answer string | choices dict | id string | kind string | lang string | question string | steps list | subject string | text string | topic string |
|---|---|---|---|---|---|---|---|---|---|
(D) Peroxisome; it oxidizes fatty acids and neutralizes alcohol and peroxides with catalase. | {
"A": "Ribosome",
"B": "Nucleolus",
"C": "Lysosome",
"D": "Peroxisome"
} | bio_0301 | multiple_choice | en | Which organelle breaks down fatty acids and detoxifies alcohol in liver cells? | [
"Eliminate protein-making and digestive organelles.",
"Recall catalase and beta-oxidation.",
"Choose peroxisome."
] | biology | Answer: (D) Peroxisomes detoxify and burn fats.
Steps:
1. Eliminate protein-making and digestive organelles.
2. Recall catalase and beta-oxidation.
3. Choose peroxisome.
Answer: (D) | cell_biology |
(B) 9:3:3:1; independent assortment of two heterozygous genes produces nine dual-dominant to three of each single to one recessive. | {
"A": "1:1",
"B": "9:3:3:1",
"C": "3:1",
"D": "1:0"
} | bio_0302 | multiple_choice | en | Which phenotypic ratio results from a dihybrid cross AaBb x AaBb with complete dominance? | [
"Eliminate monohybrid and test-cross ratios.",
"Recall 16-box Punnett outcomes.",
"Choose 9:3:3:1."
] | biology | Answer: (B) The dihybrid ratio is 9:3:3:1.
Steps:
1. Eliminate monohybrid and test-cross ratios.
2. Recall 16-box Punnett outcomes.
3. Choose 9:3:3:1.
Answer: (B) | genetics |
(A) Inheritance of acquired characteristics; Lamarck wrongly proposed that traits gained by use pass to offspring, as in stretching giraffe necks. | {
"A": "Inheritance of acquired characteristics",
"B": "Natural selection",
"C": "Genetic drift",
"D": "Gene flow"
} | bio_0303 | multiple_choice | en | Which idea did Lamarck propose to explain evolution? | [
"Eliminate Darwinian and population mechanisms.",
"Recall use-and-disuse inheritance.",
"Choose inheritance of acquired characteristics."
] | biology | Answer: (A) Lamarck proposed inheritance of acquired traits.
Steps:
1. Eliminate Darwinian and population mechanisms.
2. Recall use-and-disuse inheritance.
3. Choose inheritance of acquired characteristics.
Answer: (A) | evolution |
(C) Secondary consumers; they eat herbivores and sit on the third trophic level. | {
"A": "Producers",
"B": "Decomposers",
"C": "Secondary consumers",
"D": "Plants"
} | bio_0304 | multiple_choice | en | Which trophic level eats primary consumers? | [
"Eliminate energy capturers and recyclers.",
"Count levels upward from herbivores.",
"Choose secondary consumers."
] | biology | Answer: (C) Secondary consumers eat herbivores.
Steps:
1. Eliminate energy capturers and recyclers.
2. Count levels upward from herbivores.
3. Choose secondary consumers.
Answer: (C) | ecology |
(A) Skupina 0; její červené krvinky nenesou antigeny A ani B, takže je příjemci s jinými skupinami neodmítají. | {
"A": "0",
"B": "AB",
"C": "A",
"D": "B"
} | bio_0305 | multiple_choice | cs | Která krevní skupina je univerzálním dárcem červených krvinek? | [
"Vylučte skupiny s antigeny A nebo B.",
"Přiřaďte dárcovství chybějícím antigenům.",
"Zvolte skupinu 0."
] | biology | Answer: (A) Univerzálním dárcem je skupina 0.
Steps:
1. Vylučte skupiny s antigeny A nebo B.
2. Přiřaďte dárcovství chybějícím antigenům.
3. Zvolte skupinu 0.
Answer: (A) | physiology |
(D) Nitrogen; amino groups give proteins about 16 percent nitrogen, which sugars lack. | {
"A": "Carbon",
"B": "Hydrogen",
"C": "Oxygen",
"D": "Nitrogen"
} | bio_0306 | multiple_choice | en | Which element is found in proteins but not in carbohydrates? | [
"Eliminate elements shared by both.",
"Recall amino groups in proteins.",
"Choose nitrogen."
] | biology | Answer: (D) Proteins add nitrogen to C, H, and O.
Steps:
1. Eliminate elements shared by both.
2. Recall amino groups in proteins.
3. Choose nitrogen.
Answer: (D) | biochemistry |
About 15000 cells cover the sheet; sheet area divided by single-cell area gives the count. | null | bio_0307 | worked_example | en | An epithelial sheet measures 2 mm by 3 mm and each square cell is 20 micrometers per side. How many cells cover the sheet? | [
"Sheet area = 2000 x 3000 = 6,000,000 square micrometers.",
"Cell area = 20 x 20 = 400.",
"Divide 6,000,000 / 400 = 15000.",
"Recompute: 15000 x 400 = 6,000,000."
] | biology | Answer: About 15000 cells.
Steps:
1. Sheet area = 2000 x 3000 = 6,000,000 square micrometers.
2. Cell area = 20 x 20 = 400.
3. Divide 6,000,000 / 400 = 15000.
4. Recompute: 15000 x 400 = 6,000,000. | cell_biology |
1/64, about 1.56 percent; each gene contributes an independent 1/4 chance of homozygosity. | null | bio_0308 | worked_example | en | In a trihybrid cross AaBbCc x AaBbCc with independent assortment, what proportion of offspring is expected to be aabbcc? | [
"P(aa) = 1/4, P(bb) = 1/4, P(cc) = 1/4.",
"Multiply 1/4 x 1/4 x 1/4 = 1/64.",
"Convert to 0.015625, about 1.56 percent.",
"Recompute: 1/64 x 64 = 1."
] | biology | Answer: 1/64, about 1.56 percent.
Steps:
1. P(aa) = 1/4, P(bb) = 1/4, P(cc) = 1/4.
2. Multiply 1/4 x 1/4 x 1/4 = 1/64.
3. Convert to 0.015625, about 1.56 percent.
4. Recompute: 1/64 x 64 = 1. | genetics |
Equilibrium frequency is about 0.00316; recurrent mutation balances selection against the recessive allele. | null | bio_0309 | worked_example | en | A recessive allele arises by mutation at rate 1e-6 per generation and suffers selection coefficient 0.1. Estimate its mutation-selection balance frequency q = sqrt(mu/s). | [
"Divide mu/s = 1e-6/0.1 = 1e-5.",
"Take sqrt(1e-5) = 0.003162.",
"Recompute: 0.003162 squared x 0.1 = 1e-6."
] | biology | Answer: q about 0.00316.
Steps:
1. Divide mu/s = 1e-6/0.1 = 1e-5.
2. Take sqrt(1e-5) = 0.003162.
3. Recompute: 0.003162 squared x 0.1 = 1e-6. | evolution |
Ano, pyramidy odpovídají: býložravci drží 10 procent a masožravci 1 procento energie producentů. | null | bio_0310 | worked_example | cs | Na louce je 2000 kg producentů, 200 kg býložravců a 20 kg masožravců. Ověřte, zda platí přibližně 10procentní pravidlo přenosu. | [
"Podíl býložravců: 200/2000 = 0,1 = 10 procent.",
"Podíl masožravců: 20/200 = 0,1 = 10 procent.",
"Přepočítejte: 2000 x 0,1 x 0,1 = 20.",
"Pravidlo platí."
] | biology | Answer: Ano, každý přenos drží asi 10 procent.
Steps:
1. Podíl býložravců: 200/2000 = 0,1 = 10 procent.
2. Podíl masožravců: 20/200 = 0,1 = 10 procent.
3. Přepočítejte: 2000 x 0,1 x 0,1 = 20.
4. Pravidlo platí. | ecology |
Minute ventilation is 40 liters per minute; deep fast breathing multiplies resting airflow many times over. | null | bio_0311 | worked_example | en | During exercise tidal volume is 2000 mL and breathing rate is 20 per minute. Calculate minute ventilation. | [
"Multiply 2000 x 20 = 40000 mL per min.",
"Convert to 40 L per min.",
"Recompute: 40 x 1000 / 20 = 2000 mL."
] | biology | Answer: 40 L per min.
Steps:
1. Multiply 2000 x 20 = 40000 mL per min.
2. Convert to 40 L per min.
3. Recompute: 40 x 1000 / 20 = 2000 mL. | physiology |
A noncompetitive inhibitor binds outside the active site and deforms the enzyme so it works slower; unlike competitive inhibition, excess substrate cannot overcome it, so Vmax falls while Km stays the same. | null | bio_0312 | concept | en | Explain noncompetitive enzyme inhibition and contrast it with competitive inhibition. | [
"Bind inhibitor at a separate site.",
"Deform the enzyme and lower Vmax.",
"Contrast with substrate-overcome competition."
] | biology | Answer: Noncompetitive inhibitors deform from afar and resist substrate rescue.
Steps:
1. Bind inhibitor at a separate site.
2. Deform the enzyme and lower Vmax.
3. Contrast with substrate-overcome competition. | biochemistry |
Gap junctions are protein channels joining animal cells for ions and signals, while plasmodesmata are membrane-lined pores through plant walls linking cytoplasms; both let neighbors coordinate directly. | null | bio_0313 | concept | en | Compare gap junctions and plasmodesmata. | [
"Define animal protein channels.",
"Define plant wall pores.",
"Link both to direct coordination."
] | biology | Answer: Both bridge cytoplasms: channels in animals, pores in plants.
Steps:
1. Define animal protein channels.
2. Define plant wall pores.
3. Link both to direct coordination. | cell_biology |
Gene editing uses programmed nucleases that cut DNA at a chosen sequence, letting the cell's repair insert, delete, or swap bases there; it can correct mutations but risks off-target cuts. | null | bio_0314 | concept | en | Explain the basic idea of gene editing. | [
"Target a DNA sequence.",
"Cut with a nuclease.",
"Repair with an inserted change."
] | biology | Answer: Programmed cuts let cells rewrite chosen DNA.
Steps:
1. Target a DNA sequence.
2. Cut with a nuclease.
3. Repair with an inserted change. | genetics |
Při umělém výběru vybírá rodičovské jedince člověk podle užitku či vzhledu, například plemena psů nebo výnosné obilí; na rozdíl od přírodního výběru rozhoduje lidský záměr, nikoli přežití v divočině. | null | bio_0315 | concept | cs | Vysvětlete umělý výběr a jeho rozdíl od přírodního. | [
"Definujte výběr člověkem.",
"Uveďte plemena a odrůdy.",
"Porovnejte s přírodním výběrem."
] | biology | Answer: Umělý výběr řídí člověk, přírodní přežití.
Steps:
1. Definujte výběr člověkem.
2. Uveďte plemena a odrůdy.
3. Porovnejte s přírodním výběrem. | evolution |
Biological control deploys natural enemies such as ladybugs against aphids or parasitic wasps against caterpillars, cutting pesticide use, though introduced agents can themselves become pests. | null | bio_0316 | concept | en | Explain biological pest control with an example. | [
"Release predators or parasites.",
"Suppress pests without chemicals.",
"Watch for agent side effects."
] | biology | Answer: Natural enemies can replace chemical sprays.
Steps:
1. Release predators or parasites.
2. Suppress pests without chemicals.
3. Watch for agent side effects. | ecology |
Danger triggers adrenaline and sympathetic nerves that quicken the heart, widen airways, dilate pupils, and shunt blood to muscles while pausing digestion, readying the body to fight or flee. | null | bio_0317 | concept | en | Describe the fight-or-flight response. | [
"Release adrenaline via sympathetic nerves.",
"Boost heart, lungs, and muscles.",
"Pause digestion and rest functions."
] | biology | Answer: Adrenaline arms the body to fight or flee.
Steps:
1. Release adrenaline via sympathetic nerves.
2. Boost heart, lungs, and muscles.
3. Pause digestion and rest functions. | physiology |
Oxidation removes electrons (often with hydrogen) and reduction adds them; in respiration glucose is oxidized while NAD+ is reduced to NADH, which later pays electrons to the transport chain. | null | bio_0318 | concept | en | Explain oxidation and reduction in metabolism with an example. | [
"Define electron loss and gain.",
"Oxidize glucose to CO2.",
"Reduce NAD+ to NADH as carrier."
] | biology | Answer: Oxidation strips electrons; reduction banks them in carriers.
Steps:
1. Define electron loss and gain.
2. Oxidize glucose to CO2.
3. Reduce NAD+ to NADH as carrier. | biochemistry |
Stem cells divide into copies of themselves plus specialized daughters; totipotent zygotes make everything, pluripotent embryo cells make all tissues, and multipotent adult cells make a few, offering repair therapies. | null | bio_0319 | concept | en | What are stem cells and why do they matter medically? | [
"Define self-renewal plus differentiation.",
"Rank potency from toto to multi.",
"Apply to tissue repair."
] | biology | Answer: Stem cells self-copy and supply fresh tissues.
Steps:
1. Define self-renewal plus differentiation.
2. Rank potency from toto to multi.
3. Apply to tissue repair. | cell_biology |
Karyotyp je uspořádaný obraz chromozomů buňky; počítáním a tvarem odhalí trisomii 21, syndromy XXY či X0 a přestavby chromozomů. | null | bio_0320 | concept | cs | Vysvětlete, co je karyotyp a co se z něj pozná. | [
"Definujte seřazené chromozomy.",
"Počítejte počet a tvar.",
"Odhalte trisomie a přestavby."
] | biology | Answer: Karyotyp mapuje chromozomy a jejich poruchy.
Steps:
1. Definujte seřazené chromozomy.
2. Počítejte počet a tvar.
3. Odhalte trisomie a přestavby. | genetics |
(B) Human arm and bat wing; shared bone patterns adapted to grasping versus flying show descent with modification from a common ancestor. | {
"A": "Dolphin and shark",
"B": "Human arm and bat wing",
"C": "Cactus spine and rose thorn",
"D": "Butterfly and bird wing"
} | bio_0321 | multiple_choice | en | Which pair shows divergent evolution from a common ancestor? | [
"Eliminate convergent lookalikes.",
"Recall shared tetrapod bones.",
"Choose human arm and bat wing."
] | biology | Answer: (B) Arms and wings diverged from one limb.
Steps:
1. Eliminate convergent lookalikes.
2. Recall shared tetrapod bones.
3. Choose human arm and bat wing.
Answer: (B) | evolution |
(D) Sunlight; light, temperature, water, and minerals are nonliving factors that shape ecosystems. | {
"A": "Bacteria",
"B": "Grass",
"C": "Fungi",
"D": "Sunlight"
} | bio_0322 | multiple_choice | en | Which factor is abiotic? | [
"Eliminate living organisms.",
"Recall nonliving conditions.",
"Choose sunlight."
] | biology | Answer: (D) Sunlight is an abiotic factor.
Steps:
1. Eliminate living organisms.
2. Recall nonliving conditions.
3. Choose sunlight.
Answer: (D) | ecology |
(C) Medulla oblongata; this brainstem center drives involuntary breathing rhythm and heart rate. | {
"A": "Cerebrum",
"B": "Cerebellum",
"C": "Medulla oblongata",
"D": "Hypothalamus"
} | bio_0323 | multiple_choice | en | Which brain part controls breathing and heart rate? | [
"Eliminate thinking and coordination centers.",
"Eliminate hormone control.",
"Choose medulla oblongata."
] | biology | Answer: (C) The medulla drives breath and heartbeat.
Steps:
1. Eliminate thinking and coordination centers.
2. Eliminate hormone control.
3. Choose medulla oblongata.
Answer: (C) | physiology |
(A) Deoxyribose; DNA uses deoxyribose while RNA uses ribose, a difference of one oxygen atom. | {
"A": "Deoxyribose",
"B": "Ribose",
"C": "Glucose",
"D": "Fructose"
} | bio_0324 | multiple_choice | en | Which sugar is found in DNA? | [
"Eliminate RNA and energy sugars.",
"Recall deoxy versus oxy.",
"Choose deoxyribose."
] | biology | Answer: (A) DNA holds deoxyribose.
Steps:
1. Eliminate RNA and energy sugars.
2. Recall deoxy versus oxy.
3. Choose deoxyribose.
Answer: (A) | biochemistry |
(D) Centromera; drží sesterské chromatidy pohromadě a váže vlákna vřeténka. | {
"A": "Telomera",
"B": "Vřeténko",
"C": "Jaderná lamina",
"D": "Centromera"
} | bio_0325 | multiple_choice | cs | Která struktura spojuje sesterské chromatidy? | [
"Vylučte konce, vlákna a výstelku.",
"Přiřaďte spojení chromatid.",
"Zvolte centromeru."
] | biology | Answer: (D) Chromatidy spojuje centromera.
Steps:
1. Vylučte konce, vlákna a výstelku.
2. Přiřaďte spojení chromatid.
3. Zvolte centromeru.
Answer: (D) | cell_biology |
(C) Single-base deletion; losing one base shifts every downstream codon, unlike substitutions that alter only one. | {
"A": "Silent substitution",
"B": "Missense substitution",
"C": "Single-base deletion",
"D": "Whole-gene duplication"
} | bio_0326 | multiple_choice | en | Which mutation causes a frameshift? | [
"Eliminate one-codon substitutions.",
"Eliminate copy-number changes.",
"Choose single-base deletion."
] | biology | Answer: (C) A deleted base shifts the frame.
Steps:
1. Eliminate one-codon substitutions.
2. Eliminate copy-number changes.
3. Choose single-base deletion.
Answer: (C) | genetics |
Frequency of A jumps from 0.6 to 0.8; chance survival of mostly AA birds shows how bottlenecks shift alleles regardless of fitness. | null | bio_0327 | worked_example | en | A storm leaves 10 surviving birds: 6 AA and 4 Aa. Allele A was 0.6 before. Calculate its frequency in survivors. | [
"Count A copies: 6x2 + 4 = 16.",
"Total alleles = 10 x 2 = 20.",
"Divide 16/20 = 0.8.",
"Recompute: 16 + 4 = 20; 0.8 vs old 0.6."
] | biology | Answer: A rises to 0.8 in survivors.
Steps:
1. Count A copies: 6x2 + 4 = 16.
2. Total alleles = 10 x 2 = 20.
3. Divide 16/20 = 0.8.
4. Recompute: 16 + 4 = 20; 0.8 vs old 0.6. | evolution |
About 24000 daisies; quadrat density scales to the whole field area. | null | bio_0328 | worked_example | en | A 1 square meter quadrat holds 12 daisies. Estimate the total in a 2000 square meter field. | [
"Density = 12 per square meter.",
"Multiply 12 x 2000 = 24000.",
"Recompute: 24000/2000 = 12."
] | biology | Answer: About 24000 daisies.
Steps:
1. Density = 12 per square meter.
2. Multiply 12 x 2000 = 24000.
3. Recompute: 24000/2000 = 12. | ecology |
MAP is about 106.7 mmHg; pulse pressure of 50 adds one third onto the diastolic base of 90. | null | bio_0329 | worked_example | en | Blood pressure is 140 over 90 mmHg. Estimate mean arterial pressure (MAP) using MAP = diastolic + (systolic minus diastolic)/3. | [
"Pulse pressure = 140-90 = 50.",
"Divide 50/3 = 16.67.",
"Add 90+16.67 = 106.67.",
"Recompute: (106.67-90)x3 = 50."
] | biology | Answer: MAP about 106.7 mmHg.
Steps:
1. Pulse pressure = 140-90 = 50.
2. Divide 50/3 = 16.67.
3. Add 90+16.67 = 106.67.
4. Recompute: (106.67-90)x3 = 50. | physiology |
Je potřeba 10 ml zásobního roztoku; ředění zachovává látkové množství. | null | bio_0330 | worked_example | cs | Ze zásobního roztoku 1 mol na litr připravte 200 ml roztoku 0,05 mol na litr. Jaký objem zásobního roztoku je potřeba? | [
"Zapište V1 = C2V2 / C1.",
"Dosaďte 0,05 x 200 / 1.",
"Vypočítejte 10 / 1 = 10 ml.",
"Přepočítejte: 1 x 10 = 10 a 0,05 x 200 = 10."
] | biology | Answer: 10 ml zásobního roztoku.
Steps:
1. Zapište V1 = C2V2 / C1.
2. Dosaďte 0,05 x 200 / 1.
3. Vypočítejte 10 / 1 = 10 ml.
4. Přepočítejte: 1 x 10 = 10 a 0,05 x 200 = 10. | biochemistry |
About 1.02 meters of DNA per cell; three billion base pairs at 0.34 nanometers each stretch past a meter when unwound. | null | bio_0331 | worked_example | en | The human genome has 3 billion base pairs and each pair spans 0.34 nanometers. Estimate the total DNA length in one cell. | [
"Multiply 3e9 x 0.34e-9 m = 1.02 m.",
"Recompute: 1.02 / 0.34e-9 = 3e9 pairs.",
"Report about 1.02 m."
] | biology | Answer: About 1.02 m of DNA per cell.
Steps:
1. Multiply 3e9 x 0.34e-9 m = 1.02 m.
2. Recompute: 1.02 / 0.34e-9 = 3e9 pairs.
3. Report about 1.02 m. | cell_biology |
Imprinting silences certain genes depending on which parent contributed them, through methylation marks; losing the active paternal copy causes Prader-Willi traits while losing the maternal copy causes Angelman traits. | null | bio_0332 | concept | en | What is genomic imprinting? Give an example. | [
"Define parent-of-origin silencing.",
"Mark alleles by methylation.",
"Contrast Prader-Willi and Angelman."
] | biology | Answer: Imprinting expresses genes from one parent only.
Steps:
1. Define parent-of-origin silencing.
2. Mark alleles by methylation.
3. Contrast Prader-Willi and Angelman. | genetics |
About 540 million years ago nearly all animal body plans appeared within a few million years, likely spurred by rising oxygen, new predator-prey dynamics, and Hox gene toolkits enabling complex forms. | null | bio_0333 | concept | en | What was the Cambrian explosion and what may have caused it? | [
"Date rapid body-plan origins.",
"Name oxygen and predation drivers.",
"Credit Hox gene toolkits."
] | biology | Answer: Animal forms burst forth with oxygen and teeth.
Steps:
1. Date rapid body-plan origins.
2. Name oxygen and predation drivers.
3. Credit Hox gene toolkits. | evolution |
Sulfur dioxide and nitrogen oxides from burning fossil fuels form sulfuric and nitric acids in rain that acidify lakes, leach soil nutrients, and damage leaves and stonework. | null | bio_0334 | concept | en | Explain the causes and effects of acid rain. | [
"Burn fuels into SO2 and NOx.",
"Form acids in precipitation.",
"Acidify lakes and strip soils."
] | biology | Answer: Fossil fumes sour rain that wounds water and woods.
Steps:
1. Burn fuels into SO2 and NOx.
2. Form acids in precipitation.
3. Acidify lakes and strip soils. | ecology |
Při pohledu nablízko se řasnaté tělísko stáhne, čočka se vyklene a lomivost vzroste; při pohledu do dálky se čočka zploští a oko zaostří bez námahy. | null | bio_0335 | concept | cs | Vysvětlete, jak oko zaostřuje na různou vzdálenost. | [
"Popište uvolněnou čočku do dálky.",
"Stáhněte řasnaté tělísko nablízko.",
"Vysvětlete změnu lomivosti."
] | biology | Answer: Čočka mění tvar a tím ohnisko.
Steps:
1. Popište uvolněnou čočku do dálky.
2. Stáhněte řasnaté tělísko nablízko.
3. Vysvětlete změnu lomivosti. | physiology |
NAD+ and FAD accept electrons and hydrogen during glycolysis and the Krebs cycle, becoming NADH and FADH2 that deliver the cargo to the electron transport chain for ATP production. | null | bio_0336 | concept | en | Explain the roles of NAD+ and FAD in cellular respiration. | [
"Collect electrons from fuel.",
"Carry them as NADH and FADH2.",
"Unload at the transport chain."
] | biology | Answer: NAD+ and FAD ferry electrons to the chain.
Steps:
1. Collect electrons from fuel.
2. Carry them as NADH and FADH2.
3. Unload at the transport chain. | biochemistry |
Benign tumors grow locally inside a capsule and rarely threaten life, while malignant tumors invade nearby tissue and metastasize through blood and lymph to found distant colonies. | null | bio_0337 | concept | en | Contrast benign and malignant tumors. | [
"Define contained local growth.",
"Define invasion of neighbors.",
"Define distant metastasis."
] | biology | Answer: Benign lumps stay; malignant cancers spread.
Steps:
1. Define contained local growth.
2. Define invasion of neighbors.
3. Define distant metastasis. | cell_biology |
Germline mutations occur in gamete-forming cells and pass to offspring, while somatic mutations arise in body cells and affect only the individual, as in most skin cancers. | null | bio_0338 | concept | en | Distinguish germline from somatic mutations. | [
"Place changes in gamete lines.",
"Place changes in body cells.",
"Trace inheritance versus single-body effects."
] | biology | Answer: Germline edits heirs; somatic edits one body.
Steps:
1. Place changes in gamete lines.
2. Place changes in body cells.
3. Trace inheritance versus single-body effects. | genetics |
Predators evolve speed, venom, and sharper senses while prey evolve armor, camouflage, and toxins in return, each advance selecting counter-advances in an endless evolutionary race. | null | bio_0339 | concept | en | Explain predator-prey evolutionary arms races. | [
"Arm predators with speed and venom.",
"Shield prey with armor and toxins.",
"Escalate both sides by selection."
] | biology | Answer: Hunters and hunted escalate weapons together.
Steps:
1. Arm predators with speed and venom.
2. Shield prey with armor and toxins.
3. Escalate both sides by selection. | evolution |
Pesticidy, úbytek květů a nemoci decimují včely a motýly; bez opylovačů klesají výnosy ovoce a semen a hroutí se potravní sítě. | null | bio_0340 | concept | cs | Vysvětlete, jak člověk ohrožuje opylovače a proč to vadí. | [
"Uveďte pesticidy a úbytek květů.",
"Popište úhyn včel a motýlů.",
"Varujte před propadem úrody."
] | biology | Answer: Mizející opylovači ohrožují úrodu i přírodu.
Steps:
1. Uveďte pesticidy a úbytek květů.
2. Popište úhyn včel a motýlů.
3. Varujte před propadem úrody. | ecology |
(B) Cardiac; heart muscle is striated like skeletal muscle but contracts involuntarily under autonomic control. | {
"A": "Skeletal",
"B": "Cardiac",
"C": "Smooth",
"D": "All muscle types"
} | bio_0341 | multiple_choice | en | Which muscle type is both striated and involuntary? | [
"Eliminate voluntary striated muscle.",
"Eliminate involuntary smooth muscle.",
"Choose cardiac."
] | biology | Answer: (B) Cardiac muscle is striated yet involuntary.
Steps:
1. Eliminate voluntary striated muscle.
2. Eliminate involuntary smooth muscle.
3. Choose cardiac.
Answer: (B) | physiology |
(A) Hemoglobin; this iron-containing protein in red blood cells binds oxygen in the lungs. | {
"A": "Hemoglobin",
"B": "Starch",
"C": "Cholesterol",
"D": "Glucose"
} | bio_0342 | multiple_choice | en | Which substance is a protein? | [
"Eliminate carbohydrate and lipid options.",
"Recall oxygen transport by protein.",
"Choose hemoglobin."
] | biology | Answer: (A) Hemoglobin is a protein.
Steps:
1. Eliminate carbohydrate and lipid options.
2. Recall oxygen transport by protein.
3. Choose hemoglobin.
Answer: (A) | biochemistry |
(C) Anaphase; sister chromatids separate to opposite poles right after lining up at the equator. | {
"A": "Prophase",
"B": "Telophase",
"C": "Anaphase",
"D": "Interphase"
} | bio_0343 | multiple_choice | en | Which mitotic stage comes immediately after metaphase? | [
"Eliminate earlier and later stages.",
"Eliminate the non-dividing phase.",
"Choose anaphase."
] | biology | Answer: (C) Anaphase follows metaphase.
Steps:
1. Eliminate earlier and later stages.
2. Eliminate the non-dividing phase.
3. Choose anaphase.
Answer: (C) | cell_biology |
(D) XO; a single X chromosome in females causes short stature and infertility. | {
"A": "XXY",
"B": "XYY",
"C": "Trisomy 21",
"D": "XO (single X)"
} | bio_0344 | multiple_choice | en | Which chromosome set causes Turner syndrome? | [
"Eliminate extra-Y and trisomy options.",
"Recall one missing X.",
"Choose XO."
] | biology | Answer: (D) Turner syndrome is XO.
Steps:
1. Eliminate extra-Y and trisomy options.
2. Recall one missing X.
3. Choose XO.
Answer: (D) | genetics |
(B) Molekulární důkazy; podobnost sekvencí DNA a bílkovin odpovídá příbuznosti druhů. | {
"A": "Fosilie",
"B": "Molekulární důkazy",
"C": "Embryologie",
"D": "Biogeografie"
} | bio_0345 | multiple_choice | cs | Který důkaz evoluce porovnává sekvence DNA různých druhů? | [
"Vylučte kosti, zárodky a rozšíření.",
"Přiřaďte sekvence molekulám.",
"Zvolte molekulární důkazy."
] | biology | Answer: (B) Sekvence porovnávají molekulární důkazy.
Steps:
1. Vylučte kosti, zárodky a rozšíření.
2. Přiřaďte sekvence molekulám.
3. Zvolte molekulární důkazy.
Answer: (B) | evolution |
(A) Lichens and mosses; they colonize bare rock, trap dust, and build the first thin soil for later plants. | {
"A": "Lichens and mosses",
"B": "Oak trees",
"C": "Deer",
"D": "Mushrooms"
} | bio_0346 | multiple_choice | en | Which organisms are pioneer species on bare rock? | [
"Eliminate animals and late trees.",
"Eliminate decomposers needing soil.",
"Choose lichens and mosses."
] | biology | Answer: (A) Lichens and mosses pioneer bare rock.
Steps:
1. Eliminate animals and late trees.
2. Eliminate decomposers needing soil.
3. Choose lichens and mosses.
Answer: (A) | ecology |
About 162 bpm; 85 percent of the age-predicted maximum of 190 sets the hard-training ceiling. | null | bio_0347 | worked_example | en | Estimate the 85 percent training-zone pulse for a 30-year-old using max HR = 220 minus age. | [
"Max = 220-30 = 190 bpm.",
"Zone = 0.85 x 190 = 161.5.",
"Recompute: 161.5/190 = 0.85.",
"Report about 162 bpm."
] | biology | Answer: About 162 bpm.
Steps:
1. Max = 220-30 = 190 bpm.
2. Zone = 0.85 x 190 = 161.5.
3. Recompute: 161.5/190 = 0.85.
4. Report about 162 bpm. | physiology |
216 grams of water; each glucose yields 6 water molecules of 18 grams per mole. | null | bio_0348 | worked_example | en | Respiration makes 6 water molecules per glucose. What mass of water forms from 2 moles of glucose? | [
"Moles of water = 2 x 6 = 12.",
"Mass = 12 x 18 = 216 g.",
"Recompute: 216/18 = 12; 12/6 = 2 mol glucose."
] | biology | Answer: 216 g of water.
Steps:
1. Moles of water = 2 x 6 = 12.
2. Mass = 12 x 18 = 216 g.
3. Recompute: 216/18 = 12; 12/6 = 2 mol glucose. | biochemistry |
3000x magnification; the 15 mm image is three thousand times the 5 micrometer object. | null | bio_0349 | worked_example | en | A pollen grain is 5 micrometers wide. In a photo it appears 15 millimeters wide. Calculate the magnification. | [
"Convert 15 mm = 15000 micrometers.",
"Divide 15000 / 5 = 3000.",
"Recompute: 3000 x 5 = 15000.",
"Report 3000x."
] | biology | Answer: 3000x magnification.
Steps:
1. Convert 15 mm = 15000 micrometers.
2. Divide 15000 / 5 = 3000.
3. Recompute: 3000 x 5 = 15000.
4. Report 3000x. | cell_biology |
Fialový rodič je heterozygot Aa; poměr bílých kolem poloviny odpovídá testovacímu křížení. | null | bio_0350 | worked_example | cs | Fialový květ neznámého genotypu byl křížen s bílým (aa). Ze 200 potomků bylo 98 bílých. Určete genotyp fialového rodiče. | [
"Podíl bílých = 98/200 = 0,49, asi 1/2.",
"Poměr 1:1 znamená křížení Aa x aa.",
"Rodič je tedy Aa.",
"Přepočítejte: očekáváno 100 bílých, pozorováno 98."
] | biology | Answer: Rodič je Aa.
Steps:
1. Podíl bílých = 98/200 = 0,49, asi 1/2.
2. Poměr 1:1 znamená křížení Aa x aa.
3. Rodič je tedy Aa.
4. Přepočítejte: očekáváno 100 bílých, pozorováno 98. | genetics |
p equals 0.5 and q equals 0.5; the symmetric genotype counts split the 120 alleles evenly. | null | bio_0351 | worked_example | en | In 60 diploid beetles, genotypes are BB 15, Bb 30, and bb 15. Calculate allele frequencies p and q. | [
"Total alleles = 60 x 2 = 120.",
"B copies = 15x2 + 30 = 60, so p = 60/120 = 0.5.",
"b copies = 15x2 + 30 = 60, so q = 0.5.",
"Recompute: 60+60=120 and p+q=1.0."
] | biology | Answer: p=0.5, q=0.5.
Steps:
1. Total alleles = 60 x 2 = 120.
2. B copies = 15x2 + 30 = 60, so p = 60/120 = 0.5.
3. b copies = 15x2 + 30 = 60, so q = 0.5.
4. Recompute: 60+60=120 and p+q=1.0. | evolution |
Top-down control runs from predators that suppress herbivores and free plants, while bottom-up control runs from nutrient and plant productivity that limit everything above. | null | bio_0352 | concept | en | Contrast top-down and bottom-up control of food webs. | [
"Push control down from predators.",
"Push control up from nutrients.",
"Give wolf versus fertilizer effects."
] | biology | Answer: Predators press down; nutrients push up.
Steps:
1. Push control down from predators.
2. Push control up from nutrients.
3. Give wolf versus fertilizer effects. | ecology |
Liver enzymes oxidize alcohol and drugs into water-soluble forms excreted in bile or urine, while also clearing ammonia into urea; chronic overload scars the liver into cirrhosis. | null | bio_0353 | concept | en | Explain how the liver processes toxins and alcohol. | [
"Oxidize toxins with enzymes.",
"Excrete products in bile and urine.",
"Convert ammonia to urea."
] | biology | Answer: The liver dissolves poisons for disposal.
Steps:
1. Oxidize toxins with enzymes.
2. Excrete products in bile and urine.
3. Convert ammonia to urea. | physiology |
Buffers pair a weak acid with its base so added H+ or OH- convert into weak partners; blood bicarbonate soaks up acids from exercise and releases CO2 through the lungs. | null | bio_0354 | concept | en | Explain how buffers resist pH change, with the blood example. | [
"Pair weak acid with conjugate base.",
"Soak up added H+ or OH-.",
"Vent CO2 to reset balance."
] | biology | Answer: Buffers swallow acids and bases alike.
Steps:
1. Pair weak acid with conjugate base.
2. Soak up added H+ or OH-.
3. Vent CO2 to reset balance. | biochemistry |
Drsné retikulum s ribozomy vyrábí bílkoviny pro výdej, hladké bez ribozomů syntetizuje lipidy, odbourává jedy a hospodaří s vápníkem. | null | bio_0355 | concept | cs | Popište drsné a hladké endoplazmatické retikulum a jejich funkce. | [
"Přiřaďte drsnému syntézu bílkovin.",
"Přiřaďte hladkému lipidy a detoxikaci.",
"Uveďte hospodaření s vápníkem."
] | biology | Answer: Drsné tvoří bílkoviny, hladké lipidy a čistí.
Steps:
1. Přiřaďte drsnému syntézu bílkovin.
2. Přiřaďte hladkému lipidy a detoxikaci.
3. Uveďte hospodaření s vápníkem. | cell_biology |
Plants tolerate extra chromosome sets that can boost size and instantly isolate new species, while in humans whole-genome duplication unbalances development and is almost always lethal early. | null | bio_0356 | concept | en | Contrast polyploidy in plants versus humans. | [
"Note plant tolerance and vigor.",
"Note human dosage imbalance.",
"Link plant success to speciation."
] | biology | Answer: Plants thrive polyploid; humans cannot.
Steps:
1. Note plant tolerance and vigor.
2. Note human dosage imbalance.
3. Link plant success to speciation. | genetics |
Microevolution is allele-frequency change within populations, such as pesticide resistance, while macroevolution is the accumulation of such changes into new species and major forms over geological time. | null | bio_0357 | concept | en | Distinguish microevolution from macroevolution. | [
"Define within-population shifts.",
"Define species-level outcomes.",
"Link small steps to large patterns."
] | biology | Answer: Micro shifts alleles; macro builds species.
Steps:
1. Define within-population shifts.
2. Define species-level outcomes.
3. Link small steps to large patterns. | evolution |
Invaders escape their native enemies, outcompete locals, and rewire food webs, as zebra mussels filter lakes bare, cane toads poison predators, and kudzu smothers forests. | null | bio_0358 | concept | en | Explain how invasive species harm ecosystems, with an example. | [
"Escape enemies in new lands.",
"Outcompete native species.",
"Rewire food webs and habitats."
] | biology | Answer: Invaders conquer without enemies and starve natives.
Steps:
1. Escape enemies in new lands.
2. Outcompete native species.
3. Rewire food webs and habitats. | ecology |
Type 1 diabetes destroys pancreatic beta cells so no insulin is made and injections are required, while type 2 keeps insulin but tissues resist it, managed first by diet, exercise, and sensitizing drugs. | null | bio_0359 | concept | en | Contrast type 1 and type 2 diabetes. | [
"Destroy beta cells in type 1.",
"Resist insulin in type 2.",
"Treat with replacement versus sensitizing."
] | biology | Answer: Type 1 lacks insulin; type 2 resists it.
Steps:
1. Destroy beta cells in type 1.
2. Resist insulin in type 2.
3. Treat with replacement versus sensitizing. | physiology |
Žluč emulguje tuky na kapénky, lipázy je štěpí na mastné kyseliny a glycerol a micely je dopraví ke vstřebání do střevní stěny. | null | bio_0360 | concept | cs | Vysvětlete trávení a vstřebávání tuků. | [
"Popište emulgaci žlučí.",
"Vysvětlete štěpení lipázami.",
"Sledujte micely ke vstřebání."
] | biology | Answer: Žluč emulguje, lipázy štěpí, micely doručí.
Steps:
1. Popište emulgaci žlučí.
2. Vysvětlete štěpení lipázami.
3. Sledujte micely ke vstřebání. | biochemistry |
(A) Nuclear pore; large protein complexes in the envelope gate traffic of RNA and proteins. | {
"A": "Nuclear pore",
"B": "Nucleolus",
"C": "Ribosome",
"D": "Centriole"
} | bio_0361 | multiple_choice | en | Which structure controls what enters and leaves the nucleus? | [
"Eliminate ribosome-making and division parts.",
"Recall gated envelope channels.",
"Choose nuclear pore."
] | biology | Answer: (A) Nuclear pores gate nuclear traffic.
Steps:
1. Eliminate ribosome-making and division parts.
2. Recall gated envelope channels.
3. Choose nuclear pore.
Answer: (A) | cell_biology |
(C) Affected fathers pass to all daughters; fathers give their only X to every daughter and Y to sons. | {
"A": "All sons of carrier mothers affected",
"B": "Skips generations equally",
"C": "Affected fathers pass to all daughters",
"D": "Only males affected"
} | bio_0362 | multiple_choice | en | Which pattern shows X-linked dominant inheritance? | [
"Eliminate recessive and equal patterns.",
"Trace father X to daughters.",
"Choose affected fathers pass to all daughters."
] | biology | Answer: (C) X-linked dominant fathers affect all daughters.
Steps:
1. Eliminate recessive and equal patterns.
2. Trace father X to daughters.
3. Choose affected fathers pass to all daughters.
Answer: (C) | genetics |
(B) Dairy cows bred for milk; humans chose the highest-yielding parents each generation. | {
"A": "Antibiotic resistance",
"B": "Dairy cows bred for milk",
"C": "Peppered moths",
"D": "Darwin's finches"
} | bio_0363 | multiple_choice | en | Which is an example of artificial selection? | [
"Eliminate wild natural-selection cases.",
"Identify human-chosen breeding.",
"Choose dairy cows bred for milk."
] | biology | Answer: (B) Dairy breeding is artificial selection.
Steps:
1. Eliminate wild natural-selection cases.
2. Identify human-chosen breeding.
3. Choose dairy cows bred for milk.
Answer: (B) | evolution |
(D) Desert; scarce rain and hot days select for water-storing cacti and nocturnal animals. | {
"A": "Tundra",
"B": "Rainforest",
"C": "Taiga",
"D": "Desert"
} | bio_0364 | multiple_choice | en | Which biome is hot and dry with cacti? | [
"Eliminate cold and wet biomes.",
"Match aridity to cacti.",
"Choose desert."
] | biology | Answer: (D) Cacti mark the desert biome.
Steps:
1. Eliminate cold and wet biomes.
2. Match aridity to cacti.
3. Choose desert.
Answer: (D) | ecology |
(A) Sítnice; tyčinky a čípky v ní převádějí světlo na nervové vzruchy. | {
"A": "Sítnice",
"B": "Čočka",
"C": "Rohovka",
"D": "Skléra"
} | bio_0365 | multiple_choice | cs | Která část oka obsahuje fotoreceptory? | [
"Vylučte lomivé a obalové části.",
"Přiřaďte čípky sítnici.",
"Zvolte sítnici."
] | biology | Answer: (A) Fotoreceptory nese sítnice.
Steps:
1. Vylučte lomivé a obalové části.
2. Přiřaďte čípky sítnici.
3. Zvolte sítnici.
Answer: (A) | physiology |
(C) Dehydration synthesis; removing water links monomers into polymers. | {
"A": "Hydrolysis",
"B": "Oxidation",
"C": "Dehydration synthesis",
"D": "Fermentation"
} | bio_0366 | multiple_choice | en | Which reaction builds polymers from monomers? | [
"Eliminate breaking and energy reactions.",
"Recall water removal joins units.",
"Choose dehydration synthesis."
] | biology | Answer: (C) Dehydration synthesis builds polymers.
Steps:
1. Eliminate breaking and energy reactions.
2. Recall water removal joins units.
3. Choose dehydration synthesis.
Answer: (C) | biochemistry |
About 113.1 cubic micrometers; a 6 micrometer sphere holds roughly 36 pi. | null | bio_0367 | worked_example | en | A spherical nucleus has diameter 6 micrometers. Calculate its volume using V = 4/3 pi r cubed. | [
"Radius = 6/2 = 3.",
"Cube: 3^3 = 27.",
"Multiply 4/3 x pi x 27 = 36 pi = 113.1.",
"Recompute: 113.1/36 = 3.14 = pi."
] | biology | Answer: About 113.1 cubic micrometers.
Steps:
1. Radius = 6/2 = 3.
2. Cube: 3^3 = 27.
3. Multiply 4/3 x pi x 27 = 36 pi = 113.1.
4. Recompute: 113.1/36 = 3.14 = pi. | cell_biology |
25 percent of children are expected to be group AB; the IA allele must come from the mother and IB from the father. | null | bio_0368 | worked_example | en | Parents have genotypes IAi and IBi for ABO blood group. What proportion of children is expected to be group AB? | [
"P(IA from mother) = 1/2.",
"P(IB from father) = 1/2.",
"Multiply 1/2 x 1/2 = 1/4.",
"Recompute: 4-box square holds one IAIB."
] | biology | Answer: 1/4, or 25 percent, group AB.
Steps:
1. P(IA from mother) = 1/2.
2. P(IB from father) = 1/2.
3. Multiply 1/2 x 1/2 = 1/4.
4. Recompute: 4-box square holds one IAIB. | genetics |
About 25000 insects survive; 5 percent of half a million still leaves tens of thousands. | null | bio_0369 | worked_example | en | A pesticide kills 95 percent of 500000 insects. How many survive? | [
"Surviving fraction = 1 - 0.95 = 0.05.",
"Multiply 500000 x 0.05 = 25000.",
"Recompute: 25000/500000 = 0.05."
] | biology | Answer: About 25000 survivors.
Steps:
1. Surviving fraction = 1 - 0.95 = 0.05.
2. Multiply 500000 x 0.05 = 25000.
3. Recompute: 25000/500000 = 0.05. | evolution |
Odhad je 400 ryb; Lincoln-Petersenův index škáluje vzorek na populaci. | null | bio_0370 | worked_example | cs | V rybníce bylo označeno 80 ryb. Při dalším odlovu 60 kusů bylo 12 označených. Odhadněte velikost populace. | [
"Zapište N = M x C / R.",
"Dosaďte 80 x 60 / 12.",
"Vypočítejte 4800 / 12 = 400.",
"Přepočítejte: 400 x 12 / 60 = 80."
] | biology | Answer: N=400 ryb.
Steps:
1. Zapište N = M x C / R.
2. Dosaďte 80 x 60 / 12.
3. Vypočítejte 4800 / 12 = 400.
4. Přepočítejte: 400 x 12 / 60 = 80. | ecology |
Alveolar ventilation is 5.4 liters per minute; subtracting dead space leaves the air that reaches alveoli. | null | bio_0371 | worked_example | en | Tidal volume is 600 mL, dead space is 150 mL, and breathing rate is 12 per minute. Calculate alveolar ventilation. | [
"Per breath: 600-150 = 450 mL.",
"Multiply 450 x 12 = 5400 mL per min.",
"Convert to 5.4 L per min.",
"Recompute: 5400/12 = 450, plus 150 = 600."
] | biology | Answer: 5.4 L per min.
Steps:
1. Per breath: 600-150 = 450 mL.
2. Multiply 450 x 12 = 5400 mL per min.
3. Convert to 5.4 L per min.
4. Recompute: 5400/12 = 450, plus 150 = 600. | physiology |
Allosteric regulators bind away from the active site and reshape the enzyme to switch it on or off; activators improve substrate fit while inhibitors spoil it, tuning pathways to cell needs. | null | bio_0372 | concept | en | Explain allosteric regulation of enzymes. | [
"Bind regulators at distant sites.",
"Reshape the active site.",
"Switch pathways with cell signals."
] | biology | Answer: Distant binding reshapes enzymes on or off.
Steps:
1. Bind regulators at distant sites.
2. Reshape the active site.
3. Switch pathways with cell signals. | biochemistry |
Cells signal by direct contact through junctions, locally by paracrine molecules to neighbors, and body-wide by hormones in blood, plus neurons for millisecond precision. | null | bio_0373 | concept | en | Describe three ways cells communicate over different distances. | [
"Touch neighbors through junctions.",
"Bathe locals in paracrine signals.",
"Broadcast hormones by blood."
] | biology | Answer: Cells touch, whisper locally, or broadcast by blood.
Steps:
1. Touch neighbors through junctions.
2. Bathe locals in paracrine signals.
3. Broadcast hormones by blood. | cell_biology |
Epigenetics changes gene activity without altering DNA sequence, through methyl tags that silence genes and histone tweaks that open or close chromatin; marks can even pass to daughter cells. | null | bio_0374 | concept | en | What is epigenetics? Name two mechanisms. | [
"Define regulation beyond sequence.",
"Silence with methylation.",
"Open or close with histone marks."
] | biology | Answer: Methyl and histone marks tune genes beyond DNA.
Steps:
1. Define regulation beyond sequence.
2. Silence with methylation.
3. Open or close with histone marks. | genetics |
U člověka se vyvinula chůze po dvou končetinách, zvětšený mozek, výroba nástrojů a řeč; fosilie a DNA ukazují postupný vývoj z afrických lidoopů. | null | bio_0375 | concept | cs | Shrňte hlavní znaky evoluce člověka. | [
"Uveďte bipedii a volné ruce.",
"Popište růst mozku a nástroje.",
"Doložte fosiliemi a DNA."
] | biology | Answer: Bipedie, mozek a nástroje vedly k člověku.
Steps:
1. Uveďte bipedii a volné ruce.
2. Popište růst mozku a nástroje.
3. Doložte fosiliemi a DNA. | evolution |
Mammals hibernate through winter on stored fat, hummingbirds drop into nightly torpor, birds migrate to food, and seeds and buds wait dormant until spring warmth returns. | null | bio_0376 | concept | en | Describe four overwintering strategies of organisms. | [
"Sleep through winter by hibernation.",
"Cool down in daily torpor.",
"Leave by migration or wait dormant."
] | biology | Answer: Hibernate, cool, migrate, or sleep as seeds.
Steps:
1. Sleep through winter by hibernation.
2. Cool down in daily torpor.
3. Leave by migration or wait dormant. | ecology |
Osteoblasts lay down mineralized matrix that hardens into bone while osteoclasts dissolve old matrix, so skeletons lengthen in youth and rebuild around stresses for life. | null | bio_0377 | concept | en | Explain how bones grow and remodel. | [
"Build matrix with osteoblasts.",
"Resorb old bone with osteoclasts.",
"Shape skeleton to lifelong loads."
] | biology | Answer: Builders lay bone while dissolvers sculpt it.
Steps:
1. Build matrix with osteoblasts.
2. Resorb old bone with osteoclasts.
3. Shape skeleton to lifelong loads. | physiology |
Industrial trans fats raise LDL cholesterol while lowering HDL, stiffening arteries and promoting heart disease more than butter does, which is why many countries restrict them. | null | bio_0378 | concept | en | Why are trans fats harmful to health? | [
"Raise bad LDL cholesterol.",
"Lower good HDL cholesterol.",
"Clog arteries toward disease."
] | biology | Answer: Trans fats worsen both cholesterols at once.
Steps:
1. Raise bad LDL cholesterol.
2. Lower good HDL cholesterol.
3. Clog arteries toward disease. | biochemistry |
Penicillin blocks bacterial wall building so germs burst, while other antibiotics jam bacterial ribosomes; human cells lack walls and use different ribosomes, so they escape harm. | null | bio_0379 | concept | en | Explain how antibiotics kill bacteria without harming human cells. | [
"Target walls humans lack.",
"Jam bacterial-only ribosomes.",
"Spare human structures."
] | biology | Answer: Antibiotics hit bacterial-only targets.
Steps:
1. Target walls humans lack.
2. Jam bacterial-only ribosomes.
3. Spare human structures. | cell_biology |
Barva očí závisí na množství melaninu v duhovce a řídí ji více genů, proto se mísí odstíny hnědé, zelené a modré a děti se mohou lišit od rodičů. | null | bio_0380 | concept | cs | Vysvětlete zjednodušeně dědičnost barvy očí. | [
"Přiřaďte barvu melaninu.",
"Uveďte více genů.",
"Vysvětlete míšení odstínů."
] | biology | Answer: Barvu očí ladí více genů přes melanin.
Steps:
1. Přiřaďte barvu melaninu.
2. Uveďte více genů.
3. Vysvětlete míšení odstínů. | genetics |
(D) Biological; it defines species as groups that interbreed and produce fertile offspring in nature. | {
"A": "Morphological",
"B": "Ecological",
"C": "Phylogenetic",
"D": "Biological"
} | bio_0381 | multiple_choice | en | Which species concept defines species by interbreeding? | [
"Eliminate form, niche, and tree concepts.",
"Recall fertile-offspring criterion.",
"Choose biological."
] | biology | Answer: (D) The biological concept uses interbreeding.
Steps:
1. Eliminate form, niche, and tree concepts.
2. Recall fertile-offspring criterion.
3. Choose biological.
Answer: (D) | evolution |
(A) Mutualism; both partners gain, as in bees fed by flowers they pollinate. | {
"A": "Mutualism",
"B": "Commensalism",
"C": "Parasitism",
"D": "Competition"
} | bio_0382 | multiple_choice | en | Which interaction benefits both species? | [
"Eliminate one-sided and negative outcomes.",
"Recall plus-plus partnerships.",
"Choose mutualism."
] | biology | Answer: (A) Mutualism helps both partners.
Steps:
1. Eliminate one-sided and negative outcomes.
2. Recall plus-plus partnerships.
3. Choose mutualism.
Answer: (A) | ecology |
(B) Phagocytes; macrophages and neutrophils engulf and digest invaders within minutes of infection. | {
"A": "Plasma cells",
"B": "Phagocytes",
"C": "Memory cells",
"D": "Platelets"
} | bio_0383 | multiple_choice | en | Which cells first engulf pathogens in innate defense? | [
"Eliminate antibody and memory cells.",
"Eliminate clotting fragments.",
"Choose phagocytes."
] | biology | Answer: (B) Phagocytes eat invaders first.
Steps:
1. Eliminate antibody and memory cells.
2. Eliminate clotting fragments.
3. Choose phagocytes.
Answer: (B) | physiology |
(D) K; vitamin K enables clotting-factor production in the liver. | {
"A": "C",
"B": "D",
"C": "B12",
"D": "K"
} | bio_0384 | multiple_choice | en | Which vitamin is needed for blood clotting? | [
"Eliminate antioxidant and bone vitamins.",
"Eliminate red-cell vitamin.",
"Choose K."
] | biology | Answer: (D) Vitamin K drives clotting.
Steps:
1. Eliminate antioxidant and bone vitamins.
2. Eliminate red-cell vitamin.
3. Choose K.
Answer: (D) | biochemistry |
(C) Nukleoid; kruhová DNA bakterií leží volně v této oblasti cytoplazmy. | {
"A": "Jádro",
"B": "Ribozom",
"C": "Nukleoid",
"D": "Mitochondrie"
} | bio_0385 | multiple_choice | cs | Která struktura nese genetickou informaci v prokaryotické buňce? | [
"Vylučte jádro a organely.",
"Přiřaďte DNA nukleoidu.",
"Zvolte nukleoid."
] | biology | Answer: (C) DNA bakterií leží v nukleoidu.
Steps:
1. Vylučte jádro a organely.
2. Přiřaďte DNA nukleoidu.
3. Zvolte nukleoid.
Answer: (C) | cell_biology |
(A) AA; both alleles are the dominant form. | {
"A": "AA",
"B": "Aa",
"C": "aa",
"D": "Aa and aa"
} | bio_0386 | multiple_choice | en | Which genotype is homozygous dominant? | [
"Eliminate mixed genotypes.",
"Eliminate recessive pairs.",
"Choose AA."
] | biology | Answer: (A) AA is homozygous dominant.
Steps:
1. Eliminate mixed genotypes.
2. Eliminate recessive pairs.
3. Choose AA.
Answer: (A) | genetics |
About 5730 years old; 50 percent equals exactly one half-life of decay. | null | bio_0387 | worked_example | en | Carbon-14 has a half-life of 5730 years. A sample holds 50 percent of its original carbon-14. Estimate its age. | [
"Express 50 percent as (1/2) to the power 1.",
"Count 1 half-life.",
"Report 1 x 5730 = 5730 years.",
"Recompute: after 5730 years half remains."
] | biology | Answer: About 5730 years.
Steps:
1. Express 50 percent as (1/2) to the power 1.
2. Count 1 half-life.
3. Report 1 x 5730 = 5730 years.
4. Recompute: after 5730 years half remains. | evolution |
The lake holds 100 grams of phosphorus; concentration times volume gives the total mass. | null | bio_0388 | worked_example | en | A lake of 2 million liters holds phosphorus at 0.05 mg per liter. Calculate the total phosphorus mass in grams. | [
"Multiply 0.05 x 2,000,000 = 100,000 mg.",
"Convert to 100 g.",
"Recompute: 100,000/2,000,000 = 0.05 mg per L."
] | biology | Answer: 100 g of phosphorus.
Steps:
1. Multiply 0.05 x 2,000,000 = 100,000 mg.
2. Convert to 100 g.
3. Recompute: 100,000/2,000,000 = 0.05 mg per L. | ecology |
About 2.0 liters lost; sweat rate times duration gives fluid loss. | null | bio_0389 | worked_example | en | An athlete sweats 0.8 liters per hour for 2.5 hours. Estimate total fluid loss. | [
"Multiply 0.8 x 2.5 = 2.0 L.",
"Recompute: 2.0/2.5 = 0.8 L per h.",
"Report about 2.0 L."
] | biology | Answer: About 2.0 L lost.
Steps:
1. Multiply 0.8 x 2.5 = 2.0 L.
2. Recompute: 2.0/2.5 = 0.8 L per h.
3. Report about 2.0 L. | physiology |
Roztok obsahuje 0,2 mol látky; koncentrace krát objem dává látkové množství. | null | bio_0390 | worked_example | cs | Kolik molů látky je v 500 ml roztoku o koncentraci 0,4 mol na litr? | [
"Převeďte 500 ml = 0,5 l.",
"Vynásobte 0,4 x 0,5 = 0,2 mol.",
"Přepočítejte: 0,2 / 0,5 = 0,4 mol na litr."
] | biology | Answer: 0,2 mol.
Steps:
1. Převeďte 500 ml = 0,5 l.
2. Vynásobte 0,4 x 0,5 = 0,2 mol.
3. Přepočítejte: 0,2 / 0,5 = 0,4 mol na litr. | biochemistry |
About 32 cells are in mitosis; 8 percent of 400 gives the dividing fraction. | null | bio_0391 | worked_example | en | In an onion root tip sample of 400 cells, 8 percent are in mitosis. How many dividing cells are expected? | [
"Convert 8 percent = 0.08.",
"Multiply 400 x 0.08 = 32.",
"Recompute: 32/400 = 0.08."
] | biology | Answer: About 32 cells in mitosis.
Steps:
1. Convert 8 percent = 0.08.
2. Multiply 400 x 0.08 = 32.
3. Recompute: 32/400 = 0.08. | cell_biology |
Sperm contribute almost no cytoplasm, so the zygote's mitochondria come from the egg; mitochondrial traits and DNA therefore pass only through mothers. | null | bio_0392 | concept | en | Explain why mitochondrial DNA is inherited maternally. | [
"Note sperm shed cytoplasm.",
"Trace zygote mitochondria to egg.",
"Conclude mother-only transmission."
] | biology | Answer: Eggs supply the zygote's mitochondria.
Steps:
1. Note sperm shed cytoplasm.
2. Trace zygote mitochondria to egg.
3. Conclude mother-only transmission. | genetics |
Feathers first evolved for insulation and display in small dinosaurs, then exapted for gliding and powered flight, with Archaeopteryx preserving the reptile-bird transition. | null | bio_0393 | concept | en | Explain the evolution of flight in birds. | [
"Start with insulated feathered dinosaurs.",
"Glide before powered flight.",
"Bridge with Archaeopteryx fossils."
] | biology | Answer: Feathers warmed before they flew.
Steps:
1. Start with insulated feathered dinosaurs.
2. Glide before powered flight.
3. Bridge with Archaeopteryx fossils. | evolution |
When populations overshoot carrying capacity they consume resources faster than renewal, then crash by starvation and disease, as reindeer introduced to islands boomed and starved. | null | bio_0394 | concept | en | Explain population overshoot and dieback with an example. | [
"Exceed capacity by plenty.",
"Deplete food and shelter.",
"Crash by hunger and disease."
] | biology | Answer: Overshoot spends resources, then starves.
Steps:
1. Exceed capacity by plenty.
2. Deplete food and shelter.
3. Crash by hunger and disease. | ecology |
Při nízkém tlaku ledviny uvolní renin, ten spustí tvorbu angiotenzinu zužujícího cévy a aldosteronu zadržujícího sůl a vodu; tlak se tím zvýší zpět. | null | bio_0395 | concept | cs | Vysvětlete, jak ledviny řídí krevní tlak. | [
"Uvolněte renin při nízkém tlaku.",
"Zúžte cévy angiotenzinem.",
"Zadržte sůl a vodu aldosteronem."
] | biology | Answer: Ledviny zvyšují tlak reninem a solí.
Steps:
1. Uvolněte renin při nízkém tlaku.
2. Zúžte cévy angiotenzinem.
3. Zadržte sůl a vodu aldosteronem. | physiology |
Ice crystals hold water molecules apart in an open lattice, so ice is less dense than liquid and floats, insulating lakes so fish survive winter below. | null | bio_0396 | concept | en | Explain why ice floats and why it matters for life. | [
"Freeze water into open lattice.",
"Lower density below liquid.",
"Insulate lakes for winter life."
] | biology | Answer: Floating ice blankets winter waters.
Steps:
1. Freeze water into open lattice.
2. Lower density below liquid.
3. Insulate lakes for winter life. | biochemistry |
Microtubules form the spindle that pulls chromosomes apart, actin builds the cleavage furrow that pinches animal cells, and vesicles guided by the cytoskeleton raise plant cell plates. | null | bio_0397 | concept | en | Explain the cytoskeleton's roles in cell division. | [
"Pull chromosomes with spindle.",
"Pinch cells with actin ring.",
"Guide vesicles to cell plates."
] | biology | Answer: The cytoskeleton pulls, pinches, and builds dividers.
Steps:
1. Pull chromosomes with spindle.
2. Pinch cells with actin ring.
3. Guide vesicles to cell plates. | cell_biology |
Autosomes are the 22 non-sex chromosome pairs carrying general genes, while the X and Y sex chromosomes determine maleness or femaleness and carry sex-linked traits. | null | bio_0398 | concept | en | Distinguish autosomes from sex chromosomes. | [
"Count 22 general pairs.",
"Name X and Y as sex pair.",
"Assign general versus sex traits."
] | biology | Answer: Autosomes run the body; sex chromosomes set sex.
Steps:
1. Count 22 general pairs.
2. Name X and Y as sex pair.
3. Assign general versus sex traits. | genetics |
Selection only favors what reproduces best here and now, with no foresight; bacteria stay simple because it works, and complexity persists only where it pays. | null | bio_0399 | concept | en | Explain why evolution has no goal or direction. | [
"Reward present reproduction only.",
"Keep simple winners simple.",
"Deny progress toward perfection."
] | biology | Answer: Evolution rewards now, not some ideal future.
Steps:
1. Reward present reproduction only.
2. Keep simple winners simple.
3. Deny progress toward perfection. | evolution |
Klimax je stabilní závěrečné společenstvo sukcese, například bučina; vzniká postupným obsazováním stanoviště a vyrovnáním s podnebím a půdou. | null | bio_0400 | concept | cs | Vysvětlete, co je klimaxové společenstvo a jak vzniká. | [
"Definujte stabilní závěr sukcese.",
"Sledujte obsazování stanoviště.",
"Vyrovnejte společenstvo s podnebím."
] | biology | Answer: Klimax je vyrovnaný závěr sukcese.
Steps:
1. Definujte stabilní závěr sukcese.
2. Sledujte obsazování stanoviště.
3. Vyrovnejte společenstvo s podnebím. | ecology |
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