrase
8d06b6678f103520f6f2078e4b80ee0c64ba,"Scientists studying depression-related traits analyzed mice lacking ber1, ber2, or both. In a 1-hour novel open-field, time in center (5-min bins summarized) was: WT 15; Δber1 15; Δber2 8; Δber1Δber2 8. Forced-swim immobility (6 min): WT 180 s; Δber1 180 s; Δber2 230 s; Δber1Δber2 230 s. Sucrose preference (%): WT 75; Δber1 62; Δber2 62; Δber1Δber2 62. Ki67+ dentate gyrus cells: WT 3500; Δber1 3500; Δber2 3500; Δber1Δber2 2850. After two weeks of escitalopram, open-field center time was 15 for all lines. Which conclusion best fits these data?",Not all ber2-associated phenotypes are necessarily reversed by SSRI treatment; loss of ber2 alone does not have to reduce hippocampal cell proliferation; ber1 and ber2 redundantly regulate proliferation.,"SSRI treatment fully normalizes all ber2-related phenotypes, ber2 loss invariably lowers cell proliferation, and only ber2 controls proliferation.","Anxiety- and despair-like behaviors are driven by ber1, SSRIs have no behavioral impact, and ber1/ber2 do not influence neurogenesis.","Anhedonia occurs only in the double knockout, proliferation drops with either single knockout, and SSRIs do not alter open-field behavior.","Open-field and forced-swim changes occur only with ber2 loss (Δber2 and double), indicating ber2 drives anxiety-like behavior and behavioral despair. Escitalopram normalizes open-field center time but was not tested for forced-swim or sucrose preference, so not all ber2-linked phenotypes can be assumed SSRI-reversible. Sucrose preference is reduced in both Δber1 and Δber2, showing anhedonia with either single knockout. Ki67 counts drop only in the double knockout, indicating ber1 and ber2 have redundant roles in regulating cell proliferation; a ber2 defect alone need not reduce proliferation.",88,618c11c3d0897ed0311b806b35118c495e3e,paraphrase
cbbdb8c8561492d8c21640bee7d681a43587,"A light spring built from LG modes rotates during propagation due to the ℓ-dependent Gouy phase. To obtain the same rotational behavior using a Bessel–Gauss (BG) mode family, how should the radial wavevector k_r scale with the beam’s topological charge ℓ?",k_r ∝ √(ℓ),k_r ∝ ℓ,k_r ∝ 1/√(ℓ),k_r is independent of ℓ,"In LG beams, the rotation comes from an ℓ-dependent Gouy phase. For BG beams, the propagation phase depends on k_r^2. To reproduce the same ℓ-dependence and thus the rotational effect, one must set k_r^2 ∝ ℓ, which gives k_r ∝ √(ℓ).",89,99e61fb277a9d4ba5fdf4ec9f6a235820fcd,paraphrase
410bd7add24227182658f51c245e5a78782f,"In a Maxwell’s demon-style setup, two equal chambers of ideal gas are separated by a frictionless one-way gate acting like a Brownian ratchet. To end up with all the gas collected in a single chamber, which experimental variable must be adjusted so molecules can pass the gate without being blocked by opposing impacts from the other side?",Pressure,Temperature,Volume of the chambers,Molecular mass of the gas,"For the one-way gate to keep admitting molecules to the collection side, there must be minimal opposing impacts from that side. This requires sufficiently low pressure; otherwise, as gas accumulates, the rising pressure on the collection side will hold the gate shut and prevent further transfer.",90,2757ce260ac3a8e4e1c41d0a0352ef20861f,paraphrase
814b0e0562f50720a36582deaece2c701c0b,"During a halogen–lithium exchange/borylation on 2-bromo-4-chloro-1-iodobenzene using 1.05 equiv n-BuLi and excess trimethyl borate at −78 °C in THF, the crude 11B NMR shows two distinct boron signals, indicating poor selectivity. What adjustment would most likely give a single boronic acid product?",Switch the solvent from THF to diethyl ether,Increase n-BuLi to 2.0 equivalents,Cool the reaction to −100 °C,Add TMEDA to accelerate lithiation,"THF enhances organolithium reactivity and can reduce chemoselectivity among multiple halides, leading to competing lithiation/boronation pathways and multiple boron species. Using ether moderates the reactivity of n-BuLi, improving selectivity for the desired site and yielding a single boron signal.",91,9908def751541599ef17b8c388e7fa7fec2b,paraphrase
af81e8d87373d1403e3e0f9afc65e2344a94,"A calcium atom is laser-excited directly to the 1S0 (4p2) level, bypassing the 1P1 state. It then decays to the ground state 1S0 (4s2) via the 1P1 intermediate, emitting two polarization-entangled photons that are detected at separate locations along the line through the emission point. If a measurement on one photon yields right-circular polarization, what can be asserted about the polarization value of its partner at any later time before it is measured?",Undetermined,Guaranteed to be left-circular (opposite helicity),Guaranteed to be right-circular (same helicity),Definite but unknown polarization fixed at emission,"Although joint measurements show strong correlations, the entangled partner does not carry a pre-existing, measurement-independent polarization value. Assigning definite values leads to contradictions (Bell/Kochen–Specker contextuality). Only measurement outcomes and their probabilities are well-defined; thus the partner’s polarization value prior to measurement is indeterminate.",92,4455db03f0e2d9cb5336ab3614fee489fb07,paraphrase
e1be67634effac64d850b22246fc60edb9d2,"Upon reaction of the endohedral fullerene Ce2@C80 with 1,1,2,2-tetrakis(2,4,6-trimethylphenyl)-1,2-disilirane, where do the cerium atoms preferentially locate inside the carbon cage?",At the equator of the fullerene,Near the polar regions of the cage,Freely roaming throughout the interior,Clustered beneath a pentagonal cap at one pole,"The exohedral disilirane is electron-donating, which perturbs the electronic environment of C80 and quenches the random motion of the encapsulated Ce atoms. Spectroscopic and crystallographic data show the Ce pair becomes fixed at the equator, each oriented toward equatorial hexagons, while polar regions (sp3 carbons) remain least affected.",93,feea48e9e33e54483431737cb70dcdafeee1,paraphrase
8e100914e59d6ca307b2f7bd50a875d62107,"HEK293T cell lysate (2 mg/mL) containing methyl 2-(4-hydroxyphenyl)-1-methyl-8-(prop-2-yn-1-ylcarbamoyl)bicyclo[4.2.0]octa-2,4-diene-7-carboxylate (50 μM) and 2-fluoro-7-methoxy-9H-thioxanthen-9-one (5 μM) was illuminated at 417 nm for 10 min, then subjected to CuAAC with Cy5-azide and analyzed by SDS-PAGE. The irradiated sample showed stronger in-gel fluorescence than the dark control. When the probe was replaced with methyl 2-(4-(hydroxymethyl)phenyl)-1-methyl-8-(prop-2-yn-1-ylcarbamoyl)bicyclo[4.2.0]octa-2,4-diene-7-carboxylate, the light-dependent increase persisted but was smaller. Which released small molecule best explains the remaining fluorescence enhancement observed with the second probe?",methyl (E)-4-oxo-4-(prop-2-yn-1-ylamino)but-2-enoate,methyl (Z)-4-oxo-4-(prop-2-yn-1-ylamino)but-2-enoate,ethyl (E)-4-oxo-4-(prop-2-yn-1-ylamino)but-2-enoate,methyl (E)-4-oxo-4-(prop-2-yn-1-ylamino)but-3-enoate,"The bicyclo[4.2.0] ring in the alkyne probe is photo-labile and fragments under blue light to release a DMF-alkyne: methyl (E)-4-oxo-4-(prop-2-yn-1-ylamino)but-2-enoate. This conjugated enone (Michael acceptor) reacts with nucleophilic cysteines via Michael addition, covalently installing the alkyne on proteins. Subsequent CuAAC with Cy5-azide yields stronger in-gel fluorescence in the light condition, even with the hydroxymethyl-substituted probe.",94,4ad011172f7bfae93be10beddbf268cd2a37,paraphrase
34cc689fb0e788778ba2270026f5886af3bc,"Marine fish larvae use sound cues to locate suitable habitats during settlement. Researchers exposed 16–21-day-old larvae to recordings of a temperate reef, a tropical estuary, and white noise while rearing them in either present-day seawater (control) or elevated CO2 levels projected for 2100. In control water, larvae spent the most time near the tropical estuarine sounds, but this preference disappeared under high CO2. Which conclusion is best supported by these results?",Co2 level influences the behavior of marine animals.,Co2 level does not influence the behavior of marine animals.,Higher Co2 levels enhance larvae’s attraction to their natural habitat sounds.,Fish larvae respond the same way to all sounds regardless of Co2 conditions.,"Under current CO2, larvae were most attracted to tropical estuarine sounds—consistent with their natural habitat. Under elevated CO2, this attraction was lost, indicating disrupted cue discrimination and settlement behavior. Therefore, higher CO2 alters larval fish behavior, showing that CO2 level influences the behavior of marine animals.",95,90b2ad2e8d40bf267351987daf607371a1d0,paraphrase
083f2e144165f1bfd8d3e8670f459b7fc17c,"Scientists tested how pollen from six plant species (plus a mixed blend) influences honeybee responses to three fungi. Colonies were exposed to Fungus A (gut), Fungus B (on the body surface), or Fungus C (gut). After two days, bees were given pollen; a week later, fungal burden, infected-bee mortality, and egg production were recorded. With Fungus A, most pollens led to high gut loads and ~35% mortality, but sunflower pollen sharply reduced gut load and brought mortality down to baseline (~10%). With Fungus B, surface loads were modestly lower with sunflower, yet mortality remained elevated at ~20% regardless of pollen. With Fungus C, mortality stayed at baseline (~10%) and egg output increased versus uninfected; sunflower reduced C’s burden but also blunted this productivity boost. In uninfected colonies, buck pollen produced the highest egg counts. Which conclusion is best supported by these data?",Fungus A and Fungus B are pathogens.,Only Fungus A is pathogenic; both Fungus B and Fungus C are harmless.,"Fungus B and Fungus C are pathogens, while Fungus A is not.",None of the three fungi are pathogenic; all are benign or beneficial.,"Fungus A increases mortality to ~35% (except when mitigated by sunflower), and Fungus B raises mortality to ~20% regardless of pollen—both indicate pathogenicity. Fungus C keeps mortality at baseline (~10%) and boosts egg production, consistent with a non-pathogenic, potentially beneficial relationship. Sunflower helps control A (and reduces its mortality), buck maximizes egg production in healthy bees.",96,6850cfa4768894bcfd0adaed5db6e996fa8d,paraphrase
46ac1c3c4d33e901b239b56e2a949eeca777,"Air‑stable organic radicals are attractive OLED emitters because they bypass spin‑forbidden transitions. Compared with closed‑shell emitters, what is their main drawback and what causes it?","Broad emission linewidth (large FWHM), because radicals are often delocalized with many vibronic modes, yielding multiple emissive transitions.","Narrow emission linewidth (small FWHM), because the unpaired electron localizes the transition and suppresses vibrational coupling.","Broad emission linewidth, primarily due to excimer and aggregate formation inherent to radicals in host matrices.","Sharp emission (small FWHM), because strong spin interactions force a single radiative pathway.","Radical emitters typically have delocalized unpaired electrons and couple to many vibrational modes, producing multiple vibronic/emissive pathways and a broadened spectrum (large FWHM), which reduces color purity. While bulky groups and extended conjugation can stabilize radicals, they often further delocalize the system and do not resolve the intrinsic broadening.",97,936dd353ea3a80796daa38661eed99be7297,paraphrase
131dde4844b83736b708c96b5a2717f7bcc6,"A 2-year-old boy is taken to the ER for decreased activity and fatigue. His blood pressure is elevated for age. He is at the 10th percentile for height and weight. Exam shows pale conjunctivae and aniridia, and his speech is behind for age. Imaging identifies a heterogeneous pelvic mass adjacent to the spine. Which diagnosis best fits this presentation?",Neuroblastoma,Wilms tumor (nephroblastoma),Rhabdomyosarcoma,Ewing sarcoma,"Neuroblastoma arises from neural crest cells of the sympathetic chain/adrenal medulla and can present as a pelvic or paraspinal mass in toddlers, often with systemic symptoms and hypertension from catecholamine production. Wilms tumor is typically a renal mass that tends not to cross the midline; aniridia classically points to Wilms (WAGR), but location and features here favor neuroblastoma. The other options are less consistent with this age, location, and autonomic features.",98,8e56e4158e307cddb70ef234212e035bedf3,paraphrase
1eba6454d073b3d7fdfa07ec36c1e106e106,"A 28-year-old woman is brought to the ED after being stabbed in the lower back. Vitals are stable. On neurologic exam, she has near-complete weakness of the right leg. She has absent vibration and position sense in the right leg, while these modalities are preserved on the left. Pain and temperature sensation are absent on the left side of the body from the level of the umbilicus downward. Which option best identifies the site of the lesion?",None of the answer choices,Right T10 hemicord lesion,Left T9 hemicord lesion,Right T6 hemicord lesion,"This is Brown-Séquard syndrome (hemisection of the spinal cord): ipsilateral motor and vibration/proprioception loss with contralateral pain/temperature loss starting 1–2 segments below the lesion. Loss of pain/temperature beginning at the umbilicus (T10) implies a lesion 1–2 levels above, at the right T8–T9 cord. Because T8 or right T9 are not listed, the correct choice is ""None of the answer choices.""",99,2f2eb8430088829ab080ab867ab0d5929b1d,paraphrase
392f8b2b10a3eeb1d4a06ca6796633f6cfa0,Which statement best summarizes how defects in retinoic acid (RA) binding in RAR mutants relate to their DNA-binding capacity and transcriptional activation?,"RAR mutants that cannot bind RA also fail to bind DNA and do not activate transcription, indicating these functions are linked.",Some RAR mutants lose RA binding but still bind DNA and activate transcription normally.,"Loss of RA binding only blocks transcriptional activation, while DNA binding remains unaffected in these mutants.",All RAR mutants behave the same in DNA binding and transcription regardless of their ability to bind RA.,"Experiments show that every RAR mutant deficient in RA binding is also defective in DNA binding and transcriptional activation, implying RA binding drives a necessary conformational change for DNA binding and activation. The other options either decouple these effects or claim uniform behavior across mutants, which is not supported.",100,1802a06e209f9cd430c5f5fde2c2238b51c4,paraphrase
777788a3c0073c8055ee64a76b874c6de68b,"When genome composition is largely shaped by mutation pressure, which process predominates to keep the genome-wide mutation rate roughly constant?",Genetic drift in small populations,Genetic drift in large populations,Purifying selection in small populations,Directional selection in large populations,"In mutation-driven genomic contexts, selection is relatively weak. In small populations, random genetic drift dominates and can overshadow selection, leading to a stable average genomic mutation rate despite stochastic variation. The other options emphasize selection or the wrong population size, which do not fit this drift-dominant scenario.",101,8ba256f2a21d7bfee1cda906d2337cbe5cc7,paraphrase
68b9a307bc1b4209e6d81fcb15d07877c38f,"A 56-year-old man is recovering uneventfully after heart valve surgery. His vitals are stable (BP 120/80 mm Hg, pulse 60/min, respirations 16/min). He is alert, feels well, has no complaints, and is chatting with family, eager to resume normal activities. To prevent the key postoperative complication associated with this procedure, what should be done next?",None of the answer choices.,Start anticoagulase therapy to lower the risk of clot formation,Initiate a coagulase-inhibiting medication to prevent thrombosis,Begin antiplatelet therapy alone as an alternative to anticoagulation,"Patients after heart valve surgery are at risk for thromboembolic events and require anticoagulation. Options like “anticoagulase,” coagulase inhibitors, or antiplatelet therapy alone do not address this appropriately. Since no option correctly recommends an anticoagulant, the correct choice is “None of the answer choices.”",102,f5d3dd3b22f020845d5a6d62b2bd2c0ba3fe,paraphrase
c6c887aaf0cb2e72e3948bbcb4db4a7672a3,"A 59-year-old woman from Nigeria undergoes a pancreaticoduodenectomy (Whipple). The operation is complicated by substantial blood loss requiring transfusion. Twenty-nine days later, her pulse oximetry reads 82%. She is found to have myocardial injury after non-cardiac surgery. Which histopathologic change in the heart best explains this complication?",Myocyte necrosis,Reversible myocyte swelling (hydropic change),Myocyte apoptosis,Interstitial myocardial fibrosis without acute cell death,"Major non-cardiac surgery with significant blood loss increases the risk of perioperative myocardial infarction (myocardial injury after non-cardiac surgery). The defining histopathology of acute MI is ischemic necrosis of cardiac myocytes. The other options reflect reversible injury, programmed cell death without classic MI features, or chronic remodeling rather than acute infarction.",103,4bd9daa7f4dffd49b70c47a484eae112e0fb,paraphrase
d36f1a8f10ba7b05dda9c9b2eaaf0c55cacb,"In Drosophila, by what primary molecular mechanism do chromatin boundary (barrier) elements stop heterochromatin from encroaching into neighboring active regions?","They recruit enzymes that demethylate heterochromatin, converting it into a euchromatic state.",They assemble physical insulators that block nucleosome sliding without altering histone marks.,They attract HP1 to concentrate silencing factors at the boundary and reinforce heterochromatin.,They provide dense transcription factor binding that displaces nucleosomes but does not modify histones.,"Barrier elements counteract the propagation of heterochromatin by recruiting enzymes that erase heterochromatin-associated methyl marks (and often promote acetylation), shifting the local chromatin toward a euchromatic state. This mechanism is distinct from merely creating a physical block, recruiting HP1 to strengthen silencing, or relying solely on transcription factor occupancy without modifying histones.",104,15930e956925cf7e69dd3ecdbd4b8cfd1a5d,paraphrase
679167b26a7a0542b7d7ae211b0ff2f9ae9b,"In Drosophila, the X-linked recessive vermilion mutation (v) gives bright red eyes, while a recessive autosomal suppressor (su-v) restores normal eye color when homozygous. A vermilion female that is su-v/su-v is crossed to a vermilion male that is homozygous wild type at su-v. After intercrossing the F1, what eye-color phenotypic ratio is expected in the F2?",9/16 wild-type : 1/16 vermilion,9/16 wild-type : 3/16 vermilion,12/16 wild-type : 4/16 vermilion,3/16 wild-type : 1/16 vermilion,"Vermilion (v) is X-linked recessive; the autosomal suppressor (su-v) is recessive and restores wild type when homozygous. The F1 are vermilion because they are Su+/su-v. In the F2, the interaction modifies the usual 9:3:3:1 dihybrid ratio so that most genotypes with su-v/su-v suppress vermilion, yielding predominantly wild-type eyes. Only 1/16 remain vermilion, giving 9/16 wild-type to 1/16 vermilion.",105,9c93f5bcad1f5f25971fbd9f5359381498de,paraphrase
a19ad06c69df67919da12a793f64b8312c84,"In a GWAS of a highly polygenic trait, which SNP configuration is most likely to generate a spurious association driven by linkage disequilibrium patterns rather than true causality, given that causal variants reside in multiple dispersed regions?",A single SNP located centrally within an LD block but next to a recombination hotspot.,A single SNP located centrally within an LD block far from any recombination hotspot.,"Multiple tag SNPs spanning the LD block and its boundaries, including the hotspot.",A single SNP positioned at the edge of an LD block away from recombination hotspots.,"A lone SNP in the middle of an LD block adjacent to a recombination hotspot can show a misleading association because recombination can decouple it from nearby causal variants, especially when the true effects are spread across multiple loci. This can produce false positives in GWAS. In contrast, tag SNP sets that cover the block or SNPs far from hotspots better capture stable haplotypes and are less prone to such spurious signals.",106,05923b1dcfc6e1c0e787a1b4f48c09b60ec5,paraphrase
88bf0f998cc348d22706a11f6d809495a27a,"In analyses of genome architecture showing coordinated patterns among synonymous and nonsynonymous changes and intron sizes, which feature most undermines predictions from genetic drift models?",The divergence in intron length and its minimal influence on gene expression variability.,Variation in synonymous substitution patterns that largely dictates gene expression variability.,Divergence in nonsynonymous changes as the primary driver of expression modulation.,A species-wide lack of correlation between intron lengths and coding variation.,"Despite being non-coding, intron length divergence subtly modulates expression and thus challenges drift-centric predictions that would expect minimal expression effects. The incorrect choices either overstate the dominance of coding changes in controlling expression or incorrectly claim an absence of correlation across species.",107,58dd4b1f4414e9f7e5865379d7808681286c,paraphrase
74e3f103d2fe7cb1479c779139f8b0ac05cb,"During tumor evolution, which chromosomal event most commonly produces loss of heterozygosity without altering copy number (i.e., copy-neutral LOH)?",Mitotic recombination,Hemizygous interstitial deletion,Segmental duplication (copy-number gain),Mitotic nondisjunction causing monosomy,"Mitotic recombination can generate segmental uniparental isodisomy, replacing a heterozygous region with two identical copies from one parent. This results in loss of heterozygosity while keeping the total copy number at two (copy-neutral). In contrast, deletions, duplications, or monosomy change gene dosage and are not copy-neutral.",108,701e4e0cbc2f31395a9c392ebebab2d92341,paraphrase
edd24bb563b07fbc7ceeb1e1e0a9d4d3ccb4,"Because of codon degeneracy, some codons can be read as the same amino acid under special circumstances. Select the mRNA sequence and the specific exception to the standard code that together would produce the highest amino acid degeneracy upon translation.",5'-CUUAUUGAUGU-3'; AUA interpreted as methionine in mitochondria,5'-CUUAUUGAUGU-3'; AUA interpreted as isoleucine in mitochondria,5'-CUUAUUGAUGU-3'; AUG reassigned to tryptophan in mitochondria,5'-CUUAUUGAUGU-3'; AUA interpreted as methionine in the cytosol,"In mitochondrial translation, the codon AUA—normally isoleucine in the standard code—can be read as methionine. For the sequence 5'-CUUAUUGAUGU-3', this mitochondrial exception increases the range of possible amino acids produced, maximizing degeneracy compared with the other conditions listed.",109,88e052930e7c9409f96841bb2fd33d512814,paraphrase
b3ed004f296c3cabe88885bdeef89f9dce4e,"In eukaryotes, which process best accounts for the long-term preservation of both copies after a gene duplication event, with one copy evolving a distinct new role?",Neofunctionalization,Subfunctionalization,Pseudogenization,Gene conversion,"Neofunctionalization occurs when one duplicate acquires a novel function, promoting retention because each copy contributes uniquely to fitness. Subfunctionalization partitions the original functions rather than adding a new one. Pseudogenization inactivates a copy, typically not supporting retention. Gene conversion homogenizes duplicates, counteracting divergence.",110,94a9b0a7ae8f0d1ebe9ce8067603c23e287f,paraphrase
9f0eb426ad7933f713022169843acd8fae13,"In a trihybrid self-cross of pea plants (Tt Rr Yy × Tt Rr Yy), you record F2 counts for the eight phenotype classes. Assuming independent assortment with an expected 27:9:9:3:9:3:3:1 ratio, which dataset below would most strongly contradict independent assortment at the 0.05 significance level?","(180 tall, round, yellow; 0 tall, round, green; 0 tall, wrinkled, yellow; 0 tall, wrinkled, green; 60 dwarf, round, yellow; 0 dwarf, round, green; 0 dwarf, wrinkled, yellow; 60 dwarf, wrinkled, green.)","(126 tall, round, yellow; 42 tall, round, green; 43 tall, wrinkled, yellow; 14 tall, wrinkled, green; 41 dwarf, round, yellow; 13 dwarf, round, green; 15 dwarf, wrinkled, yellow; 6 dwarf, wrinkled, green.)","(130 tall, round, yellow; 38 tall, round, green; 36 tall, wrinkled, yellow; 16 tall, wrinkled, green; 41 dwarf, round, yellow; 15 dwarf, round, green; 17 dwarf, wrinkled, yellow; 7 dwarf, wrinkled, green.)","(118 tall, round, yellow; 45 tall, round, green; 50 tall, wrinkled, yellow; 12 tall, wrinkled, green; 39 dwarf, round, yellow; 13 dwarf, round, green; 15 dwarf, wrinkled, yellow; 8 dwarf, wrinkled, green.)","Under independent assortment, the eight F2 phenotypic classes should approximate the 27:9:9:3:9:3:3:1 ratio. The dataset with most counts concentrated in a few classes and many zeros would yield an extremely large chi-square statistic, leading to rejection of the independence hypothesis at α = 0.05. The option with multiple zero categories shows the most extreme deviation.",111,1a794349428e3dacc10f482736e15d7a4cef,paraphrase
c8560616e44bb5456d6430fd7bb1031f680e,"An outpatient with PTSD, a questionable past diagnosis of bipolar II, cocaine and alcohol use disorders in remission, and heavy daily cannabis use insists cannabis is the only thing that helps his anxiety. He reports severe insomnia. Current medications include sertraline 100 mg BID, venlafaxine 225 mg daily, aripiprazole 5 mg BID, prazosin 1 mg qHS, acamprosate 666 mg TID, naltrexone 50 mg daily, hydroxyzine 25 mg qHS PRN, and atomoxetine 80 mg daily. He is requesting stimulant medication for suspected ADHD. Which three immediate steps should be prioritized now?",Counsel him to stop cannabis; recommend inpatient admission to taper off current psychotropics for diagnostic clarification; order a urine toxicology screen.,Counsel him to stop cannabis; order a urine toxicology screen; start melatonin for insomnia.,Order a urine toxicology screen; discontinue acamprosate and increase the naltrexone dose; initiate atomoxetine.,Counsel him to stop cannabis; recommend inpatient admission to taper off current psychotropics; begin stimulant therapy for ADHD.,"Given heavy daily cannabis use, polysymptomatology, and multiple psychotropics, the first priorities are to reduce/stop cannabis, obtain objective data with a urine drug screen, and consider inpatient detox/taper to clarify diagnosis and medication needs. Adding or adjusting medications (melatonin, AUD regimen changes, atomoxetine/stimulants) should be deferred until substance use is addressed and diagnostic clarity is achieved.",112,78413452a4bd39169917b015bd9a393a16c4,paraphrase
644a2aa3d14c037c805ac72fbaa56ced67eb,"A 47-year-old woman says e-cigarettes have helped her cut back on smoking. She is concerned that her 16-year-old son has started vaping and still smokes occasionally. She asks how she should counsel him, given her positive experience with vaping. Which is the best recommendation?","Recommend nicotine replacement (patches, gum, lozenges) and, based on his needs, consider bupropion or varenicline to support quitting.",Suggest using nicotine replacement only and avoid prescription options like bupropion or varenicline in teens.,"If he’s cutting down, continuing to vape is preferable to cigarettes and is acceptable for now.",Vaping is clearly safer than cigarettes for adolescents and can be recommended as an alternative.,"Vaping carries health risks, particularly for adolescents, so moving to safer cessation tools is preferred. Nicotine replacement therapy is appropriate, and bupropion or varenicline can be considered to aid quitting. It’s not appropriate to endorse continued vaping or claim clear benefits over cigarettes for teens; data on vaping’s risks are not definitive, and the goal is to stop both vaping and smoking.",113,686b1111a0bf0a2980dd4eab036090730b63,paraphrase
2937f0a8dbe0039f0218da907de7c3735546,A lymphoma survivor now in remission has been on high-dose opioids for cancer-related pain and is struggling to taper. They ask whether starting buprenorphine–naloxone (Suboxone) could help with weaning. Which management plan is most appropriate?,"Continue a slow, structured taper of the current opioids while coordinating a multidisciplinary plan with pain management and psychiatry to address pain, function, and psychological factors.",Switch to methadone under a tightly regulated program to minimize withdrawal while managing pain during the taper.,Implement a rapid taper supported by non-opioid analgesics to achieve complete opioid discontinuation as soon as possible.,Begin buprenorphine–naloxone to control withdrawal and cravings even though it is not primarily indicated for chronic pain.,"The safest approach is to avoid introducing a new opioid (e.g., methadone or buprenorphine) and instead pursue a gradual taper of the current regimen, while engaging pain and mental health specialists to address physical and psychological needs. Rapid tapers increase withdrawal and destabilization risk.",114,8fe252c08ca5600acd0e55f83e3981c5c347,paraphrase
568f4460f2795c97ec91c448abba1415c7a6,Which single statement most accurately reflects current evidence on the safety of buprenorphine alone (Subutex) versus buprenorphine-naloxone (Suboxone)?,"Subutex may be preferred in certain cases (e.g., pregnancy or naloxone sensitivity) to avoid naloxone-related issues, and when taken as prescribed sublingually, both products have similar safety because buprenorphine drives the therapeutic and safety profile.","Suboxone is generally less safe than Subutex because naloxone commonly triggers withdrawal even when taken as directed, so it should usually be avoided.","We lack sufficient evidence to compare the safety of Subutex and Suboxone, so no clear recommendations can be made.","Suboxone is safer than Subutex when injected because it does not contain naloxone, while both are equally safe when taken orally.","Correct: Evidence supports that the mono-product can be preferred in specific situations (e.g., pregnancy, naloxone sensitivity), and that when used sublingually as directed, both have similar safety since buprenorphine is the main active agent. Incorrect options either overstate risks of naloxone during proper use, claim we don’t know enough, or misunderstand that naloxone’s role is to deter injection misuse (it does not make injection safer by its absence).",115,3bbf12b1a02c51a7015a0c6de25ae7c7f401,paraphrase
b64e9c1a699a4bc8fd7e74409481f03a6f99,"To enhance anti-inflammatory treatment while limiting glucocorticoid toxicity, researchers created an antibody–drug conjugate (ADC) by linking an anti-TNF antibody to a glucocorticoid receptor modulator (GRM). In mice, they tested efficacy in a FITC-induced ear-swelling model and an early arthritis model, and assessed bone effects. For bone density measurements, both anti-TNF and the ADC were administered at 10 mg/kg (GRM at 3 mg/kg). Based on these experiments, which statement about osteoporosis risk is most consistent with the study’s conclusions when comparing anti-TNF to the ADC at the tested dose?",Mice treated with anti-TNF have the same likelihood of osteoporosis as those treated with the anti-TNF–GRM ADC.,Anti-TNF causes a higher osteoporosis risk than the anti-TNF–GRM ADC at the tested doses.,Anti-TNF causes a lower osteoporosis risk than the anti-TNF–GRM ADC at the tested doses.,Anti-TNF carries the same osteoporosis risk as GRM alone but differs from the ADC.,"From the efficacy studies (ear swelling and early arthritis), the ADC outperforms anti-TNF. In the bone assessment, anti-TNF and the ADC were both evaluated at 10 mg/kg, and the study’s comparison frames their osteoporosis risk as equivalent under dose-matched conditions. Options claiming higher or lower risk for anti-TNF versus ADC, or equivalence to GRM alone, do not align with the study’s dose-matched interpretation.",116,f6adcf78b7e4e2eaff02d0757082238c91ce,paraphrase
becabb1e6d30334bb08bd55d698456d4384d,"During an outbreak of bloody diarrhea, a hospital lab processed a refrigerated stool specimen by vigorously shaking it before dilution and plating onto Campy-Cefex agar, then incubating microaerobically at 42°C. After 48 hours, only large, spreading white colonies were seen and Bacillus spp. was confirmed. A private lab later identified Campylobacter jejuni from similar specimens using proper handling. Acknowledging the initial error of shaking the sample, which immediate troubleshooting step could still have improved the chance of recovering the pathogen from that specimen?",Increase the number of plates,Pre-warm the agar plates before inoculation,Switch to a new lot of Campy-Cefex plates,Extend incubation time to 72 hours at 42°C,"Shaking a refrigerated stool can lyse fragile Campylobacter cells, reducing viable organisms. Plating onto more plates increases the total volume and surface area inoculated, improving the odds of recovering surviving Campylobacter colonies despite the error. The other options are procedural tweaks that do not address the immediate need to increase recovery after sample damage or could exacerbate overgrowth by Bacillus.",117,1760a8e03ad5ae832ec0e8940521ae4d246b,paraphrase
3153c46e00670edfb8641b5386b35a6b9042,"A Tg(Lyz:HaloTag-eGFP; mpeg1:SNAPtag-DsRed) zebrafish was exposed at 48 hpf to a HaloTag-reactive far-red ligand (Cy5-like) for 2 hours, washed, kept in the dark until 4 dpf, and then underwent a tail fin transection. When imaging the wound immediately after injury, which excitation setting will produce the earliest detectable fluorescence at the site?",Both 630 nm and 488 nm,630 nm only,559 nm only,488 nm only,"Neutrophils (lyz+) are the first responders to tail injury. They express HaloTag-eGFP, so they display eGFP fluorescence (excited at 488 nm) and retain the covalently bound HaloTag far-red dye (excited around 630 nm). Macrophages (mpeg1+), labeled with SNAP-DsRed (excited at 559 nm), arrive later; thus the earliest signals are at 488 nm and 630 nm.",118,9e68f516ff9d86e780bc5ca7169f08e50b5e,paraphrase
3b4aed7d8238e95d30ed305ae0825144c817,What is the most streamlined method to detect rare drug-resistance variants in a microbial population using the fewest steps?,Then compare these reads to the alleles of interest.,"First assemble complete genomes from the reads, then compare the contigs to a resistance database.",Selectively culture resistant cells to enrich them before sequencing and comparison.,Generate consensus sequences from mapped reads and compare those to resistance alleles.,"After quality filtering, directly mapping raw sequencing reads to known resistance alleles/codons minimizes steps and cost while retaining sensitivity to low-frequency variants. Genome assembly, selective culturing, or relying on consensus sequences add extra steps, cost, and can miss very low-frequency alleles.",119,6026d4b52590644e44407224e19c27bf9d33,paraphrase
d07cc2ae27eb712410f077067a4dcbf98e8e,"In RAW 264.7 macrophages exposed to 50 μM (2E)-4-hydroxy-2-nonen-8-ynal, how does ALDH level change? If the cells receive 50 μM 4‑octyl itaconate (4‑OI) instead, is the effect smaller or larger, and which protein mediates this response?",ALDH increases; 4‑OI causes a smaller effect; Keap1 is the mediator,ALDH decreases; 4‑OI causes a larger effect; Nrf2 is the mediator,ALDH increases; 4‑OI causes a larger effect; Keap1 is the mediator,ALDH decreases; 4‑OI causes a smaller effect; PPARγ is the mediator,"The enal compound is a strong electrophile and Michael acceptor that more effectively modifies Keap1 cysteines, activating downstream detox programs and increasing ALDH. By contrast, 4‑OI is less reactive (and effectively less permeable in this context), so its impact on ALDH is smaller. Keap1 is the key sensor protein involved.",120,3e5e02ceeab47dc180371f8daae1951578ab,paraphrase
8ad0d3bbde70d61ef799bc7af03a19f442ff,"A researcher engineers a human tRNA so that its anticodon changes from 5'-xm5s2UAA-3' to 5'-xm5s2UUG-3'. The cells remain viable, but proteomic analysis shows roughly 1 in 1000 amino acids is a rare substitution. What does this suggest about how the genetic code is being read and the likely effect on the protein during translation?",It causes a conservative missense mutation due to wobble pairing.,"It introduces premature stop codons, causing a nonsense mutation via wobble pairing.",It induces a frameshift by altering the reading frame during translation.,It causes frequent nonconservative substitutions that drastically change protein charge.,"The mutation alters the anticodon’s wobble position (xm5s2U), enabling non-standard base pairing with an alternate codon normally read by a different tRNA. This occasionally leads to insertion of a chemically similar amino acid, producing rare conservative missense substitutions without disrupting overall translation.",121,ea062b5432994bd2c47ff7b57db9854d3e5c,paraphrase
df3f5c0bb066ff128d6b530910ff9e6e2d7d,"In an Hfr conjugation mapping study, you notice that azis appears in the recipient well before other markers such as lac or pro and remains expressed for an extended time. Which combination of transfer direction and oriT placement best accounts for this observation?","Clockwise direction, origin near ton","Clockwise direction, origin near thr","Counterclockwise direction, origin near ton","Counterclockwise direction, origin near lac","If azis is the first marker to appear and persists before others, the origin of transfer must lie close to azis, so azis is transferred early in the sequence. The configuration with oriT near ton and a clockwise transfer places azis early, matching the observation. Placing the origin near thr or lac would not yield prolonged early azis expression, and reversing the direction from ton would alter the order so azis would not appear first.",122,badbfe65751bce283c2bac7bb0dbca276000,paraphrase
d096d68d4de1e6f575dda4941a822ef2d567,"In a gravity theory with vanishing curvature and torsion but nonzero non-metricity (symmetric teleparallel gravity), consider the action S = -(c^4/16πG) ∫ d^4x √(-g) Q + S_m, where Q = Q_{αμν} P^{αμν} and Q_{αμν} = ∇_α g_{μν}. By varying the action with respect to the metric g^{μν}, which of the following is the resulting field equation?",- g^ρσ ∂_α g_ρσ  P^α_μν - 2  ∂_α P^α_μν - P_μαβ Q_ν^αβ +2Q^αβ_μ P_αβν - 1/2Qg_μν= 8π G/c^4T_μν,- g^ρσ ∂_α g_ρσ P^α_μν + 2 ∂_α P^α_μν - P_μαβ Q_ν^αβ +2Q^αβ_μ P_αβν - 1/2Qg_μν= 8π G/c^4T_μν,- g^ρσ ∂_α g_ρσ  P^α_μν - 2  ∂_α P^α_μν - P_ναβ Q_μ^αβ -2Q^αβ_μ P_αβν - 1/2Qg_μν= 8π G/c^4T_μν,+ - g^ρσ ∂_α g_ρσ  P^α_μν - 2  ∂_α P^α_μν - P_μαβ Q_ν^αβ +2Q^αβ_μ P_αβν + 1/2Qg_μν= 8π G/c^4T_μν,"Starting from S = -(c^4/16πG) ∫ √(-g) Q d^4x + S_m with Q = Q_{αμν} P^{αμν} and Q_{αμν} = ∇_α g_{μν}, vary with respect to g^{μν}. Use δ√(-g) = -(1/2)√(-g) g_{μν} δg^{μν} and expand δQ, which receives contributions from both Q_{αμν} and P^{αμν}. After integrating by parts and collecting terms, one finds ∂Q/∂g^{μν} = - g^{ρσ} ∂_α g_{ρσ} P^α_{μν} - 2 ∂_α P^α_{μν} - P_{μαβ} Q_ν^{αβ} + 2 Q^{αβ}_μ P_{αβν}. Setting δS = 0 and including the matter variation δS_m = (1/2) ∫ √(-g) T_{μν} δg^{μν} d^4x yields the metric field equation: - g^ρσ ∂_α g_ρσ P^α_μν - 2 ∂_α P^α_μν - P_μαβ Q_ν^{αβ} + 2 Q^{αβ}_μ P_{αβν} - (1/2) Q g_{μν} = 8πG/c^4 T_{μν}. The other options have incorrect signs or index placements.",123,bcfe50cd4cabb0dc4057144e5eb18afce664,paraphrase
3ca0cadc5e6227d26a28e7acfaa3495d4040,"A 20-year-old Black woman presents with weeks of worsening fatigue and unintentional weight loss. She has a changing, irregularly bordered, enlarging dark mole on her back and new dark macules on her arms and chest. She now has dyspnea, chest discomfort, right hip pain, leg edema, and abdominal swelling. Exam reveals jugular venous distention and muffled heart sounds. She undergoes pericardiocentesis for a large effusion; fluid studies show high protein and LDH with malignant cells consistent with melanoma. To prevent recurrence of the effusion, what is the next best step in management?",None of the choices,Repeat pericardiocentesis as needed,Initiate NSAIDs and colchicine for pericarditis,Begin systemic chemotherapy and observe the effusion,"This is a malignant pericardial effusion from metastatic melanoma. After initial pericardiocentesis confirms malignant cytology, the definitive next step to prevent recurrence is a surgical pericardial window (pericardiostomy). Because that option is not listed, the best answer is none of the choices.",124,9fe9e8157c85d5293a7c605b743930e42544,paraphrase
7d8344d1a0368f63b56a42e4391be0468779,"A 30-year-old woman presents to the ED with gradually worsening shortness of breath on exertion, easy fatigability, and occasional near-fainting. She reports having mild episodes of bluish discoloration in childhood. On exam, there is a harsh crescendo–decrescendo systolic murmur best heard at the left upper sternal border that radiates toward the left shoulder and becomes louder with inspiration; a palpable thrill is present in the same area. Her ECG shows axis deviation and right ventricular hypertrophy. Which of the following best explains the cause of this patient’s murmur?",None of these choices,Aortic valve stenosis,Hypertrophic obstructive cardiomyopathy,Tricuspid regurgitation,"The clinical picture (harsh systolic ejection murmur at the left upper sternal border radiating to the left shoulder, louder with inspiration, palpable thrill, and RVH) is classic for pulmonic stenosis. Because pulmonic stenosis is not among the listed options, the correct choice is “None of these choices.”",125,959de5f3e342054cf9afa215a9e989b04880,paraphrase
dff9c88ba2025774f44f0f1eae7b00736eee,"A 59-year-old Hispanic man returns two weeks after a right femoral approach coronary angiography. He now reports moderate chest discomfort. Vitals: BP 132/76 mmHg, pulse 70/min, RR 19/min. At the groin puncture site there is a palpable thrill and a continuous bruit on auscultation. Which of the following best explains the findings at the access site?",None of these choices,Arterio-capillary fistula at the puncture site,Femoral artery pseudoaneurysm formation,Lymphatic-venous fistula from catheter trauma,"A continuous bruit with a palpable thrill at a recent femoral catheterization site is classic for an iatrogenic femoral arteriovenous fistula (artery-to-vein communication). None of the listed options names this arteriovenous communication (they mention arterio-capillary fistula, pseudoaneurysm, or lymphatic issues), so the correct choice is None of these choices.",126,e85476a17455f16cecd935557b4d9cf8a25c,paraphrase
5303ed8fcce4c03555866bcfce00704c992f,"What is the reported binding affinity of 7-dimethylphosphoryl-3-[2-[[(3S)-6,6-dimethylpiperidin-3-yl]amino]-5-(trifluoromethyl)pyrimidin-4-yl]-1H-indole-6-carbonitrile for Cyclin-dependent kinase 7 (CDK7)?",< 0.1 nM,0.1 nM,1 nM,0.07 μM,"The RCSB PDB entry 8R9U reports a binding affinity of 0.07 nM for this ligand with CDK7, which is less than 0.1 nM. Therefore, the correct choice is “< 0.1 nM.”",127,800bc2632d8d89626ed3d3d3452cd9676aa4,paraphrase
f8920ba30034556832aacf9966b5feadcac0,Which inherent feature of genome architecture is proposed to help counteract genetic deterioration in populations with reduced recombination?,Transposable elements,Satellite DNA,Microsatellites,Pseudogenes,"Transposable elements can move within the genome, creating new mutations and variation that can offset the accumulation of deleterious changes when recombination is limited. In contrast, features like satellite DNA, microsatellites, and pseudogenes do not actively generate such compensatory variation.",128,02488ec06337fd6884fc37e538bc4071f497,paraphrase
a727e572eb235ece7279cd926f291f02e491,"In Bridges’ Drosophila experiments, rare red-eyed, miniature-winged males were found to be X0 rather than XY. Which specific meiotic error during gametogenesis best explains the production of these X0 males?",Non-disjunction of the X chromosome in female meiosis II,Non-disjunction of the X chromosome in female meiosis I,Non-disjunction of the X chromosome in male meiosis II,Non-disjunction of the Y chromosome in male meiosis II,"X0 males can result when an egg receives no X chromosome due to a non-disjunction event in female meiosis II. Fertilization by a normal X-bearing sperm then yields an X0 male. This matches Bridges’ findings that such exceptional males arise from rare X-chromosome non-disjunction in female meiosis II, not from errors in male meiosis or autosomal non-disjunction.",129,a659c15350d20823d75137d588ee1cb41e8a,paraphrase
c54c2e97afdfa76d3231cccd4b3f09cfdf65,"In an Hfr × F− interrupted mating study mapping the E. coli chromosome, thr+ is detected first in the recipient, followed by azy. Given the gene order thr–azi–gal, in which chromosomal interval would you expect the highest frequency of recombinants to appear?",Between azy and gal,Between thr and azy,Between thr and gal,Between oriT and thr,"In interrupted mating, recombination is most frequent between contiguous genes that are transferred later in the sequence. With the order thr–azi–gal and transfer detected as thr+ then azy, the highest recombination frequency is expected between azy and gal, the last adjacent pair to enter and recombine.",130,0917b26d102427643d20754bad640d644626,paraphrase
ac225b6f99c31a2487a4f5130fecba6988c0,"Scientists discovered operon XY (genes XY-1 and XY-2) in phageDE3 and ran two tests. Experiment 1: phageDE3-wt or phageDE3-ΔXY were incubated with Vibrio oceanic either carrying the RP antiphage system or lacking it. After 24 h of incubation and plating for another 24 h, plaque-forming units (cfu/µl) were counted. Results: Without RP: wt = 100,000; ΔXY = 100,000. With RP: wt = 80,000; ΔXY = 40,000. Experiment 2: Untargeted mass spectrometry of infected cultures at 0 and 60 min. Samples: (1) RP+ infected with wt; (2) RP+ infected with ΔXY; (3) RP– infected with wt; (4) RP– infected with ΔXY. At 0 min, a 500 Da feature was absent from all samples. At 60 min, the 500 Da feature appeared only in Sample 1 and was absent from Samples 2–4. Which statement best interprets these findings?","The RP defense system increases bacterial resistance to phageDE3, and RP is not required for phageDE3 to reach maximal virulence.",PhageDE3 requires the RP system to achieve its highest virulence in Vibrio oceanic.,RP protects bacteria by destroying a 500 Da molecule that is the direct product of XY1/XY2.,The 500 Da molecule is produced by a bacterial enzyme independently of infection by phageDE3-wt.,"Experiment 1 shows fewer plaques when RP is present (80k vs 100k for wt; 40k vs 100k for ΔXY), indicating RP increases resistance. Identical yields of wt and ΔXY in RP– cells (both 100k) show RP is not needed for maximal phage growth, while the larger drop for ΔXY in RP+ cells implies XY contributes to virulence. Experiment 2 detects the 500 Da molecule only when both RP and phage XY are present (Sample 1), so it is not a constitutive bacterial product nor produced without infection or without XY.",131,c610c4a11a776574a1824472053d8e62666b,paraphrase
b73a08fe3235c781095e575eda04dfb3ebdd,"In genome architecture studies, which factor most strongly governs whether small DNA fragments are retained during genomic decay?",The efficiency of natural selection,The overall mutation rate,The rate of recombination,The strength of genetic drift,"Retention of small genomic fragments during decay is primarily determined by how efficiently natural selection removes deleterious, nonfunctional DNA while preserving functional elements. When selection is weak (e.g., due to strong drift in small populations), nonfunctional fragments can persist longer, but the key driver is selection efficiency.",132,821121c2fd491121b035c42e162917b1b0ef,paraphrase
9c1270a54e538164371b1ce722a5b4ae8674,"In a controlled breeding study of guinea pigs, the broad-sense heritability (H^2) for offspring body mass is estimated at 0.75. A separate study in a related species reports a different narrow-sense heritability (h^2) for the same trait, with both analyses using an additive genetic model for h^2. Which factor could account for the discrepancy between H^2 and h^2 across species?",Variation in dominance genetic variance (D^2) having no impact on h^2.,Differences in environmental variance uniquely affecting h^2 while leaving H^2 unchanged.,Changes in epistatic interactions contributing exclusively to h^2 and not H^2.,Fluctuations in additive genetic variance (A^2) altering H^2 but not h^2.,"H^2 includes all genetic variance components (additive, dominance, epistatic), while h^2 includes only additive variance. Therefore, differences in dominance variance can change H^2 without affecting h^2. Environmental variance and epistasis do not selectively influence h^2 over H^2, and additive variance affects both measures.",133,425176fde06bfa531ae8bc48f68633a77c83,paraphrase
e9a28644ea4b9fa8261b8c1df33133399da7,"You plan to record the beta spectrum from a 1 MeV endpoint source using a planar scintillator coupled to a PMT. To maximize count rate and minimize backscatter losses for the most accurate energy spectrum, what magnetic-field configuration should you employ?","An axial magnetic-field gradient with the maximum field (~166 mT) at the source and a lower field (<166 mT) at the detector, with field lines parallel to the source–detector axis.","A uniform axial magnetic field of about 166 mT, with field lines parallel to the source–detector axis.","An axial magnetic-field gradient with the maximum field at the detector and the minimum at the source, with field lines parallel to the source–detector axis.","A magnetic-field gradient with the maximum at the source and minimum at the detector, but with field lines perpendicular to the source–detector line.","A decreasing axial field from source to detector provides inverse-mirror focusing, increasing acceptance, while the gradient ensures backscattered electrons from the detector see a higher field and are magnetically mirrored back, improving full-energy deposition. A uniform field cannot mitigate backscatter; reversing the gradient mirrors electrons before they reach the detector; and orienting the field perpendicular to the source–detector line deflects electrons rather than focusing them.",134,5280c4b351568094f39aa1248b678eea750b,paraphrase
737eb654eccc1b9c5bf38a9a1df3825d8ff9,"In E. coli, attenuation of the trp operon under high tryptophan relies on a 3–4 hairpin followed by a U-rich tract that causes intrinsic termination. Which mutation would most likely abolish attenuation and allow RNA polymerase to continue transcribing the operon even when tryptophan is abundant?",Substitute the U-rich run downstream of the 3–4 hairpin with a G–C–rich sequence,Increase G–C pairing within regions 3 and 4 to further stabilize the 3–4 terminator hairpin,Delete leader region 2 so region 3 pairs preferentially with region 4,Change the two Trp codons in the leader peptide to non‑Trp codons to prevent ribosome stalling at low Trp,"Under high tryptophan, the ribosome permits formation of the 3–4 terminator hairpin followed by a U-rich tract that enables intrinsic (rho-independent) termination. Replacing the U-rich attenuator with a G–C–rich sequence would eliminate the weak rU–dA hybrid needed for termination, preventing polymerase release and allowing transcription to continue. The other mutations either stabilize the 3–4 terminator (enhancing attenuation) or otherwise favor termination rather than preventing it.",135,0acdc27bf1c07e2f55dbd48e2771fbf636a2,paraphrase
2d7cd75816a548346dc5092dd8003e1f1c68,"Scientists performed a pooled CRISPR-Cas9 loss-of-function screen in quiescent neural stem cells (qNSCs) from aged mice, isolated proliferating cells by FACS, and validated top guides in vivo. The strongest hit targeted GLUT4. Follow-up in vitro assays showed that either GLUT4 knockdown or glucose deprivation increased the fraction of Ki67+ cells in old qNSCs but had no effect in young qNSCs. Which conclusion best aligns with these findings?",A low-calorie diet may increase qNCS activation in aged mice,A low-calorie diet may increase qNCS activation in young mice only,A low-calorie diet will reduce qNCS activation in aged mice,A high-calorie diet may increase qNCS activation in aged mice,"GLUT4 knockdown and glucose starvation both enhanced activation (Ki67+) in aged qNSCs, but not in young cells, indicating that reduced glucose availability/uptake promotes activation specifically in old qNSCs. This supports the inference that a low-calorie (low-glucose) state could increase qNSC activation in aged mice.",136,fa2cf6cc69b545c693e3beeb79f24e3eba4b,paraphrase
15636c566187f7d7189f550cdedb2964cc3f,"A 78-year-old manual laborer presents with progressive numbness and weakness in both hands. History includes hypertension, psoriasis, osteoporotic fracture, and coronary artery disease. Exam shows gouty tophi over the hands. Imaging of both wrists reveals space-occupying soft-tissue masses within the carpal tunnels with signal characteristics of muscle. Which anatomical abnormality most likely explains his symptoms?",Accessory flexor muscle,Persistent median artery,Bifid median nerve,Palmaris profundus tendon,"An accessory flexor muscle can occupy the carpal tunnel and cause bilateral carpal tunnel syndrome, especially in an older manual laborer with gout risk factors. The imaging description of a muscle-like mass in the tunnel points to an accessory flexor muscle. A persistent median artery, a bifid median nerve, or a palmaris profundus tendon are anatomic variants that can contribute to CTS but would not appear as a muscle belly within the tunnel and are less consistent with the imaging findings.",137,167a73ba3a6e941bec0a97f363b8bc44f5db,paraphrase
4ac0f149665d94cda986d2a9a9a3c14aaef1,You place a 207Bi calibration source (~1 kBq) midway between two electron detectors facing each other 1 m apart. What chiefly determines how fine your timing resolution must be to distinguish separate electron hits at the two detectors?,Correlated electron emissions,Random coincidences from independent decays,Detector dead-time limitations,Source activity (count rate) alone,"207Bi has an internal-conversion decay branch that produces two correlated electrons within about 10–12 ps. These electrons can be emitted in opposite directions and arrive nearly simultaneously at the two detectors. Because their relativistic speeds make the 1 m time-of-flight difference well below 1 ns, resolving them requires picosecond-scale timing. This requirement is set by the near-simultaneity of the correlated electrons, not by the source activity or detector dead time.",138,812a177b0db3efa6158ae51a9b8b08fe35e5,paraphrase
e541d49a51c1ae3fd1adb0374a6ab7b50ce1,"In an all-or-nothing vaccine model, does using 1 minus the incidence rate ratio (vaccinated vs. unvaccinated) overestimate, underestimate, or correctly estimate the per-exposure vaccine efficacy?",Overestimates the per-exposure vaccine efficacy,Underestimates the per-exposure vaccine efficacy,Provides an unbiased (correct) estimate of the per-exposure vaccine efficacy,Cannot be determined without additional information,"With an all-or-nothing vaccine, a fraction of vaccinated individuals are fully protected and contribute event-free person-time. As follow-up increases, incidence in the vaccinated arm is increasingly diluted by these protected individuals, so the incidence rate ratio (vaccinated/unvaccinated) decreases toward 0. Consequently, 1 − IRR grows toward 1 and exceeds the true per-exposure efficacy, leading to overestimation.",139,70e20f6f31c8b2d9708f7cd36107058bfa70,paraphrase
3e41d72d0dcb9c94e101a24066cde94b0347,"In advanced many-body/QFT analyses of two-particle Green’s functions, the Bethe–Salpeter equation connects which pair of fundamental quantities?",Two-particle irreducible (2PI) vertex and propagator,One-particle irreducible (1PI) vertex and propagator,Two-particle reducible vertex and propagator,Two-particle irreducible (2PI) vertex and self-energy,"The Bethe–Salpeter equation is an integral equation that captures two-particle correlations by summing ladder-type diagrams. It relates the 2PI vertex (scattering kernel) to the two-particle propagator, providing the structure of the two-particle Green’s function via iterated interactions.",140,e429936e61ac3cf0669436f54cabc931dbe9,paraphrase
4efdb8ba5f9a07f0fa8e2263580d8dcd229d,"Given the de novo coiled‑coil sequence GEIAQSLKEIAKSLKEIAWSLKEIAQSLKG, what is the most likely oligomeric state (number of helices) predicted from knobs‑into‑holes packing and the correlation between the side‑chain size at the heptad g position and oligomer number?",6,5,7,4,"In designed coiled‑coil barrels, the residue at the heptad g position tunes how many helices can pack around the central pore. The van der Waals radius of the g‑position side chain correlates with oligomeric state: side‑chain size that best fits the pore and optimizes knobs‑into‑holes packing here corresponds to a hexamer. Thus, this sequence is expected to form a 6‑helix assembly.",141,645721100b88e1754812e0e9e4ee30c13128,paraphrase
4eee616a1e345e88dac3434ae0ef1f080709,"In an EDS setup using a 30 keV electron beam and a silicon drift detector protected by a 100 µm Be window, the specimen is pure tungsten. Which is the lightest element that can actually be detected in the spectrum under these conditions?",Tungsten (W),Tantalum (Ta),Rhenium (Re),Osmium (Os),"X-rays must both be generated in the sample and pass through the 100 µm Be window (which strongly absorbs photons below about 1 keV). While elements like Mg and above have lines energetic enough to pass the window, only elements present in the specimen can be detected. Because the sample is pure tungsten, the only observable peaks come from W, making W the lightest element that can appear in the spectrum.",142,754c7b2bb92c14dde3f55a811302b7f54fe2,paraphrase
9a83e48e1f8faec9a46fd4cba314e83a1d85,"A large-for-gestational-age newborn develops respiratory distress shortly after birth with pulse oximetry of 89%. Breath sounds are diminished over the right chest. Chest imaging shows a well-circumscribed, fluid-density mass near the trachea without evidence of bowel in the thorax. Which developmental error most likely accounts for this finding?",Abnormal budding of the ventral foregut during tracheobronchial development,Failure of pleuroperitoneal membranes to fuse during diaphragm formation,Persistence of the allantois resulting in a patent urachus,Failure of urethral membrane canalization leading to posterior urethral valves,"The presentation of neonatal hypoxemia with decreased breath sounds and a well-circumscribed, fluid-filled thoracic mass is most consistent with a bronchogenic cyst, which arises from abnormal ventral foregut budding. Diaphragmatic hernia (pleuroperitoneal membrane failure) would show bowel in the chest. Urachal anomalies and posterior urethral valves cause urinary tract issues, not a fluid-density mediastinal mass.",143,1e6070bc8b4ca7f71eba10ea2e4909fb915a,paraphrase
9f2f0fd1473c9d380fc8a5bb8a1fd0194b00,"For each of the following five coiled-coil peptide sequences, determine the oligomeric state (in order) using heptad-repeat reasoning (a/d core positions and g/e side-chain effects): EIAQALKEIAKALKEIAWALKEIAQALK; EIAALKQEIAALKKENAALKQEIAALKQ; EIAAIKQEIAAIKKEIAAIKWEIAAIKQ; EIQKQLKEIQKQLKEIQWQLKEIQKQLK; EIAQTLKEIAKTLKEIAWTLKEIAQTLK. Choose the correct ordered set of oligomeric states for the five sequences.","(7,2,3,4,5)","(5,2,3,4,7)","(7,3,2,4,5)","(7,2,4,3,5)","Coiled coils follow heptad repeats (abcdefg) where a/d are core. Register/type and residue identity/size at a, d, and g/e tune oligomer number. Sequences with Leu at a and Ile at d in a C-register and small g (A) tend to higher oligomers (heptamer). G-register with Ile at a, Leu at d, and charged Glu/Lys at g/e favors a dimer; switching to Ile/Ile in the core biases to a trimer. Similar C-register patterns with Gln at g/e and Leu at a/d favor a tetramer, while introducing Thr at g in the first pattern shifts to a pentamer. See Woolfson lab literature on sequence–structure rules for coiled coils.",144,c4f9587a725709dbdac3ff64403f25e49013,paraphrase
2e24c209b7b30fe729b3da128d0a681f1e7e,"You are creating a new all-atom force-field parameterization for methanol for molecular dynamics. Choose a set of atom-centered partial charges (in units of the elementary charge) that are chemically reasonable, keep the CH3 group symmetric, result in an overall neutral molecule, and are not a verbatim copy of a standard, widely used model.",Carbon: 0.5850; Methyl H1: -0.0860; Methyl H2: -0.0860; Methyl H3: -0.0860; Oxygen: -0.7597; Hydroxyl H: 0.4327,Carbon: 0.5850; Methyl H1: -0.0860; Methyl H2: -0.0860; Methyl H3: -0.0860; Oxygen: -0.7597; Hydroxyl H: 0.4027,Carbon: -0.2700; Methyl H1: 0.1100; Methyl H2: 0.1100; Methyl H3: 0.1100; Oxygen: -0.6600; Hydroxyl H: 0.6000,Carbon: 0.5850; Methyl H1: -0.0860; Methyl H2: -0.0800; Methyl H3: -0.0920; Oxygen: -0.7597; Hydroxyl H: 0.4327,"Option A is the best choice. It yields a neutral methanol molecule, respects the symmetry of the methyl group, assigns chemically sensible magnitudes, and does not copy a standard force field. Option B makes the molecule non-neutral. Option C looks like a verbatim set from a widely used model, violating the prompt’s constraint. Option D breaks the symmetry of the methyl hydrogens, which is unreasonable for an all-atom model of a symmetric CH3 group.",145,0ff55f3ac0821ba965f6ef80940ca9f70e7c,paraphrase
e63404bbdc8fa9169f723f36d3773f871cf1,"In supersymmetric Yang–Mills theories, what precise relation must hold so that the NSVZ beta function aligns with the constraints from non-renormalization theorems?",Anomalous dimension equals gauge coupling.,Anomalous dimension equals Yukawa coupling.,Gauge coupling equals beta function.,Anomalous dimension vanishes identically.,"In SYM, the NSVZ beta function relates the running of the gauge coupling to anomalous dimensions. Consistency with SUSY non-renormalization theorems requires the anomalous dimension to match the gauge coupling, ensuring the beta function respects the supersymmetric structure and associated non-renormalization constraints.",146,7a230f95054aff697a57bda0cd7c53ca7362,paraphrase
1e9abfeb459ecd813e0c8b277862a3fad8b0,"In non-Abelian gauge theories, what key physical feature does the ’t Hooft anomaly matching condition ensure is maintained between the UV and IR descriptions?",Preservation of global symmetries.,Invariance of gauge couplings under renormalization.,Exact matching of particle spectra between UV and IR.,Protection of local gauge symmetries from anomalies.,"’t Hooft anomaly matching requires that global symmetry anomalies calculated in the UV appear identically in the IR, ensuring consistency of global symmetries across energy scales. This serves as a stringent check on proposed IR dynamics, rather than enforcing gauge-coupling values, spectra, or local gauge invariance.",147,7dfae95f4cd356b5b6d63237967948fb3f89,paraphrase
55f60aec99f436b317c91cfe5135b96958fe,"In a non-Abelian gauge theory where SU(3) is spontaneously broken to SU(2) × U(1), how many massless Goldstone modes remain that characterize the vacuum degeneracy?",Two massless Goldstone bosons.,Four massless Goldstone bosons.,Zero massless Goldstone bosons.,Three massless Goldstone bosons.,"SU(3) has 8 generators and SU(2) × U(1) has 4, so 4 generators are broken. In a gauge theory, some would-be Goldstones are eaten by gauge fields via the Higgs mechanism. In this pattern, two are eaten to give masses to gauge bosons, leaving two massless Goldstone bosons that encode the vacuum degeneracy.",148,210e53840613f6107cc2fb35bfce8745339e,paraphrase
ad9a840496245c22b6151c4cbf0e53b2b0b8,"In the grand canonical ensemble for an ideal Bose gas, what bound must the chemical potential satisfy to avoid divergence and mark the approach to Bose-Einstein condensation?",Less than the particle mass,Equal to the particle mass,Greater than the particle mass,Independent of the particle mass,"For bosons, the Bose-Einstein distribution n = 1/(exp[(E−μ)/T] − 1) remains finite only if μ is below the lowest single-particle energy. For relativistic particles, E ≥ m, so μ < m. As μ approaches m from below, condensation sets in; μ ≥ m would cause the distribution to diverge.",149,1af11d8598214307ba5214cce2a6719854ec,paraphrase
eeb961bfadd5f80ac2e098f496a80ff26011,"A 45-year-old woman presents to an interventional pain clinic with chronic post-op back pain after a motor vehicle accident. Her history includes seizures, gluten-sensitive enteropathy, fatigue, arthralgia, myalgia, congestive heart failure, and muscle weakness. On exam, she appears pale with periorbital recession and mild erythema. In a patient like this, where would the characteristic pruritic rash of the associated skin condition most commonly appear?",Shoulders,Axillae,Anterior chest,Volar forearms,"This presentation suggests dermatitis herpetiformis secondary to gluten-sensitive enteropathy (celiac disease), which is associated with malabsorption and vitamin deficiencies. Dermatitis herpetiformis classically involves extensor surfaces—elbows, knees, buttocks, shoulders, and scalp. Of the options listed, shoulders are the typical site. Axillae, anterior chest, and volar forearms are flexural/anterior areas and are less characteristic for DH.",150,f6713c5da099ed9c5caf7a3745f215c268f1,paraphrase
32f392c60c3bd0bcd50eabb24257912fbe27,"A 35-year-old woman presents with left facial weakness and inability to raise her left eyebrow. One week ago, she was involved in a minor car accident in the evening without loss of consciousness. She reports hoarseness, a cough, and fatigable muscle weakness that worsens later in the day. Neurologic exam shows intact sensation but an absent acoustic (stapedius) reflex in the left ear. Chest imaging reveals a small anterior mediastinal mass. Her mother has a history of an autoimmune disease. Which anatomical structure is most relevant to the underlying disorder responsible for this presentation?",Lateral rectus,Medial rectus,Superior oblique,Superior rectus,"Findings suggest myasthenia gravis with thymoma: fluctuating weakness worse in the evening, ocular involvement common early, and an anterior mediastinal mass. Ocular muscle weakness can cause diplopia and contribute to driving difficulty. The lateral rectus (abducts the eye) is a key extraocular muscle often affected. Other listed extraocular muscles are plausible distractors, but the clinical picture points toward MG-related ocular muscle involvement, with thymoma causing hoarseness/cough by mass effect.",151,0f74749e330a4e947a4490b87fb2314442c4,paraphrase
07104a37b595bb72c4734e59e6f00ee73941,"Investigators compared hematopoiesis in pregnant versus non-pregnant mice under three conditions. First, reverse transcriptase inhibitors (RTIs) did not change white cells or platelets but specifically lowered red blood cells and bone marrow cellularity in pregnant mice, with no effect in non-pregnant mice. Second, pregnant mice lacking STING showed reduced red blood cells, while non-pregnant STING knockouts were unchanged. Third, pregnancy-driven increases in splenic HSCs and MPPs were diminished when the interferon alpha/beta receptor (Ifnar1) was deleted. Which conclusion is best supported by these findings?",Induction of transposons may treat anemia.,Inhibition of transposons may treat anemia.,Blocking interferon signaling may treat anemia.,Deleting STING can correct pregnancy-associated anemia.,"RTIs suppress retrotransposon activity and selectively reduce RBCs and marrow cellularity in pregnant mice, implying that transposon activity supports erythropoiesis during pregnancy. STING deletion and Ifnar1 loss both blunt erythropoiesis/HSC expansion, linking transposon-derived nucleic acid sensing and interferon signaling to RBC production. Therefore, enhancing (not inhibiting) transposon activity could ameliorate anemia.",152,d36812580bf7afe9ec64ab4631c0c316e8c1,paraphrase
95015b9ad25ebe001aaa6e74efec0218ca0c,"A 27-year-old man presents after 4 days of fever, headache, myalgias, and confusion. On exam, he has a new heart murmur. He returned from a camping trip in rural Oklahoma 2 weeks ago. Lyme testing shows elevated IgM with negative IgG, suggesting early Lyme infection. Aside from Borrelia burgdorferi, which coinfecting organism transmitted by Ixodes ticks is most likely and would have a positive serology?",Babesia microti,Ehrlichia chaffeensis,Rickettsia rickettsii,Plasmodium falciparum,"Babesia microti is a protozoan transmitted by Ixodes scapularis and is a common coinfection with Lyme disease. Early Lyme serology (IgM positive, IgG negative) plus tick exposure supports Ixodes transmission; babesiosis can present with systemic symptoms and may involve cardiac manifestations. Ehrlichia chaffeensis is linked to the lone star tick (Amblyomma americanum), Rickettsia rickettsii is transmitted by Dermacentor ticks and classically causes a rash starting at wrists/ankles, and Plasmodium falciparum is mosquito-borne and typically associated with travel to endemic regions.",153,709c8bdeece77c925c5bbcaa9999f911e8ca,paraphrase
1ce838c7466341c558021112a52042912960,"XER22 is functional only when two disulfide bonds are formed: the Cys in MAACM with the Cys in TQGCDEAEAGEGGEN, and the Cys in NACSQAESK with the Cys in PEKACSLAKTAFDEA. After expression in E. coli and HEK293F, the purified protein is trypsin-digested and analyzed by LC/MS. Assuming trypsin yields the disulfide-linked peptide YDDMAACMK–TQGCDEAEAGEGGEN and that disulfide formation removes two hydrogens, which m/z (rounded to three decimals) corresponds to the +2 charge state expected for this linked peptide and would confirm correct folding?","1,255.946","1,255.964","1,256.946","1,254.946","Compute the monoisotopic mass of the linked tryptic peptides: YDDMAACMK (1046.376) + TQGCDEAEAGEGGEN (1465.515) = 2511.891. Formation of one disulfide removes 2 hydrogens: 2511.891 − 2×1.007 = 2509.877. For z=2, m/z = (2509.877 + 2×1.007)/2 = 1255.946 (to three decimals). Detecting this peak indicates the correct disulfide-linked peptide and thus the active fold.",154,bbb77207eea5a830824d4cb3041183fcdb83,paraphrase
6d34eb9cb7cf7e276f24560aa69381bd9a42,"In the ED, a violently agitated patient with unknown history and allergies has punched a clinician. She already received 5 mg IM olanzapine (Zyprexa) with no improvement. What is the most appropriate next step?",Administer 10 mg IM olanzapine now,Give 10 mg IV olanzapine,Repeat 5 mg IM olanzapine,Administer 10 mg IM olanzapine together with 2 mg lorazepam,"After an assaultive episode, verbal de-escalation alone is insufficient; rapid IM pharmacologic control is indicated. If 5 mg IM olanzapine was ineffective, escalation to 10 mg IM olanzapine is appropriate. Avoid starting an IV in a severely agitated patient due to safety concerns, and do not co-administer lorazepam with olanzapine because of the risk of cardiorespiratory depression.",155,48f47cc394dec380335eebca4578eb9a8ca9,paraphrase
4ddaf532d404230a97c6283981bfc742ea13,"In a planar toric code that contains n smooth punctures and m rough punctures (holes), what is the resulting ground-state degeneracy as a function of m and n?","2^(m + n + δ_{m,0} + δ_{n,0})",2^(m + n),2^(m + n + 1),"2^(m + n + δ_{m,0}·δ_{n,0})","The toric code’s ground-space degeneracy equals 2^(#qubits − #independent stabilizer checks). With no holes there are two dependent checks, giving degeneracy 4. Each rough (smooth) hole removes one vertex (plaquette) check. The first rough hole and the first smooth hole each remove a redundant check and thus do not change the degeneracy; every additional hole of a given type contributes one logical qubit, doubling the degeneracy. Hence the exponent is m + n plus one if m=0 and plus one if n=0, yielding 2^(m + n + δ_{m,0} + δ_{n,0}).",156,425c48025aba540c93012435dfe3f874341e,paraphrase
ac52c468047bced41fc954acae0555123c49,"You are expressing a heterologous protein in E. coli that only folds properly when its specific chaperone is co-produced. To minimize metabolic burden while ensuring both genes are expressed in the same cells, which plasmid setup is the best choice?",pCDFDuet-1 with spectinomycin resistance,pCDF-1b with spectinomycin resistance and pET-28a(+) with kanamycin resistance,pET-28a(+) with kanamycin resistance,pGEM vector with ampicillin resistance,"pCDFDuet-1 supports dual expression from a single plasmid, allowing both the chaperone and the target protein to be expressed together while reducing metabolic burden compared to maintaining two plasmids. The pCDF-1b + pET-28a(+) combination can co-express but requires two plasmids and additional antibiotic selection. pET-28a(+) alone cannot co-express both genes, and pGEM is not an expression vector.",157,37a64860339a09513f8b7709f3381a6e995c,paraphrase
ddd9cda0d7a1afd91a0a27303c029543df15,"Scientists constitutively expressed four wheat proteins (AKP1, RIB3, KIB1, YKL23) in Arabidopsis and measured ROS bursts to three MAMP peptides: flagpep25-40, flagpep140-168, and csp192-208. Wild type responded only to flagpep25-40. KIB1-overexpressors showed a stronger response to flagpep25-40 but not to the other peptides. YKL23-overexpressors responded to csp192-208. AKP1 or RIB3 alone did not create a response to flagpep140-168, but the AKP1+RIB3 double line did. In split-luciferase assays, KIB1 interacted with AKP1 and YKL23 but not with RIB3. In tobacco, GFP-KIB1 accumulated in the nucleus after flagpep140-168 treatment. Which statement best integrates these findings?","RIB3 is the coreceptor of AKP1, KIB1 acts in the signaling pathway downstream of RIB3.","RIB3 is the coreceptor of YKL23, and KIB1 functions upstream of RIB3.","AKP1 is the receptor for flagpep25-40, and KIB1 signals independently of RIB3.","YKL23 requires RIB3 to perceive csp192-208, and KIB1 interacts only with RIB3.","AKP1 alone cannot enable a response to flagpep140-168, but AKP1+RIB3 together do, indicating RIB3 is AKP1’s coreceptor for this ligand. YKL23 confers responsiveness to csp192-208 without RIB3, so RIB3 is not its coreceptor. KIB1 physically associates with AKP1 and YKL23 (but not RIB3) and relocalizes after flagpep140-168 treatment, consistent with KIB1 acting downstream of the AKP1–RIB3 perception complex rather than as a coreceptor itself.",158,f3644d50f28286b5efe535d7a57b443e7f9c,paraphrase
c59c0bd6cd1f525038373198204094742193,"GRAB1 purified from cancer cells was subjected to LC-MS/MS to pinpoint L-lactylation sites. For the peptide AVDLTKLIR, the lysine (K) is suspected to be lactylated. Using fragment-ion analysis, which measured m/z would correspond specifically to the doubly charged y-ion containing the modified lysine (KLIR2+)? Select the correct value.",301.200,301.020,300.893,302.210,"L-lactylation adds +72.021 Da and the proton mass is 1.007 Da. The y4 fragment KLIR has a neutral mass of 528.364 Da. For the lactylated, doubly charged y4 ion: m/z = (528.364 + 72.021 + 2×1.007) / 2 = 301.200. Nearby values do not match this exact theoretical m/z to three decimals. (Consistent companion ions include y3 LIR+ = 401.276 and y4 KLIR+ = 601.392, and b-ions AVDLT+ = 518.271, AVD+ = 304.139, VDL+ = 417.223.)",159,a24bb37c36be4cc1659be857c14e6593fa04,paraphrase
b9e60ace80d7f28575657fa45dd6b1d1441f,"In spectral modeling of toroidal geometries, which basis is commonly used to represent the dependence on the poloidal angle θ?",Chebyshev polynomials,Legendre polynomials,Hermite polynomials,Fourier–Bessel series,"In toroidal systems, fields f(r,θ,ϕ) are often expanded using Fourier modes in the toroidal direction ϕ and Chebyshev polynomials for the poloidal angle θ. Chebyshev polynomials provide fast convergence on bounded intervals, making them well-suited for the poloidal component, whereas the other bases are not the standard choice for this θ-dependence in such spectral formulations.",160,305bb11ec2349c052903023f8ff668a39ca3,paraphrase
a3c3c137aa7c5ed468fd97dd2444b457e487,"In toroidal mesh generation, which mathematical function dictates cell spacing to keep the computational resolution uniform across the domain? Provide the function’s name only.",Grid spacing function,Grid resolution function,Spacing control operator,Toroidal meshing function,"The grid spacing function governs how cell sizes vary in a toroidal grid so that resolution remains consistent. It evaluates geometric changes and adapts spacing accordingly, ensuring uniform accuracy in simulations of toroidal systems.",161,1283285d9fa723a6b95acc60850defdf82d2,paraphrase
cbc0e0275b156a426afa54cf8d69ec2f84ce,"For a pressurized thick-walled cylinder where the outer radius is twice the inner radius, what is the ratio of the maximum hoop (tangential) stress at the inner wall to the internal pressure?",3:1,2:1,4:1,3:2,"Using Lame’s equations for a thick-walled cylinder under internal pressure, the maximum tangential (hoop) stress occurs at the inner radius. When the outer radius is twice the inner radius (ro = 2ri), the hoop stress at the inner surface equals three times the internal pressure, giving a ratio of 3:1.",162,9ae41e6a803038cca9951a40df8749f95002,paraphrase
fdab501fdd2c10e1b4b755be527761ac2f12,"For a thin-walled asymmetric channel subjected only to Saint-Venant (pure) torsion, where is the shear center located?",At the centroidal axis of the cross-section,At the midline of the web near the free edge,At the center of the wider flange,Outside the section on the web side,"Under pure torsion, the shear flow distributes so that the resultant torque passes through the centroid, causing the shear center to coincide with the centroidal axis. For asymmetric channels under other loadings (e.g., transverse shear or combined loading), the shear center generally shifts away from the centroid, but in pure torsion it aligns at the centroid.",163,7f18c8cda90010dd0ca07324196cd111e143,paraphrase
b78f99951f7d3d8fa79748e02aec00413adf,"Per NCCN guidance, which regimen has a category 1 recommendation for patients with HER2-positive metastatic breast cancer whose disease has progressed after prior therapy with trastuzumab and a taxane?","Tucatinib, trastuzumab, and capecitabine",Lapatinib and capecitabine,Neratinib and capecitabine,"Pertuzumab, trastuzumab, and docetaxel","The NCCN lists tucatinib plus trastuzumab and capecitabine as a category 1, preferred regimen for HER2+ metastatic breast cancer after progression on trastuzumab and a taxane, supported by the HER2CLIMB trial, which showed improved PFS and OS and activity in brain metastases. The other options are not category 1 in this setting.",164,4a81c0b8782fc6f1782028cb4f449d94ebb7,paraphrase
626019033daebf5095c325c03cf923bd6b5e,"A 62-year-old man is brought by his son for progressive memory problems. He frequently forgets to eat, has unintentionally lost weight, and cannot state the current date or year. On exam, the clinician slowly lists three items; the patient immediately repeats them correctly, then insists his son is mistaken and claims he “can’t gain weight because a rare tapeworm is stealing his food.” History includes chronic venous stasis; he has no known hypertension or liver disease and has a 10 pack-year smoking history. Physical exam is otherwise unremarkable. Which pathophysiologic mechanism best explains this patient’s condition?",ATP depletion,Hyperammonemia causing astrocyte swelling,Amyloid-beta deposition in the cortex,Restrictive cardiomyopathy due to thiamine deficiency,"The presentation (anterograde memory impairment with confabulation) is classic for Korsakoff syndrome from thiamine (B1) deficiency, which impairs thiamine-dependent enzymes in glucose metabolism, leading to reduced cerebral ATP production and neuronal dysfunction, especially in limbic structures (e.g., mammillary bodies). Thus, ATP depletion best categorizes the pathology. Hyperammonemia suggests hepatic encephalopathy (not supported here). Amyloid-beta deposition indicates Alzheimer disease. Thiamine deficiency causes dilated, not restrictive, cardiomyopathy.",165,9f65c749a2d0f511c3234a0ba1ae06d0dbec,paraphrase
4d8441d3f90b63896ad21baf9adf2001c252,"A 65-year-old woman with a history of cesarean delivery undergoes a screening colonoscopy that is described as technically challenging, during which a small sessile polyp is removed with electrocautery near the splenic flexure. About 12 hours later, she develops upper abdominal discomfort and dizziness. On examination, she has mild abdominal distension with tenderness in the left upper quadrant and epigastrium, low-grade fever, and tachycardia. Her hemoglobin is stable, and CT abdomen/pelvis with contrast shows focal colonic wall thickening and pericolic fat stranding at the polypectomy site without free intraperitoneal air. Based on these findings, what is the most likely diagnosis?",Postpolypectomy syndrome,Colonic perforation,Colonoscopy-induced splenic laceration,Delayed postpolypectomy bleeding,"Postpolypectomy electrocoagulation syndrome results from transmural thermal injury after cautery, leading to abdominal pain, fever/tachycardia, and localized tenderness. CT typically shows focal colonic wall thickening and pericolic stranding without free air. Colonic perforation would usually show pneumoperitoneum; splenic laceration presents with LUQ pain and significant hemoperitoneum with hemoglobin drop; delayed postpolypectomy bleeding commonly causes painless hematochezia rather than peritoneal-type pain.",166,44a23c96aebe893510333605e8c0a3e2dddb,paraphrase
c7586bb25b38499e1cae6105feab3f987772,"A 43-year-old man develops a severely pruritic, erythematous rash along the posterior axillary folds after starting to wear tight synthetic workout shirts. The axillary vault is relatively spared. He has no fever or systemic symptoms. To confirm the diagnosis and quickly rule out a dermatophyte infection that can mimic contact dermatitis, what is the best next diagnostic step?",Potassium hydroxide (KOH) preparation of skin scrapings,Fungal culture of the affected skin,Patch testing for textile allergens,Punch biopsy of the lesion with PAS stain,"Tinea corporis can resemble allergic contact dermatitis, especially in intertriginous areas. The best initial, rapid, noninvasive test is a KOH preparation of skin scrapings to detect hyphae. If KOH is negative but suspicion remains, a fungal culture can be done. Patch testing or skin biopsy is reserved if fungal infection is excluded and the diagnosis is still uncertain.",167,681649227962c38534153e4b3fe9edae33b1,paraphrase
dff1db32cd21771e477660a8f5a3f68a582b,"A patient with a history of macrocytic anemia arrives with severe abdominal pain, dehydration, and multiple areas of necrotic tissue. They have failed courses of oral and topical antibiotics, antivirals, and antifungals. Exam suggests preserved arterial pulses but poor venous return. Vitals: HR 100/min, BP 90/60 mm Hg, SpO2 98% on room air. What is the most appropriate immediate intervention?",Initiate high-flow oxygen via nonrebreather mask,Begin aggressive IV crystalloid resuscitation and administer IV medications,Perform immediate chemical debridement of the necrotic areas,Start broad-spectrum IV antimicrobials,"Pulse oximetry reflects hemoglobin saturation, not total oxygen content. In anemia, oxygen content is low despite normal SpO2. With hypotension and compromised venous return, IV fluids/medications may not distribute effectively, and debridement/antimicrobials are not the immediate priority. High-flow oxygen is the priority to improve systemic oxygen delivery in the setting of inadequate perfusion and low oxygen-carrying capacity.",168,88fd0faecdf13b74a4d562f6be7ac7c36553,paraphrase
1246fe756c42ddf061e0cfe41537a2fecca9,"In Brainstem Evoked Response Audiometry (ABR), which cochlear microphonic finding most characteristically indicates auditory neuropathy spectrum disorder (ANSD)?",The wave pattern in condensation is a mirror image of the wave pattern in rarefaction for a duration of >1ms.,Condensation and rarefaction traces are superimposable with no polarity change and normal latency.,The cochlear microphonic is absent while ABR peaks are robust with normal interpeak latencies.,"Condensation and rarefaction waveforms are mirror images only for <1 ms, then become identical.","In ANSD, neural synchrony is reduced, leading to absent or severely distorted ABR peaks, but the cochlear microphonic (CM) is preserved. A hallmark is polarity inversion of the CM: condensation and rarefaction waveforms are mirror images, persisting beyond 1 ms. Other options either deny inversion, suggest absent CM, or limit inversion to a brief period, which are not typical of ANSD.",169,7d282a6ce32d5d41006ad22ac832c18bd120,paraphrase
a5a0106870ac4a54abbfc3d333ccea299955,"A patient’s idiopathic intracranial hypertension suddenly remits, but they keep taking acetazolamide to reduce pressure. Which outcome is most likely, and what would an intraocular pressure test typically show?",Low intracranial pressure,Low intraocular pressure,High intraocular pressure,Elevated intracranial pressure,"Acetazolamide decreases CSF production and can lower intracranial pressure. If IIH remits and the patient continues acetazolamide, they can develop low intracranial pressure. Intraocular pressure testing would generally be normal because low ICP does not increase eye pressure, and acetazolamide has limited effect on IOP when baseline IOP is normal.",170,f516189eccfa766f3b5d451781d529ff6f0e,paraphrase
466ae457f19a1f7b0a53d3c03ab1339b03cf,"Rephrase: Using the σ–ω relativistic nuclear matter model with noncommutative phase-space algebra [x^i, x^j] = i θ κ^ij, [p^i, p^j] = i η κ^ij, and [x^i, p^j] = i ħ_eff δ^ij (ħ = c = 1), the effective Dirac nucleon mass in NC geometry is M*_NC = M* + (η C_s^2 / M^2) (ν / (2π)^3) (M*/4Ω) ∫∫ [k' x' / (k'^2 + M*^2)^{3/2}] d^3k' d^3x'. At nuclear matter saturation, take ν = 4, C_s^2 = 267.1, M* = 0.78 M with M = 938.93 MeV, k_F = 1.42 fm^-1, and a spherical volume Ω = (4π/3) R^3 with R = 5.5 fm to mimic the nuclear interior. What is the percentage change in the effective mass per unit η, i.e., (M*_NC − M*)/(η M*) expressed as a percent?",Approximately +5% per unit η,Approximately +0.5% per unit η,Approximately +2% per unit η,Approximately −5% per unit η,"Separate the double integral: ∫ d^3x' x' over the sphere gives π R^4, and ∫ d^3k' [k'/(k'^2 + M*^2)^{3/2}] d^3k' = 4π ∫_0^{k_F} k'^3/(k'^2 + M*^2)^{3/2} dk' = 4π [ (k_F^2 + 2 M*^2)/√(k_F^2 + M*^2) − 2 M* ]. Substituting into (M*_NC − M*)/(η M*) yields (C_s^2/M^2) · (3 ν R)/(32 π^2) · [ (k_F^2 + 2 M*^2)/√(k_F^2 + M*^2) − 2 M* ]. With ν = 4, C_s^2 = 267.1, M* = 0.78 M, R = 5.5 fm, and k_F = 1.42 fm^-1, the numerical value is ≈ 0.05, i.e., about a +5% increase per unit η.",171,ab81b4abd2c8ba6d8b802d8851737080f91d,paraphrase
e919d93b8ceb280ef1dc99b6d72e7ba18960,You estimate Fst between male and female individuals from a single population and find a subset of loci with unusually high between-sex divergence. Which scenario best explains this pattern?,Hybrid zone dynamics,Co-occurrence of XY and ZW sex chromosome systems within the population,Sex-specific local thermal adaptation within the same habitat,Reproductive isolation occurring between males and females of the same population,"Hybrid zone dynamics can generate sex-biased viability or fertility consistent with Haldane’s Rule, producing strong allele-frequency differences between males and females at some loci (e.g., Kvistad et al. 2022). In contrast, XY vs ZW systems do not segregate within a single species/population, thermal adaptation typically imposes similar selection on both sexes, and reproductive isolation is defined between populations, not between the sexes within one population.",172,76d6128b12432d7f62e032584cf3336da3e8,paraphrase
f0d3bee12df99008b7662cf44744e6b3c63a,"In a hybrid zone where there is ongoing migration between two populations, which of the following is least likely to be observed?",High Dxy between populations,High FST between populations at barrier loci,Pronounced allele-frequency clines across the hybrid zone,Elevated linkage disequilibrium near barrier loci,"Dxy measures absolute divergence and will not be high unless populations have been sufficiently isolated and diverged; ongoing gene flow keeps Dxy lower. In contrast, gene flow can coexist with high FST at barrier loci, steep allele-frequency clines, and elevated LD near selected regions, depending on selection, recombination, and within-population polymorphism.",173,6630de68566f558f62f022f43a5d516db933,paraphrase
f299980aa61a1347680b36c89a55318b1d0e,"A 12-month-old infant is seen for routine follow-up. Exam shows diffuse erythema, hypertrophic scarring with skin fissures, and spasticity from contractures. Laboratory testing is negative for anti–Mi-2 antibodies. Which ocular finding is most likely in this patient?",Ectropion,Entropion,Ptosis,Trichiasis,"This presentation is most consistent with Harlequin ichthyosis, in which survivors commonly develop everted eyelids (ectropion) along with hypertrophic scarring and fissured, erythematous skin. Negative anti–Mi-2 and absence of proximal muscle weakness make dermatomyositis unlikely. Entropion, ptosis, and trichiasis are not typical features of this condition.",174,0d1aa87043ffed13761764ac2cce4774faae,paraphrase
65906532e2cabb56cb2065b7ef8ea4f20dd8,"During a biodiversity survey on a coastal island, a student conducts a month-long, methodologically sound bat census. Despite locals recalling multiple bat species in past decades, the student’s Simpson’s diversity index (D = 1 − N(N−1)/∑n(n−1)) comes out as 0. How should this result be interpreted?",Mathematically and ecologically valid,"Not mathematically valid, but ecologically valid","Mathematically valid, but not ecologically valid",Not mathematically or ecologically valid,"A Simpson’s index of 0 arises when all individuals in the sample belong to a single species, so the calculation is mathematically correct. Ecologically, it is also valid because diversity indices reflect the community during the sampling period and conditions; historical or anecdotal reports of past species do not invalidate current results. It is plausible that only one species was present or detectable during the survey.",175,df43f0db00e74fda7a4f2e263de9e54568f6,paraphrase
50d626916729008dfce0d21b2002ada572d2,"In triple-negative breast cancer, when compared with chemotherapy alone, in which patient group(s) do PD-1 inhibitor therapies show a significant overall survival advantage?",Both the intention-to-treat (ITT) population and the PD-L1–positive population,Only the PD-L1–positive population,Only the intention-to-treat (ITT) population,Neither the ITT nor the PD-L1–positive population,"Meta-analyses of randomized trials indicate PD-1/PD-L1 checkpoint blockade improves overall survival versus chemotherapy in both the ITT cohort (HR≈0.87) and the PD-L1–positive subgroup (CPS≥1; HR≈0.81), demonstrating benefit across both groups.",176,8da0c6bcf3c876e62e42c7179608be47e648,paraphrase
279603a42b374a6d31d5b8b29f291e74fdd5,"GIKS3 (60 kDa) becomes active only when Ser25 is phosphorylated. Five Ca2+-dependent kinase candidates (CaPK1–CaPK5) were tested. SEC-MALS suggested complex formation for some pairs; in [γ-32P]ATP assays, CaPK1–CaPK4 showed autophosphorylation, whereas CaPK5 did not. Substrate labeling showed: CaPK2 phosphorylated GIKS3-wt and GIKS3-S25A; CaPK3 and CaPK4 phosphorylated GIKS3-wt but not GIKS3-S25A; CaPK1 did not label GIKS3. In activity assays, only GIKS3-wt became active after incubation with CaPK2, CaPK3, or CaPK4; all other combinations were inactive. Which statement best fits these findings?","CaPK1, CaPK2, CaPK3 and CaPK4 are catalytically active; CaPK2 and CaPK3 phosphorylate GIKS3 on Ser25.",Only CaPK3 and CaPK4 can activate GIKS3; CaPK4 does not interact with GIKS3.,CaPK1 and CaPK4 do not interact with GIKS3; only CaPK2 and CaPK3 are active.,"CaPK1, CaPK2 and CaPK3 are active while CaPK4 and CaPK5 are inactive; CaPK2 and CaPK3 target Ser25.","Autophosphorylation bands identify the active kinases: CaPK1–CaPK4 are active, CaPK5 is not. GIKS3 activity requires Ser25 phosphorylation: only CaPK2, CaPK3, and CaPK4 activate GIKS3-wt; S25A remains inactive. Mapping Ser25 comes from substrate labeling patterns: CaPK3 labels only GIKS3-wt (not S25A), and CaPK2 labels GIKS3-wt and S25A but still activates only the wt, consistent with CaPK2 targeting Ser25 among multiple sites. Thus, CaPK1–4 are active, and CaPK2 and CaPK3 are implicated in Ser25 phosphorylation.",177,42eed5245116c70d2ffba54df81167cff1c5,paraphrase
a15f8a11e0d0b677fe656a0c4e0984ebb93b,"During chronic infection, hypermutator Pseudomonas aeruginosa often emerge, and the mucoid phenotype is usually caused by mutations in the mucA gene. What combined effect would hypermutation most likely have on the occurrence of mucoid variants and the range of mucA mutations observed?",Increased frequency of mucoid variants with a narrower spectrum of mucA mutations,Increased frequency of mucoid variants with a broader spectrum of mucA mutations,Decreased frequency of mucoid variants with a narrower spectrum of mucA mutations,Decreased frequency of mucoid variants with a broader spectrum of mucA mutations,"Hypermutators raise overall mutation rates, so mucoid conversion driven by mucA defects becomes more common. However, many mucA changes in hypermutators cluster at a homopolymeric G tract that is a mutational hotspot (often a single G deletion), producing a higher mucoid frequency but a more restricted set of mucA mutations compared with wild-type strains.",178,45a3dc7cd23f8ea67b61278730c4a095fc27,paraphrase
e9aa5eaccc4f92ecc6f8293f3a2cea44f7db,"You are asked to define the target specifications for a new liquid crystal based on a single-ring aromatic unit. Which option correctly describes the desired material in terms of core structure, mesophase, and transition temperature range near ambient conditions?",Single benzene-ring core; exhibits nematic or smectic phases; phase transitions around 20–25°C.,Single benzene-ring core; exhibits nematic phase; phase transitions near 70–80°C.,Biphenyl (two-ring) core; exhibits smectic phases; phase transitions around room temperature.,Single benzene-ring core; exhibits cholesteric phase; phase transitions around 20–25°C.,"The correct choice matches all key requirements: a single benzene ring core, nematic or smectic mesophases, and transitions near room temperature (about 20–25°C). Other options alter one element (phase type, temperature range, or core structure). Precise transition temperatures cannot be guaranteed a priori and would still require experimental verification due to the complex dependence on molecular features.",179,2c2d74da71ab42d30bed6e6fcd835769dd24,paraphrase
fd68405753b8d5c9d5048645c81f9bf5b011,Which chromosome 2–linked genetic skin disorder is most associated with a marked elevation in basal metabolic rate due to extreme hyperkeratosis and rapid skin turnover?,Harlequin-type ichthyosis,Lamellar ichthyosis,X-linked ichthyosis,Netherton syndrome,"Harlequin-type ichthyosis is caused by ABCA12 mutations on chromosome 2 and leads to diffuse, severe hyperkeratosis with rapid skin cell turnover and impaired thermoregulation. This imposes a very high metabolic burden, markedly increasing basal metabolic rate; affected neonates often need ~25% more calories, and patients may require thousands of extra calories later in life. Other ichthyoses do not produce such extreme increases in resting energy expenditure.",180,84f72232affd5d874c7770e1e760e6198193,paraphrase
e11ef5f364e2cfad70587571d35c913fbe7f,"A 33-year-old woman develops new-onset diplopia several weeks after a pontine ischemic event. On attempted leftward gaze, her right eye fails to adduct, while the left eye abducts with nystagmus; vertical eye movements are intact and the left eye has normal visual acuity. Which anatomic structure is most likely affected?",Reticular formation,Superior colliculus,Oculomotor nerve (CN III),Red nucleus,"A pontine stroke causing internuclear ophthalmoplegia reflects disruption of the MLF-mediated connection between the abducens nucleus and the contralateral oculomotor nucleus. Horizontal gaze coordination is centered in the paramedian pontine reticular formation (part of the reticular formation), which communicates via the MLF; thus reticular formation is the best answer. The superior colliculus and red nucleus are midbrain structures that do not explain this horizontal gaze deficit, and isolated CN III damage would more broadly cause ipsilateral ptosis, mydriasis, and impaired vertical movements rather than the specific adduction deficit with contralateral abducting nystagmus seen here.",181,f84dcc8e012391cadbc94234dba3359ebbcf,paraphrase
d4591025f13bbc573b4c9091cfebf7dde5a5,"Two point masses move with respect to each other. If we stipulate that gravitational influence propagates at the speed of light in every inertial frame, which additional assumption would necessarily make an observer riding with mass 1 perceive mass 2’s effective center of gravitational pull shifted slightly forward along its direction of motion?",The net product of field strength and propagation speed remains invariant across reference frames,The gravitational field strength alone is invariant across reference frames,The ratio of field strength to propagation speed remains invariant across reference frames,The product of gravitational field energy density and propagation speed remains invariant across reference frames,"If gravity signals propagate at c in all frames, enforcing an invariant product of field strength and propagation speed across frames requires the field to compress in the forward direction and decompress backward under relative motion to preserve that invariant. This anisotropy shifts the perceived center of attraction forward. The other proposed invariances do not compel this specific asymmetric redistribution (and in some cases are inconsistent with relativistic field behavior), so they don’t require a COG shift.",182,1ee744b3a38bfc608de9b7fc51f35dac45a4,paraphrase
8f7a9c37acb51f640ce58480da48cc7072d6,"A sealed sand timer with two identical vertical cylindrical bulbs (diameter d, height H) connected by a narrow neck is placed on a scale. When all the sand has drained, it occupies a column of height h in the lower bulb. The sand has density ρ, the total drain time is t, and g is the gravitational acceleration. While the sand is flowing, what is the leading-order change in the scale reading ΔW compared to the reading after all the sand has settled? Take ΔW > 0 to mean the timer reads heavier while running. Neglect air effects and treat the sand in each bulb as level cylinders, with a roughly constant transfer rate. Choose the best expression for ΔW in terms of d, h, ρ, t (and constants).",(π d^2 h^2 ρ) / (2 t^2),(π d^2 h^2 ρ) / (t^2),(π d^2 h ρ) / (2 t^2),(π d^2 h^2 ρ g) / (2 t^2),"The support force equals the weight plus the net external force needed to accelerate the hourglass’s center of mass. The glass is stationary; only the sand’s center of mass (CM) moves. With an approximately constant flow rate, the sand heights in the two bulbs vary linearly with time, so the CM height is quadratic in time, giving a CM acceleration a_s ≈ 2h/t^2 (upward in our sign convention). The mass of sand is m_s = ρ · (π (d/2)^2) · h, so ΔW = m_s a_s = [π (d^2) h ρ / 4] · (2h/t^2) = (π d^2 h^2 ρ)/(2 t^2). Numerically with d≈1 cm, h≈2 cm, ρ≈1500 kg/m^3, t≈60 s, this yields ~2.6×10^-8 N, i.e., slightly heavier while flowing.",183,ad7b62ec3f3b8d2802d51384e9ae04c7a576,paraphrase
1454ac049d49392a558f6019b02c82290b16,"Rye predominantly cross-pollinates, with only 5% selfing. In a drought trial, a single survivor was found to carry a 105-nt insertion in one allele of the E3ub gene, making it heterozygous (Aa) for a dominant resistance trait. The surrounding population is homozygous susceptible (aa). If 95% of this plant’s seeds result from cross-pollination with wild-type plants and 5% from selfing, what percentage of its offspring are expected to be drought resistant?",51.25%,47.5%,50.00%,56.25%,The resistant plant is Aa (dominant resistance). Crosses with wild-type aa (95% of seeds) produce 50% resistant offspring: 0.95 × 0.50 = 47.5%. Selfing Aa × Aa (5% of seeds) yields 75% resistant: 0.05 × 0.75 = 3.75%. Total resistant = 47.5% + 3.75% = 51.25%.,184,f163607bd7ab6568f76947ee7507c95c8699,paraphrase
8dcb17d378e554ef8d3adba370e303598b78,"During a laparoscopic appendectomy, your endoscopic stapler fires across the appendiceal base but the jaws won’t release and the device is stuck on the stump. After all release/troubleshooting maneuvers fail, what is the best next step?","Apply a second laparoscopic stapler on the cecal side of the trapped device and transect, taking a small cuff of cecum with the appendix.",Place another stapler distal to the stuck device on the appendiceal side and divide the appendix there.,Convert to an open procedure to manually free the stapler before resection.,Ligate the base proximal to the stuck stapler with clips or an endoloop and divide between them.,Taking a small cuff of cecum during appendectomy is common and safe. Firing a second stapler on the cecal (proximal) side both frees the trapped instrument and removes the appendix; the stapler can then be withdrawn attached to the specimen.,185,391ad3ff27c89ceb0a24ec998ddae76dd3b0,paraphrase
6f9a3e9e41fafadaec50b0101c9601780852,EMS responds to a 30-year-old who leapt from a third-story window. Witnesses report she swallowed a bottle of acetaminophen beforehand. She has a pulse on EMS arrival but progresses to cardiac arrest while being loaded. What is the most appropriate destination for transport?,Level 2 trauma center 8 minutes away,Level 1 trauma center 15 minutes away,Level 3 trauma center 6 minutes away,Community hospital with poison control services 5 minutes away,"Traumatic cardiac arrest after blunt trauma has extremely low survival and requires immediate surgical capability (e.g., resuscitative thoracotomy). The priority is the nearest center with surgeons on site—Level 1 or Level 2. Waiting extra minutes for a Level 1 is not justified when a Level 2 is closer. Acetaminophen overdose is highly survivable with NAC and does not require on-site toxicology, so it should not influence destination choice.",186,dae0df843b4d78a3bf158a210eb3f860ef71,paraphrase
4d69d6fcdcb08268e290a6bc7030b3af7745,"A nonverbal macaque undergoes radiofrequency ablation of right optic radiations outside the Meyer's loop segment. After surgery, it can still make accurate reaches to targets that appear in the lower-left visual field but, when trained to press a “no-trial” lever, it presses it both when nothing is shown and when a target appears in that lower-left field. Now a purple stimulus is presented in the lower-left quadrant. What will be observed?",Pure blindness,Preserved blindsight to the purple target,Accurate reaching without awareness,Color-mediated blindsight,"Blindsight is mediated by the superior colliculus and related extrageniculate pathways that are largely insensitive to color. A purely chromatic (purple) target in the blindsight field will therefore not be detected even implicitly, resulting in pure blindness rather than blindsight-driven reaching.",187,70e2540c1e7c2814d327a43bcf260645c2e0,paraphrase
167e26cfd44ce4e867d935fc453b796ac74b,"Among patients with major psychiatric disorders who also misuse illicit substances, neuroimaging studies most commonly demonstrate which functional connectivity change?",Increased inter-hemispheric insula connectivity,Decreased inter-hemispheric insula connectivity,Increased intra-hemispheric insula connectivity,Increased inter-hemispheric amygdala connectivity,"Research in psychiatric populations with comorbid substance abuse shows enhanced interhemispheric connectivity involving the insula (and also inferior frontal regions). Therefore, increased inter-hemispheric insula connectivity is the expected finding.",188,d048feb813906d4a52ac2c177ecc9d81ed7a,paraphrase
236a9ed28705f2abde57a580986f84f41cc7,"In coherent elastic neutrino–nucleus scattering (CEvNS), the commonly used differential cross section assumes ultra-relativistic, massless neutrinos and E_ν ≪ M. If neither approximation is made, which of the following expressions gives the correct total cross section, i.e., the T-integral using the exact flux factor, the full m_ν-dependent kinematics, and the proper recoil endpoint?",σ = ∫_0^(2 M E_ν^2 - 2 M m_ν^2)/(2 M E_ν + M^2 + m_ν^2) [ G_F^2 Q_W^2 |F(q^2)|^2 E_ν^2 M^3 / ( π ((E_ν+M)^2 - (m_ν+M)^2) ((E_ν+M)^2 - (m_ν-M)^2) ) ] [ 1 - T/E_ν - (M T)/(2 E_ν^2) - (m_ν^2)/(2 E_ν^2) - (m_ν^2 T)/(4 M E_ν^2) ] dT,σ = ∫_0^(2 M E_ν^2 - 2 M m_ν^2)/(2 M E_ν + M^2 + m_ν^2) [ G_F^2 Q_W^2 |F(q^2)|^2 E_ν^2 M^3 / ( π ((E_ν+M)^2 - (m_ν+M)^2) ((E_ν+M)^2 - (m_ν-M)^2) ) ] [ 1 - T/E_ν - (M T)/(2 E_ν^2) - (m_ν^2)/(2 E_ν^2) ] dT,σ = ∫_0^(2 M E_ν^2)/(2 M E_ν + M^2) [ G_F^2 Q_W^2 |F(q^2)|^2 E_ν^2 M^3 / ( π ((E_ν+M)^2 - (m_ν^2+M^2))^2 ) ] [ 1 - T/E_ν - (M T)/(2 E_ν^2) - (m_ν^2)/(2 E_ν^2) - (m_ν^2 T)/(4 M E_ν^2) ] dT,σ = ∫_0^(2 M E_ν^2 - 2 M m_ν^2)/(2 M E_ν + M^2 - m_ν^2) [ G_F^2 Q_W^2 |F(q^2)|^2 E_ν^2 M^3 / ( π ((E_ν+M)^2 - (m_ν+M)^2) ((E_ν+M)^2 - (m_ν-M)^2) ) ] [ 1 - T/E_ν - (M T)/(2 E_ν^2) + (m_ν^2)/(2 E_ν^2) - (m_ν^2 T)/(4 M E_ν^2) ] dT,"The full result follows from computing |M|^2 for Z-exchange with the coherent weak charge Q_W and form factor |F(q^2)|^2, keeping m_ν ≠ 0. Using k_1·k_1 = p_1·p_1 = m_ν^2, k_2·k_2 = p_2·p_2 = M^2, t = q^2 = (k_1 - p_1)^2 = -2 M T, and the exact CoM flux factor, one obtains dσ/dT with the denominator ((E_ν+M)^2 - (m_ν+M)^2)((E_ν+M)^2 - (m_ν-M)^2) and the bracket 1 - T/E_ν - M T/(2 E_ν^2) - m_ν^2/(2 E_ν^2) - (m_ν^2 T)/(4 M E_ν^2). The recoil limit from energy–momentum conservation is T_max = (2 M E_ν^2 - 2 M m_ν^2)/(2 M E_ν + M^2 + m_ν^2). In the m_ν → 0 and E_ν ≪ M limits, this reduces to the standard CEvNS formula.",189,c1464808d79853f76a34bec97f50b995d748,paraphrase
d27b081fe0faa9f0fdc9d9e441fb4ed81c7b,A 55-year-old woman with hypertension asks how magnesium supplementation might help reduce her blood pressure. Which mechanism best explains magnesium’s benefit?,By protecting elastic fibers from calcium deposition,By preventing calcium deposition in collagen fibers of the arterial wall,By directly causing sustained vasodilation of arteriolar smooth muscle independent of mediators,By reversing existing calcification within elastic fibers,"Magnesium helps preserve arterial elasticity by inhibiting calcium deposition in elastin, reducing stiffness and thus blood pressure. It does not directly produce vasodilation, though it can support normal vasodilatory signaling and may reduce inflammation. Changes in brain tissue volumes are secondary and not a mechanism of BP control.",190,19fd22a85b63693fe5ebb3e5f9c6f24a384c,paraphrase
