answer string | choices dict | id string | kind string | lang string | question string | steps list | subject string | text string | topic string |
|---|---|---|---|---|---|---|---|---|---|
(B) Mitochondrion; it runs the citric acid cycle in the matrix and the electron transport chain on cristae to generate a proton gradient that drives ATP synthase. | {
"A": "Ribosome",
"B": "Mitochondrion",
"C": "Golgi apparatus",
"D": "Lysosome"
} | bio_0001 | multiple_choice | en | Which organelle produces most ATP in an animal cell by oxidative phosphorylation? | [
"Eliminate ribosomes, Golgi, and lysosomes as protein-making, sorting, and digestive roles.",
"Recall mitochondria hold the respiratory chain and ATP synthase.",
"Choose mitochondrion."
] | biology | Answer: (B) Mitochondrion; it makes ATP by oxidative phosphorylation.
Steps:
1. Eliminate ribosomes, Golgi, and lysosomes as protein-making, sorting, and digestive roles.
2. Recall mitochondria hold the respiratory chain and ATP synthase.
3. Choose mitochondrion.
Answer: (B) | cell_biology |
(B) 25%; a Punnett square gives 1 AA : 2 Aa : 1 aa, so one quarter are homozygous recessive. | {
"A": "0%",
"B": "25%",
"C": "50%",
"D": "100%"
} | bio_0002 | multiple_choice | en | In a monohybrid cross Aa x Aa with complete dominance, what proportion of offspring is expected to be genotype aa? | [
"List gametes A and a from each parent.",
"Fill the 2x2 Punnett square.",
"Count 1 of 4 boxes as aa."
] | biology | Answer: (B) 25% are expected to be aa.
Steps:
1. List gametes A and a from each parent.
2. Fill the 2x2 Punnett square.
3. Count 1 of 4 boxes as aa.
Answer: (B) | genetics |
(B) Phenotype; individuals with advantageous traits survive and reproduce better, which indirectly changes allele frequencies. | {
"A": "DNA sequence",
"B": "Phenotype",
"C": "Fossil record",
"D": "Mutation rate"
} | bio_0003 | multiple_choice | en | Natural selection acts most directly on which level of biological organization? | [
"Recall selection favors differential survival and reproduction.",
"Note survival depends on expressed traits in an environment.",
"Choose phenotype."
] | biology | Answer: (B) Phenotype is the direct target of selection.
Steps:
1. Recall selection favors differential survival and reproduction.
2. Note survival depends on expressed traits in an environment.
3. Choose phenotype.
Answer: (B) | evolution |
(C) Green plants; they convert sunlight to chemical energy by photosynthesis and support higher trophic levels. | {
"A": "Herbivores",
"B": "Carnivores",
"C": "Green plants",
"D": "Decomposers"
} | bio_0004 | multiple_choice | en | Which group forms the first trophic level as primary producers in most terrestrial ecosystems? | [
"Recall trophic level 1 captures external energy.",
"Eliminate consumers and decomposers.",
"Choose green plants."
] | biology | Answer: (C) Green plants are the primary producers.
Steps:
1. Recall trophic level 1 captures external energy.
2. Eliminate consumers and decomposers.
3. Choose green plants.
Answer: (C) | ecology |
(B) Srdce; rytmickými stahy vytváří tlak, který pohání krev tepnami k tkáním a žilami zpět. | {
"A": "Játra",
"B": "Srdce",
"C": "Ledvina",
"D": "Plíce"
} | bio_0005 | multiple_choice | cs | Který orgán je hlavní pumpou lidské oběhové soustavy? | [
"Vylučte játra, ledvinu a plíce s jinou hlavní funkcí.",
"Přiřaďte srdeční sval k tvorbě tlaku.",
"Zvolte srdce."
] | biology | Answer: (B) Srdce je hlavní pumpou oběhu.
Steps:
1. Vylučte játra, ledvinu a plíce s jinou hlavní funkcí.
2. Přiřaďte srdeční sval k tvorbě tlaku.
3. Zvolte srdce.
Answer: (B) | physiology |
(B) Amino acids; they link by peptide bonds into polypeptides that fold into functional proteins. | {
"A": "Nucleotides",
"B": "Amino acids",
"C": "Fatty acids",
"D": "Monosaccharides"
} | bio_0006 | multiple_choice | en | Which molecules are the building blocks of proteins? | [
"Recall proteins are polypeptides.",
"Eliminate nucleic acid, lipid, and carbohydrate monomers.",
"Choose amino acids."
] | biology | Answer: (B) Amino acids build proteins.
Steps:
1. Recall proteins are polypeptides.
2. Eliminate nucleic acid, lipid, and carbohydrate monomers.
3. Choose amino acids.
Answer: (B) | biochemistry |
1000x magnification; the image is one thousand times larger than the real cell after converting to common units. | null | bio_0007 | worked_example | en | A cheek cell is 20 micrometers wide. In a drawing it appears 20 millimeters wide. Calculate the magnification. | [
"Convert image size: 20 mm = 20000 micrometers.",
"Divide image by real size: 20000 / 20 = 1000.",
"Recompute: 1000 x 20 = 20000, matching the image.",
"Report 1000x."
] | biology | Answer: 1000x magnification.
Steps:
1. Convert image size: 20 mm = 20000 micrometers.
2. Divide image by real size: 20000 / 20 = 1000.
3. Recompute: 1000 x 20 = 20000, matching the image.
4. Report 1000x. | cell_biology |
q equals 0.3, p equals 0.7, and carriers equal 42 percent; the square root gives q and 2pq gives heterozygotes. | null | bio_0008 | worked_example | en | In a population, 9 percent show a recessive phenotype, so q squared equals 0.09. Assuming Hardy-Weinberg equilibrium, calculate q, p, and carrier frequency. | [
"Compute q = sqrt(0.09) = 0.3.",
"Compute p = 1 - 0.3 = 0.7.",
"Compute 2pq = 2 x 0.7 x 0.3 = 0.42.",
"Recompute: 0.49+0.09+0.42=1.0."
] | biology | Answer: q=0.3, p=0.7, carriers=42%.
Steps:
1. Compute q = sqrt(0.09) = 0.3.
2. Compute p = 1 - 0.3 = 0.7.
3. Compute 2pq = 2 x 0.7 x 0.3 = 0.42.
4. Recompute: 0.49+0.09+0.42=1.0. | genetics |
p equals 0.6 and q equals 0.4; counting alleles across 200 gene copies gives 120 A and 80 a. | null | bio_0009 | worked_example | en | In 100 diploid individuals, genotypes are AA 36, Aa 48, and aa 16. Calculate allele frequencies p for A and q for a. | [
"Total alleles = 100 x 2 = 200.",
"A copies = 36x2 + 48 = 120, so p = 120/200 = 0.6.",
"a copies = 16x2 + 48 = 80, so q = 80/200 = 0.4.",
"Recompute: p+q=1.0 and 120+80=200."
] | biology | Answer: p=0.6, q=0.4.
Steps:
1. Total alleles = 100 x 2 = 200.
2. A copies = 36x2 + 48 = 120, so p = 120/200 = 0.6.
3. a copies = 16x2 + 48 = 80, so q = 80/200 = 0.4.
4. Recompute: p+q=1.0 and 120+80=200. | evolution |
Hustota je 40 jedinců na hektar; celkový počet dělený plochou dává průměrnou hustotu. | null | bio_0010 | worked_example | cs | Na louce o rozloze 6 hektarů bylo napočítáno 240 hrabošů. Vypočítejte populační hustotu. | [
"Zapište počet 240 a plochu 6 hektarů.",
"Vydělte 240 / 6 = 40.",
"Přepočítejte: 40 x 6 = 240.",
"Uveďte 40 jedinců na hektar."
] | biology | Answer: 40 jedinců na hektar.
Steps:
1. Zapište počet 240 a plochu 6 hektarů.
2. Vydělte 240 / 6 = 40.
3. Přepočítejte: 40 x 6 = 240.
4. Uveďte 40 jedinců na hektar. | ecology |
Cardiac output is 4.9 liters per minute; heart rate times stroke volume gives minute volume. | null | bio_0011 | worked_example | en | Heart rate is 70 beats per minute and stroke volume is 70 milliliters. Calculate cardiac output in liters per minute. | [
"Multiply 70 x 70 = 4900 mL per min.",
"Convert 4900 mL to 4.9 L.",
"Recompute: 4.9 x 1000 = 4900, divided by 70 is 70.",
"Report 4.9 L per min."
] | biology | Answer: 4.9 L per min.
Steps:
1. Multiply 70 x 70 = 4900 mL per min.
2. Convert 4900 mL to 4.9 L.
3. Recompute: 4.9 x 1000 = 4900, divided by 70 is 70.
4. Report 4.9 L per min. | physiology |
Activation energy is the minimum energy needed to reach the transition state; enzymes lower it by stabilizing that state, so more molecules react at body temperature without changing overall free energy. | null | bio_0012 | concept | en | What is activation energy and how do enzymes affect it? | [
"Define activation energy as the barrier to reaction.",
"Explain enzyme-substrate binding and transition-state stabilization.",
"Note unchanged delta G and reusable enzyme."
] | biology | Answer: Enzymes lower activation energy by stabilizing the transition state.
Steps:
1. Define activation energy as the barrier to reaction.
2. Explain enzyme-substrate binding and transition-state stabilization.
3. Note unchanged delta G and reusable enzyme. | biochemistry |
The membrane is a fluid phospholipid bilayer with embedded proteins, cholesterol, and carbohydrates; lipids and many proteins move laterally, creating a mosaic that controls transport, signaling, and adhesion. | null | bio_0013 | concept | en | Describe the fluid mosaic model of the plasma membrane. | [
"Recall bilayer with hydrophilic heads and hydrophobic tails.",
"Add integral and peripheral proteins plus cholesterol.",
"Explain fluidity and mosaic functions."
] | biology | Answer: A fluid bilayer mosaic of lipids, proteins, and carbohydrates.
Steps:
1. Recall bilayer with hydrophilic heads and hydrophobic tails.
2. Add integral and peripheral proteins plus cholesterol.
3. Explain fluidity and mosaic functions. | cell_biology |
Each individual carries two alleles per gene that separate into different gametes; homologous chromosomes separate in anaphase I of meiosis, so each gamete gets one homolog. | null | bio_0014 | concept | en | State Mendel's law of segregation and explain its chromosomal basis. | [
"State two alleles separate 1:1 into gametes.",
"Link alleles to loci on homologs.",
"Map separation to anaphase I."
] | biology | Answer: Paired alleles segregate as homologs separate in meiosis I.
Steps:
1. State two alleles separate 1:1 into gametes.
2. Link alleles to loci on homologs.
3. Map separation to anaphase I. | genetics |
Genetický drift je náhodné kolísání frekvencí alel mezi generacemi; nejsilněji působí v malých populacích a po efektu hrdla či zakladatele, kde může alely ztratit i fixovat bez ohledu na selekci. | null | bio_0015 | concept | cs | Vysvětlete, co je genetický drift a kdy má největší účinek. | [
"Definujte drift jako náhodný výběr gamet.",
"Vysvětlete vliv malé velikosti populace.",
"Uveďte efekt hrdla a zakladatele."
] | biology | Answer: Genetický drift je náhodná změna frekvencí alel, nejsilnější v malých populacích.
Steps:
1. Definujte drift jako náhodný výběr gamet.
2. Vysvětlete vliv malé velikosti populace.
3. Uveďte efekt hrdla a zakladatele. | evolution |
A trophic cascade is an indirect effect where changes at one trophic level propagate downward; removing sea otters lets sea urchins overgraze kelp forests, while restoring wolves in Yellowstone reduced elk browsing and reshaped vegetation. | null | bio_0016 | concept | en | What is a trophic cascade? Give a classic example. | [
"Define top-down indirect effects across levels.",
"Give otter-urchin-kelp as marine example.",
"Give wolf-elk-vegetation as terrestrial example."
] | biology | Answer: Top predators indirectly shape lower levels, as in otter-urchin-kelp.
Steps:
1. Define top-down indirect effects across levels.
2. Give otter-urchin-kelp as marine example.
3. Give wolf-elk-vegetation as terrestrial example. | ecology |
Rising temperature is detected by hypothalamic and skin receptors, which trigger sweating and vasodilation to lose heat; falling temperature triggers shivering and vasoconstriction, returning the body toward its set point. | null | bio_0017 | concept | en | Explain negative feedback control of human body temperature. | [
"Identify sensor, control center, and effectors.",
"Trace heat-loss responses to cooling.",
"Trace heat-gain responses to warming."
] | biology | Answer: The hypothalamus reverses deviations with sweating or shivering.
Steps:
1. Identify sensor, control center, and effectors.
2. Trace heat-loss responses to cooling.
3. Trace heat-gain responses to warming. | physiology |
A competitive inhibitor resembles the substrate and occupies the active site reversibly; raising substrate concentration outcompetes it, restoring Vmax while apparent Km increases. | null | bio_0018 | concept | en | Explain competitive enzyme inhibition and how it can be overcome. | [
"Describe inhibitor binding at the active site.",
"Contrast with noncompetitive binding elsewhere.",
"Explain reversal by excess substrate."
] | biology | Answer: Competitive inhibitors block the active site but lose to excess substrate.
Steps:
1. Describe inhibitor binding at the active site.
2. Contrast with noncompetitive binding elsewhere.
3. Explain reversal by excess substrate. | biochemistry |
Mitosis yields two genetically identical diploid cells for growth and repair, while meiosis yields four genetically diverse haploid gametes through one DNA replication and two divisions with crossing over. | null | bio_0019 | concept | en | Compare the products of mitosis and meiosis in a diploid organism. | [
"Count divisions and chromosome number changes.",
"Contrast genetic identity versus diversity.",
"Link outcomes to somatic versus reproductive roles."
] | biology | Answer: Mitosis gives 2 identical diploid cells; meiosis gives 4 diverse haploid cells.
Steps:
1. Count divisions and chromosome number changes.
2. Contrast genetic identity versus diversity.
3. Link outcomes to somatic versus reproductive roles. | cell_biology |
Při neúplné dominanci má heterozygot přechodný fenotyp mezi oběma homozygoty, protože jedna funkční alela nestačí na plný projev; klasickým příkladem je hledík, kde křížení červené a bílé formy dává růžové potomstvo. | null | bio_0020 | concept | cs | Vysvětlete neúplnou dominanci a uveďte příklad. | [
"Definujte přechodný fenotyp heterozygota.",
"Odlište od kodominance.",
"Uveďte hledík s růžovými heterozygoty."
] | biology | Answer: Heterozygot má přechodný fenotyp, například růžový hledík.
Steps:
1. Definujte přechodný fenotyp heterozygota.
2. Odlište od kodominance.
3. Uveďte hledík s růžovými heterozygoty. | genetics |
(A) Bat wing and human forelimb; they share the same tetrapod bone pattern despite different functions, while the other pairs are analogous results of convergence. | {
"A": "Bat wing and human forelimb",
"B": "Shark fin and dolphin fin",
"C": "Butterfly wing and bird wing",
"D": "Cactus spine and rose prickle"
} | bio_0021 | multiple_choice | en | Which pair is an example of homologous structures indicating common ancestry? | [
"Recall homology means shared structure from common ancestry.",
"Eliminate functionally similar but structurally different pairs.",
"Choose bat wing and human forelimb."
] | biology | Answer: (A) Bat wing and human forelimb are homologous.
Steps:
1. Recall homology means shared structure from common ancestry.
2. Eliminate functionally similar but structurally different pairs.
3. Choose bat wing and human forelimb.
Answer: (A) | evolution |
(B) Tropical rainforest; stable warmth, high rainfall, and structural complexity support many niches and year-round productivity. | {
"A": "Tundra",
"B": "Tropical rainforest",
"C": "Desert",
"D": "Taiga"
} | bio_0022 | multiple_choice | en | Which terrestrial biome typically has the highest species richness? | [
"Compare productivity and climatic stability across biomes.",
"Eliminate cold or arid low-diversity biomes.",
"Choose tropical rainforest."
] | biology | Answer: (B) Tropical rainforest has the highest richness.
Steps:
1. Compare productivity and climatic stability across biomes.
2. Eliminate cold or arid low-diversity biomes.
3. Choose tropical rainforest.
Answer: (B) | ecology |
(B) Alveoli; their thin walls, huge surface area, and dense capillaries allow rapid diffusion of oxygen and carbon dioxide. | {
"A": "Trachea",
"B": "Alveoli",
"C": "Bronchi",
"D": "Diaphragm"
} | bio_0023 | multiple_choice | en | Where does most gas exchange occur in the human lung? | [
"Eliminate conducting airways and muscle.",
"Recall alveolar surface and capillary network.",
"Choose alveoli."
] | biology | Answer: (B) Alveoli are the main gas-exchange surface.
Steps:
1. Eliminate conducting airways and muscle.
2. Recall alveolar surface and capillary network.
3. Choose alveoli.
Answer: (B) | physiology |
(A) Cytosol; the ten glycolytic enzymes are soluble in cytoplasm and split glucose to pyruvate without oxygen requirement. | {
"A": "Cytosol",
"B": "Mitochondrial matrix",
"C": "Nucleus",
"D": "Thylakoid lumen"
} | bio_0024 | multiple_choice | en | In which cellular compartment does glycolysis occur in eukaryotes? | [
"Recall glycolysis precedes organelle pathways.",
"Eliminate matrix, nucleus, and thylakoid.",
"Choose cytosol."
] | biology | Answer: (A) Glycolysis occurs in the cytosol.
Steps:
1. Recall glycolysis precedes organelle pathways.
2. Eliminate matrix, nucleus, and thylakoid.
3. Choose cytosol.
Answer: (A) | biochemistry |
(B) Plazmatická membrána; fosfolipidová dvojvrstva s přenašeči a kanály určuje, které látky projdou. | {
"A": "Ribozom",
"B": "Plazmatická membrána",
"C": "Cytoskelet",
"D": "Nukleolus"
} | bio_0025 | multiple_choice | cs | Která struktura buňky selektivně řídí vstup a výstup látek? | [
"Vylučte struktury pro syntézu, oporu a tvorbu ribozomů.",
"Přiřaďte transport polopropustné membráně.",
"Zvolte plazmatickou membránu."
] | biology | Answer: (B) Plazmatická membrána řídí transport látek.
Steps:
1. Vylučte struktury pro syntézu, oporu a tvorbu ribozomů.
2. Přiřaďte transport polopropustné membráně.
3. Zvolte plazmatickou membránu.
Answer: (B) | cell_biology |
(A) All daughters are carriers and all sons are normal; fathers give X to daughters only, and the mother gives only normal X alleles. | {
"A": "All daughters are carriers and all sons are normal",
"B": "All children are color-blind",
"C": "All daughters are color-blind",
"D": "Half of sons are color-blind"
} | bio_0026 | multiple_choice | en | Red-green color blindness is X-linked recessive. A color-blind father and a normal homozygous mother have children. Which statement is correct? | [
"Write father XcY and mother XNXN.",
"Cross to get daughters XNXc and sons XNY.",
"Choose carriers for daughters and normal for sons."
] | biology | Answer: (A) Daughters are carriers and sons are normal.
Steps:
1. Write father XcY and mother XNXN.
2. Cross to get daughters XNXc and sons XNY.
3. Choose carriers for daughters and normal for sons.
Answer: (A) | genetics |
About 11460 years old; 25 percent equals two half-lives of decay. | null | bio_0027 | worked_example | en | Carbon-14 has a half-life of 5730 years. A fossil contains 25 percent of its original carbon-14. Estimate its age. | [
"Express 25 percent as 1/4 = (1/2) squared.",
"Count 2 half-lives.",
"Multiply 2 x 5730 = 11460.",
"Recompute: after 5730 years 50 percent remains, after 11460 years 25 percent remains."
] | biology | Answer: About 11460 years.
Steps:
1. Express 25 percent as 1/4 = (1/2) squared.
2. Count 2 half-lives.
3. Multiply 2 x 5730 = 11460.
4. Recompute: after 5730 years 50 percent remains, after 11460 years 25 percent remains. | evolution |
Estimated N equals 200 fish; the ratio of marked animals in the second catch scales to the whole population. | null | bio_0028 | worked_example | en | In a mark-recapture study, 50 fish are marked and released (M). Later 40 are caught (C), of which 10 are marked (R). Estimate population size N with the Lincoln-Petersen index. | [
"Write N = M x C / R.",
"Substitute 50 x 40 / 10.",
"Compute 2000 / 10 = 200.",
"Recompute: 200 x 10 / 40 = 50, matching M."
] | biology | Answer: N=200 fish.
Steps:
1. Write N = M x C / R.
2. Substitute 50 x 40 / 10.
3. Compute 2000 / 10 = 200.
4. Recompute: 200 x 10 / 40 = 50, matching M. | ecology |
Minute ventilation is 6.0 liters per minute; volume per breath times rate gives volume per minute. | null | bio_0029 | worked_example | en | Tidal volume is 500 mL and breathing rate is 12 breaths per minute. Calculate minute ventilation. | [
"Multiply 500 x 12 = 6000 mL per min.",
"Convert to 6.0 L per min.",
"Recompute: 6.0 x 1000 / 12 = 500 mL."
] | biology | Answer: 6.0 L per min.
Steps:
1. Multiply 500 x 12 = 6000 mL per min.
2. Convert to 6.0 L per min.
3. Recompute: 6.0 x 1000 / 12 = 500 mL. | physiology |
Koncentrace je 0,25 mol na litr; látkové množství dělené objemem dává molaritu. | null | bio_0030 | worked_example | cs | Ve 2 litrech roztoku je rozpuštěno 0,5 mol glukózy. Vypočítejte molární koncentraci. | [
"Zapište n = 0,5 mol a V = 2 l.",
"Vypočítejte c = 0,5 / 2 = 0,25.",
"Přepočítejte: 0,25 x 2 = 0,5.",
"Uveďte 0,25 mol na litr."
] | biology | Answer: 0,25 mol na litr.
Steps:
1. Zapište n = 0,5 mol a V = 2 l.
2. Vypočítejte c = 0,5 / 2 = 0,25.
3. Přepočítejte: 0,25 x 2 = 0,5.
4. Uveďte 0,25 mol na litr. | biochemistry |
Ratios are 6 to 1 versus 3 to 1; the smaller cube has twice the surface per unit volume. | null | bio_0031 | worked_example | en | Compare surface-area-to-volume ratio for cubes of side 1 micrometer and 2 micrometers. Show the calculation. | [
"Use SA = 6s squared and V = s cubed.",
"For s=1: SA=6, V=1, ratio=6.",
"For s=2: SA=24, V=8, ratio=3.",
"Recompute: 24/8=3 and 6/1=6."
] | biology | Answer: 6:1 for 1 micrometer and 3:1 for 2 micrometers.
Steps:
1. Use SA = 6s squared and V = s cubed.
2. For s=1: SA=6, V=1, ratio=6.
3. For s=2: SA=24, V=8, ratio=3.
4. Recompute: 24/8=3 and 6/1=6. | cell_biology |
Epistasis is interaction between different genes where one gene masks another, such as coat-color C masking pattern gene B in mice; dominance is interaction between alleles of the same gene. | null | bio_0032 | concept | en | What is epistasis and how does it differ from dominance? | [
"Define epistasis as between-gene masking.",
"Contrast with within-gene dominance.",
"Give coat color as example."
] | biology | Answer: Epistasis masks across genes; dominance acts within one gene.
Steps:
1. Define epistasis as between-gene masking.
2. Contrast with within-gene dominance.
3. Give coat color as example. | genetics |
A barrier splits one population into isolated groups that accumulate different mutations and experience different selection; reduced gene flow lets reproductive isolation evolve until they can no longer interbreed. | null | bio_0033 | concept | en | Explain allopatric speciation through geographic isolation. | [
"Start with one population split by a barrier.",
"Describe divergent selection, drift, and mutation.",
"End with reproductive isolation."
] | biology | Answer: Barriers split populations until they become reproductively isolated species.
Steps:
1. Start with one population split by a barrier.
2. Describe divergent selection, drift, and mutation.
3. End with reproductive isolation. | evolution |
Primary succession starts on lifeless substrate such as bare lava with pioneers like lichens building soil, while secondary succession restarts after disturbance in existing soil, such as forest regrowth after fire. | null | bio_0034 | concept | en | Distinguish primary from secondary ecological succession with examples. | [
"Define starting substrate for each type.",
"Give lava versus post-fire examples.",
"Compare time scale and pioneer roles."
] | biology | Answer: Primary starts on bare substrate; secondary restarts in existing soil.
Steps:
1. Define starting substrate for each type.
2. Give lava versus post-fire examples.
3. Compare time scale and pioneer roles. | ecology |
Nefron filtruje krev v glomerulu, poté v tubulech zpětně vstřebává vodu, glukózu a ionty a vylučuje odpadní látky, čímž vzniká koncentrovaná moč a udržuje se vodní a iontová rovnováha. | null | bio_0035 | concept | cs | Vysvětlete základní funkce nefronu při tvorbě moči. | [
"Popište filtraci v glomerulu.",
"Vysvětlete reabsorpci v proximálním tubulu a Henleově kličce.",
"Vysvětlete sekreci a zahuštění moči."
] | biology | Answer: Nefron filtruje, vstřebává a vylučuje za vzniku moči.
Steps:
1. Popište filtraci v glomerulu.
2. Vysvětlete reabsorpci v proximálním tubulu a Henleově kličce.
3. Vysvětlete sekreci a zahuštění moči. | physiology |
Hydrolysis of ATP to ADP releases about 30 kJ per mol that can drive unfavorable reactions when they share an intermediate, often by phosphorylation that raises reactant energy and lowers the barrier. | null | bio_0036 | concept | en | Explain how ATP hydrolysis is coupled to endergonic reactions. | [
"Recall ATP to ADP is exergonic.",
"Describe shared-intermediate coupling.",
"Give phosphorylation as mechanism."
] | biology | Answer: Exergonic ATP breakdown pays for endergonic work via coupling.
Steps:
1. Recall ATP to ADP is exergonic.
2. Describe shared-intermediate coupling.
3. Give phosphorylation as mechanism. | biochemistry |
Mitochondria and chloroplasts descend from engulfed prokaryotes; they have their own circular DNA, bacterial-size ribosomes, double membranes, and divide by binary fission. | null | bio_0037 | concept | en | Summarize the endosymbiotic theory and two lines of evidence for it. | [
"State engulfment of aerobic and photosynthetic bacteria.",
"Cite circular DNA and 70S ribosomes.",
"Cite double membranes and fission."
] | biology | Answer: Organelles arose from engulfed bacteria with their own DNA and membranes.
Steps:
1. State engulfment of aerobic and photosynthetic bacteria.
2. Cite circular DNA and 70S ribosomes.
3. Cite double membranes and fission. | cell_biology |
Homologous chromatids exchange segments at chiasmata during prophase I, creating new allele combinations on each chromatid that increase genetic diversity beyond independent assortment. | null | bio_0038 | concept | en | Explain crossing over in prophase I and why it matters for variation. | [
"Place event in pachytene of prophase I.",
"Describe chiasma formation and exchange.",
"Link new combinations to variation."
] | biology | Answer: Crossing over shuffles alleles between homologs to create diversity.
Steps:
1. Place event in pachytene of prophase I.
2. Describe chiasma formation and exchange.
3. Link new combinations to variation. | genetics |
Convergent evolution is independent origin of similar traits in distant lineages facing similar pressures; dolphins and sharks both evolved streamlined bodies and fins despite mammal versus fish ancestry. | null | bio_0039 | concept | en | What is convergent evolution? Give one example. | [
"Define independent similarity under similar selection.",
"Contrast with homology.",
"Give dolphin-shark streamlining."
] | biology | Answer: Distant lineages independently evolve similar traits, as in dolphins and sharks.
Steps:
1. Define independent similarity under similar selection.
2. Contrast with homology.
3. Give dolphin-shark streamlining. | evolution |
Vzdušný dusík fixují hlízkové a volně žijící bakterie na amonné ionty, nitrifikační bakterie je mění na dusičnany přijatelné rostlinami a denitrifikační bakterie vrací dusík zpět do atmosféry. | null | bio_0040 | concept | cs | Vysvětlete koloběh dusíku včetně role bakterií. | [
"Popište fixaci vzdušného dusíku.",
"Vysvětlete nitrifikaci k dusičnanům.",
"Vysvětlete denitrifikaci zpět do vzduchu."
] | biology | Answer: Bakterie dusík fixují, nitrifikují a denitrifikují v uzavřeném cyklu.
Steps:
1. Popište fixaci vzdušného dusíku.
2. Vysvětlete nitrifikaci k dusičnanům.
3. Vysvětlete denitrifikaci zpět do vzduchu. | ecology |
(B) Insulin; secreted by pancreatic beta cells, it increases glucose uptake, glycogenesis, and fat storage. | {
"A": "Glucagon",
"B": "Insulin",
"C": "Cortisol",
"D": "Thyroxine"
} | bio_0041 | multiple_choice | en | Which hormone primarily lowers blood glucose by promoting uptake and storage? | [
"Recall pancreatic control of glucose.",
"Eliminate glucose-raising and metabolic-rate hormones.",
"Choose insulin."
] | biology | Answer: (B) Insulin lowers blood glucose.
Steps:
1. Recall pancreatic control of glucose.
2. Eliminate glucose-raising and metabolic-rate hormones.
3. Choose insulin.
Answer: (B) | physiology |
(C) 100 times; each pH unit is a tenfold change, so two units give 10 squared. | {
"A": "2 times",
"B": "20 times",
"C": "100 times",
"D": "1000 times"
} | bio_0042 | multiple_choice | en | Solution A has pH 3 and solution B has pH 5. How much higher is the hydrogen ion concentration in A? | [
"Recall pH is minus log of H concentration.",
"Compute difference 5-3=2.",
"Raise 10 to the power 2 = 100."
] | biology | Answer: (C) 100 times higher H concentration.
Steps:
1. Recall pH is minus log of H concentration.
2. Compute difference 5-3=2.
3. Raise 10 to the power 2 = 100.
Answer: (C) | biochemistry |
(B) Thylakoid membranes; chlorophyll and electron carriers there capture light to make ATP and NADPH. | {
"A": "Stroma",
"B": "Thylakoid membranes",
"C": "Cytosol",
"D": "Mitochondrial matrix"
} | bio_0043 | multiple_choice | en | Where do the light-dependent reactions of photosynthesis occur in chloroplasts? | [
"Separate light reactions from Calvin cycle in stroma.",
"Eliminate non-chloroplast compartments.",
"Choose thylakoid membranes."
] | biology | Answer: (B) Light reactions occur on thylakoid membranes.
Steps:
1. Separate light reactions from Calvin cycle in stroma.
2. Eliminate non-chloroplast compartments.
3. Choose thylakoid membranes.
Answer: (B) | cell_biology |
(B) Methionine and start; AUG initiates translation and encodes methionine at internal positions. | {
"A": "Stop",
"B": "Methionine and start",
"C": "Alanine",
"D": "Tryptophan only"
} | bio_0044 | multiple_choice | en | In the standard genetic code, what does the mRNA codon AUG specify? | [
"Recall start codon identity.",
"Eliminate stop and other amino acids.",
"Choose methionine and start."
] | biology | Answer: (B) AUG means methionine and start.
Steps:
1. Recall start codon identity.
2. Eliminate stop and other amino acids.
3. Choose methionine and start.
Answer: (B) | genetics |
(A) Fosilní přechodné formy a homologie končetin; sdílená stavba kostry a postupné fosilní řady ukazují na společné předky. | {
"A": "Fosilní přechodné formy a homologie končetin",
"B": "Podobné prostředí",
"C": "Stejná potrava",
"D": "Stejná velikost těla"
} | bio_0045 | multiple_choice | cs | Který z následujících důkazů nejlépe podporuje společný původ obratlovců? | [
"Odlište dědičné znaky od ekologických shod.",
"Přiřaďte homologii společnému původu.",
"Zvolte fosilie a homologii."
] | biology | Answer: (A) Fosilie a homologie končetin dokládají společný původ.
Steps:
1. Odlište dědičné znaky od ekologických shod.
2. Přiřaďte homologii společnému původu.
3. Zvolte fosilie a homologii.
Answer: (A) | evolution |
(B) Disproportionate collapse or restructuring; keystone species hold food webs together beyond their abundance. | {
"A": "No community change",
"B": "Disproportionate collapse or restructuring",
"C": "Uniform increase in all prey fitness",
"D": "Immediate stable increase in species richness"
} | bio_0046 | multiple_choice | en | What is the expected effect of removing a keystone predator? | [
"Recall keystone impact exceeds biomass.",
"Eliminate no-effect and uniformly positive options.",
"Choose collapse or restructuring."
] | biology | Answer: (B) Removal causes disproportionate restructuring.
Steps:
1. Recall keystone impact exceeds biomass.
2. Eliminate no-effect and uniformly positive options.
3. Choose collapse or restructuring.
Answer: (B) | ecology |
BMI is about 22.9; weight in kilograms divided by height squared gives the index. | null | bio_0047 | worked_example | en | A person weighs 70 kg and is 1.75 m tall. Calculate body mass index (BMI). | [
"Square height: 1.75 x 1.75 = 3.0625.",
"Divide 70 / 3.0625 = 22.86.",
"Recompute: 22.86 x 3.0625 = 70.0 within rounding.",
"Report about 22.9."
] | biology | Answer: BMI about 22.9.
Steps:
1. Square height: 1.75 x 1.75 = 3.0625.
2. Divide 70 / 3.0625 = 22.86.
3. Recompute: 22.86 x 3.0625 = 70.0 within rounding.
4. Report about 22.9. | physiology |
50 mL of stock is needed; dilution conserves moles of solute. | null | bio_0048 | worked_example | en | Using C1V1 equals C2V2, what volume of 10 M stock is needed to prepare 1.0 L of 0.50 M solution? | [
"Write V1 = C2V2 / C1.",
"Substitute 0.50 x 1000 / 10.",
"Compute 500 / 10 = 50 mL.",
"Recompute: 10 x 50 = 500 and 0.50 x 1000 = 500."
] | biology | Answer: 50 mL of 10 M stock.
Steps:
1. Write V1 = C2V2 / C1.
2. Substitute 0.50 x 1000 / 10.
3. Compute 500 / 10 = 50 mL.
4. Recompute: 10 x 50 = 500 and 0.50 x 1000 = 500. | biochemistry |
Net water movement is outward and cells crenate; the external solution is hypertonic so osmosis draws water out. | null | bio_0049 | worked_example | en | Red blood cells are isotonic in 0.9 percent NaCl. Predict net water movement when placed in 2.0 percent NaCl and the resulting shape. | [
"Compare 2.0 percent to isotonic 0.9 percent.",
"Identify external solution as hypertonic.",
"Predict water leaves by osmosis.",
"Conclude cells shrink and crenate."
] | biology | Answer: Water leaves and cells crenate.
Steps:
1. Compare 2.0 percent to isotonic 0.9 percent.
2. Identify external solution as hypertonic.
3. Predict water leaves by osmosis.
4. Conclude cells shrink and crenate. | cell_biology |
Podíl je 1/16, tedy 6,25 procenta; pravděpodobnosti pro oba geny se násobí. | null | bio_0050 | worked_example | cs | Při křížení AaBb x AaBb s nezávislou segregací vypočítejte očekávaný podíl genotypu aabb. | [
"Určete P(aa) = 1/4 a P(bb) = 1/4.",
"Vynásobte 1/4 x 1/4 = 1/16.",
"Převeďte na 0,0625, tedy 6,25 procenta.",
"Přepočítejte: 16 x 1/16 = 1."
] | biology | Answer: 1/16, tedy 6,25 procenta.
Steps:
1. Určete P(aa) = 1/4 a P(bb) = 1/4.
2. Vynásobte 1/4 x 1/4 = 1/16.
3. Převeďte na 0,0625, tedy 6,25 procenta.
4. Přepočítejte: 16 x 1/16 = 1. | genetics |
20 individuals are expected; genotype frequency q squared times population size gives the count. | null | bio_0051 | worked_example | en | If recessive allele frequency q equals 0.20 in a Hardy-Weinberg population of 500 diploid individuals, how many homozygous recessive individuals are expected? | [
"Compute q squared = 0.20 x 0.20 = 0.04.",
"Multiply 0.04 x 500 = 20.",
"Recompute: 20 / 500 = 0.04 and sqrt(0.04) = 0.20."
] | biology | Answer: 20 homozygous recessive individuals.
Steps:
1. Compute q squared = 0.20 x 0.20 = 0.04.
2. Multiply 0.04 x 500 = 20.
3. Recompute: 20 / 500 = 0.04 and sqrt(0.04) = 0.20. | evolution |
Two species competing for the same limiting resource cannot coexist indefinitely, one will exclude the other; they can coexist by resource partitioning, such as feeding at different depths or times. | null | bio_0052 | concept | en | State the competitive exclusion principle and one way competing species can coexist. | [
"State complete competitors cannot share one niche.",
"Explain partitioning of food, space, or time.",
"Give warblers feeding in different tree zones."
] | biology | Answer: Complete competitors exclude each other unless niches partition.
Steps:
1. State complete competitors cannot share one niche.
2. Explain partitioning of food, space, or time.
3. Give warblers feeding in different tree zones. | ecology |
Myosin heads bind actin and pull thin filaments toward the sarcomere center using ATP, shortening the I band and H zone while filament lengths stay constant. | null | bio_0053 | concept | en | Outline the sliding filament theory of muscle contraction. | [
"Recall sarcomere with actin and myosin.",
"Describe cross-bridge cycle powered by ATP.",
"Explain band shortening without filament shortening."
] | biology | Answer: Myosin pulls actin inward to shorten sarcomeres.
Steps:
1. Recall sarcomere with actin and myosin.
2. Describe cross-bridge cycle powered by ATP.
3. Explain band shortening without filament shortening. | physiology |
Two antiparallel sugar-phosphate backbones wind around stacked base pairs, with adenine pairing thymine by two hydrogen bonds and guanine pairing cytosine by three. | null | bio_0054 | concept | en | Describe the Watson-Crick double helix: backbones, pairing, and orientation. | [
"Identify deoxyribose-phosphate backbones.",
"State A-T and G-C pairing rules.",
"Explain antiparallel 5 to 3 orientation."
] | biology | Answer: Antiparallel backbones with A-T and G-C pairs form the helix.
Steps:
1. Identify deoxyribose-phosphate backbones.
2. State A-T and G-C pairing rules.
3. Explain antiparallel 5 to 3 orientation. | biochemistry |
Prokaryotická buňka nemá jádro ani membránové organely, má kruhovou DNA v nukleoidu a menší ribozomy, zatímco eukaryotická buňka má jádro, lineární chromozomy a organely jako mitochondrie či chloroplasty. | null | bio_0055 | concept | cs | Popište hlavní rozdíly mezi prokaryotickou a eukaryotickou buňkou. | [
"Porovnejte přítomnost jádra a organel.",
"Porovnejte DNA a ribozomy.",
"Uveďte příklady bakterií a živočišných buněk."
] | biology | Answer: Prokaryota nemají jádro ani organely, eukaryota ano.
Steps:
1. Porovnejte přítomnost jádra a organel.
2. Porovnejte DNA a ribozomy.
3. Uveďte příklady bakterií a živočišných buněk. | cell_biology |
Without lactose, the repressor blocks transcription; lactose inactivates the repressor, and low glucose raises cAMP-CAP activation, so full expression occurs only with lactose present and glucose absent. | null | bio_0056 | concept | en | Explain how the lac operon is regulated by lactose and glucose. | [
"Describe repressor binding to operator.",
"Explain allolactose induction.",
"Explain cAMP-CAP catabolite activation."
] | biology | Answer: Lactose removes repression and low glucose boosts activation.
Steps:
1. Describe repressor binding to operator.
2. Explain allolactose induction.
3. Explain cAMP-CAP catabolite activation. | genetics |
Gradualism predicts slow continuous change accumulating over time, while punctuated equilibrium predicts long stasis interrupted by rapid speciation events, often linked to small isolated populations. | null | bio_0057 | concept | en | Contrast gradualism and punctuated equilibrium as models of evolutionary tempo. | [
"Define steady incremental change.",
"Define stasis plus rapid bursts.",
"Compare fossil expectations."
] | biology | Answer: Gradualism is slow and steady; punctuated equilibrium is stasis plus bursts.
Steps:
1. Define steady incremental change.
2. Define stasis plus rapid bursts.
3. Compare fossil expectations. | evolution |
Excess nitrogen and phosphorus from fertilizer or sewage fuel algal blooms that block light and, when decomposed by bacteria, deplete oxygen and create hypoxic dead zones that kill fish. | null | bio_0058 | concept | en | Explain eutrophication: causes, algal blooms, and hypoxia. | [
"Identify nutrient inputs.",
"Trace bloom growth and shading.",
"Trace decomposition to oxygen loss."
] | biology | Answer: Nutrients drive blooms whose decay depletes oxygen.
Steps:
1. Identify nutrient inputs.
2. Trace bloom growth and shading.
3. Trace decomposition to oxygen loss. | ecology |
Depolarization opens voltage-gated sodium channels to spike the membrane positive, then sodium inactivates while potassium efflux repolarizes, followed by brief hyperpolarization before the resting potential restores. | null | bio_0059 | concept | en | Describe the phases of a neuronal action potential. | [
"Start from resting minus 70 mV.",
"Trace Na influx and rising phase.",
"Trace K efflux, falling phase, and undershoot."
] | biology | Answer: Na influx depolarizes, then K efflux repolarizes with undershoot.
Steps:
1. Start from resting minus 70 mV.
2. Trace Na influx and rising phase.
3. Trace K efflux, falling phase, and undershoot. | physiology |
Ve světelné fázi na thylakoidech se energie světla mění na ATP a NADPH za uvolnění kyslíku, v temnostní fázi Calvinova cyklu ve stromatu se oxid uhličitý pomocí ATP a NADPH fixuje na cukry. | null | bio_0060 | concept | cs | Vysvětlete dvě fáze fotosyntézy: světelnou a temnostní. | [
"Popište záchyt světla a fotolýzu vody.",
"Popište vznik ATP a NADPH.",
"Popište fixaci CO2 v Calvinově cyklu."
] | biology | Answer: Světelná fáze tvoří ATP a NADPH, temnostní fáze fixuje CO2.
Steps:
1. Popište záchyt světla a fotolýzu vody.
2. Popište vznik ATP a NADPH.
3. Popište fixaci CO2 v Calvinově cyklu. | biochemistry |
(C) Intermediate filaments; these rope-like keratin and lamin fibers provide tensile strength and anchor the nucleus and organelles. | {
"A": "Microfilaments",
"B": "Microtubules",
"C": "Intermediate filaments",
"D": "Bacterial flagella"
} | bio_0061 | multiple_choice | en | Which cytoskeletal element mainly resists mechanical stress and anchors organelles in place? | [
"Recall microfilaments drive movement and microtubules guide transport.",
"Assign tensile strength and anchoring to intermediate filaments.",
"Choose intermediate filaments."
] | biology | Answer: (C) Intermediate filaments resist stress and anchor organelles.
Steps:
1. Recall microfilaments drive movement and microtubules guide transport.
2. Assign tensile strength and anchoring to intermediate filaments.
3. Choose intermediate filaments.
Answer: (C) | cell_biology |
(A) 1 dominant : 1 recessive; half the offspring receive A and show the dominant trait, half receive a and show the recessive trait. | {
"A": "1 dominant : 1 recessive",
"B": "3 dominant : 1 recessive",
"C": "9:3:3:1",
"D": "All dominant"
} | bio_0062 | multiple_choice | en | In a test cross Aa x aa with complete dominance, what phenotypic ratio is expected in the offspring? | [
"List gametes A,a from one parent and a,a from the other.",
"Fill the Punnett square to get 2 Aa and 2 aa.",
"Convert to 1:1 phenotypes."
] | biology | Answer: (A) A 1:1 phenotypic ratio is expected.
Steps:
1. List gametes A,a from one parent and a,a from the other.
2. Fill the Punnett square to get 2 Aa and 2 aa.
3. Convert to 1:1 phenotypes.
Answer: (A) | genetics |
(D) Large size, random mating, no selection; equilibrium also requires no mutation, no migration, and no drift. | {
"A": "Strong directional selection",
"B": "High mutation rate",
"C": "Frequent migration",
"D": "Large size, random mating, no selection"
} | bio_0063 | multiple_choice | en | Which set of conditions is required for a population to stay in Hardy-Weinberg equilibrium? | [
"Recall equilibrium means constant allele frequencies.",
"Eliminate selection, mutation, and migration.",
"Choose large random-mating population."
] | biology | Answer: (D) Large random-mating populations without selection stay in equilibrium.
Steps:
1. Recall equilibrium means constant allele frequencies.
2. Eliminate selection, mutation, and migration.
3. Choose large random-mating population.
Answer: (D) | evolution |
(B) Largest sustainable population; limited resources cap growth so the population levels off near K. | {
"A": "Maximum per-capita growth rate",
"B": "Largest population size the environment can sustain",
"C": "Minimum viable population",
"D": "Number of predators at equilibrium"
} | bio_0064 | multiple_choice | en | What does the carrying capacity (K) of an environment describe? | [
"Recall K is set by limiting resources.",
"Eliminate rate, minimum, and predator options.",
"Choose largest sustainable population."
] | biology | Answer: (B) K is the largest sustainable population size.
Steps:
1. Recall K is set by limiting resources.
2. Eliminate rate, minimum, and predator options.
3. Choose largest sustainable population.
Answer: (B) | ecology |
(D) Mozeček; přijímá informace o poloze těla a jemně dolaďuje svalové povely. | {
"A": "Prodloužená mícha",
"B": "Hypotalamus",
"C": "Talamus",
"D": "Mozeček"
} | bio_0065 | multiple_choice | cs | Která část mozku je hlavním centrem rovnováhy a koordinace pohybů? | [
"Vylučte centra dýchání, hormonů a čití.",
"Přiřaďte koordinaci mozečku.",
"Zvolte mozeček."
] | biology | Answer: (D) Mozeček koordinuje rovnováhu a pohyby.
Steps:
1. Vylučte centra dýchání, hormonů a čití.
2. Přiřaďte koordinaci mozečku.
3. Zvolte mozeček.
Answer: (D) | physiology |
(A) Phospholipids; their bilayers form the membrane matrix, with cholesterol and proteins embedded. | {
"A": "Phospholipids",
"B": "Triglycerides",
"C": "Waxes",
"D": "Carotenoids"
} | bio_0066 | multiple_choice | en | Which lipid class is the main structural component of cell membranes? | [
"Recall membranes are bilayers.",
"Eliminate storage, coating, and pigment lipids.",
"Choose phospholipids."
] | biology | Answer: (A) Phospholipids build membrane bilayers.
Steps:
1. Recall membranes are bilayers.
2. Eliminate storage, coating, and pigment lipids.
3. Choose phospholipids.
Answer: (A) | biochemistry |
46 chromosomes and 92 chromatids; DNA replication doubles chromatids while chromosome number stays constant until anaphase. | null | bio_0067 | worked_example | en | A diploid human cell has 46 chromosomes. After S phase (in G2), how many chromosomes and how many chromatids does it contain? | [
"Recall chromosome counts centromeres: still 46 after S.",
"Each chromosome now has 2 sister chromatids: 46x2=92.",
"Recompute: 92/2=46 chromosomes.",
"Report 46 chromosomes, 92 chromatids."
] | biology | Answer: 46 chromosomes and 92 chromatids.
Steps:
1. Recall chromosome counts centromeres: still 46 after S.
2. Each chromosome now has 2 sister chromatids: 46x2=92.
3. Recompute: 92/2=46 chromosomes.
4. Report 46 chromosomes, 92 chromatids. | cell_biology |
1/4, or 25 percent; the two genes segregate independently so probabilities multiply. | null | bio_0068 | worked_example | en | In a test cross AaBb x aabb with independent assortment, what proportion of offspring is expected to be genotype AaBb? | [
"P(Aa from Aa x aa) = 1/2.",
"P(Bb from Bb x bb) = 1/2.",
"Multiply 1/2 x 1/2 = 1/4.",
"Recompute: 4 equally likely gamete combos from AaBb, one is AB."
] | biology | Answer: 1/4, or 25 percent, are expected to be AaBb.
Steps:
1. P(Aa from Aa x aa) = 1/2.
2. P(Bb from Bb x bb) = 1/2.
3. Multiply 1/2 x 1/2 = 1/4.
4. Recompute: 4 equally likely gamete combos from AaBb, one is AB. | genetics |
160 AA, 80 Aa, and 10 aa; genotype frequencies p squared, 2pq, and q squared scale to the population. | null | bio_0069 | worked_example | en | A gene has dominant allele frequency p = 0.8 in Hardy-Weinberg equilibrium. In 250 diploid individuals, how many of each genotype are expected? | [
"Compute q = 1-0.8 = 0.2.",
"AA = 0.64x250=160; Aa = 0.32x250=80; aa = 0.04x250=10.",
"Recompute: 160+80+10=250; 0.64+0.32+0.04=1.0."
] | biology | Answer: 160 AA, 80 Aa, 10 aa.
Steps:
1. Compute q = 1-0.8 = 0.2.
2. AA = 0.64x250=160; Aa = 0.32x250=80; aa = 0.04x250=10.
3. Recompute: 160+80+10=250; 0.64+0.32+0.04=1.0. | evolution |
Odhad je 120 ryb; podíl označených ve vzorku odpovídá podílu v populaci. | null | bio_0070 | worked_example | cs | Při sčítání ryb bylo označeno a vypuštěno 30 kusů. Při dalším odlovu 60 kusů bylo 15 označených. Odhadněte velikost populace Lincoln-Petersenovým indexem. | [
"Zapište N = M x C / R.",
"Dosaďte 30 x 60 / 15.",
"Vypočítejte 1800 / 15 = 120.",
"Přepočítejte: 120 x 15 / 60 = 30."
] | biology | Answer: N=120 ryb.
Steps:
1. Zapište N = M x C / R.
2. Dosaďte 30 x 60 / 15.
3. Vypočítejte 1800 / 15 = 120.
4. Přepočítejte: 120 x 15 / 60 = 30. | ecology |
MAP is about 93.3 mmHg; diastole dominates the cardiac cycle so MAP sits closer to diastolic pressure. | null | bio_0071 | worked_example | en | Blood pressure is 120 over 80 mmHg. Estimate mean arterial pressure (MAP) using MAP = diastolic + (systolic minus diastolic)/3. | [
"Pulse pressure = 120-80 = 40.",
"Divide 40/3 = 13.33.",
"Add 80+13.33 = 93.33.",
"Recompute: (93.33-80)x3 = 40."
] | biology | Answer: MAP about 93.3 mmHg.
Steps:
1. Pulse pressure = 120-80 = 40.
2. Divide 40/3 = 13.33.
3. Add 80+13.33 = 93.33.
4. Recompute: (93.33-80)x3 = 40. | physiology |
Denaturation is loss of three-dimensional shape from broken hydrogen bonds and hydrophobic interactions by heat, pH extremes, or chemicals; the active site deforms so substrates no longer bind, while the amino acid sequence stays intact. | null | bio_0072 | concept | en | What is protein denaturation and why does it destroy enzyme function? | [
"Define loss of secondary and tertiary structure.",
"Name heat, pH, and solvents as causes.",
"Link deformed active site to lost function."
] | biology | Answer: Denaturation deforms the active site so substrates no longer bind.
Steps:
1. Define loss of secondary and tertiary structure.
2. Name heat, pH, and solvents as causes.
3. Link deformed active site to lost function. | biochemistry |
Facilitated diffusion moves molecules down their concentration gradient through channels or carriers without ATP, while active transport pumps substances against the gradient using ATP, as in the sodium-potassium pump. | null | bio_0073 | concept | en | Contrast facilitated diffusion and active transport across membranes. | [
"Compare direction relative to gradient.",
"Compare ATP requirement.",
"Give glucose channel versus Na-K pump examples."
] | biology | Answer: Facilitated diffusion is passive downhill flow; active transport pumps uphill with ATP.
Steps:
1. Compare direction relative to gradient.
2. Compare ATP requirement.
3. Give glucose channel versus Na-K pump examples. | cell_biology |
Genes on different chromosomes assort independently because homolog pairs align randomly at metaphase I; genes close together on the same chromosome are linked and violate the law unless crossing over separates them. | null | bio_0074 | concept | en | State Mendel's law of independent assortment and explain its chromosomal basis and one exception. | [
"State random homolog alignment in meiosis I.",
"Explain 9:3:3:1 dihybrid outcome.",
"Name linkage as the exception."
] | biology | Answer: Unlinked genes assort independently; linked genes are the exception.
Steps:
1. State random homolog alignment in meiosis I.
2. Explain 9:3:3:1 dihybrid outcome.
3. Name linkage as the exception. | genetics |
Podmínkami jsou proměnlivost znaků mezi jedinci, dědičnost této proměnlivosti a rozdílná zdatnost, tedy že některé varianty přežívají a množí se lépe; bez kterékoli z nich se frekvence alel selekcí nemění. | null | bio_0075 | concept | cs | Vysvětlete tři podmínky nutné pro působení přírodního výběru. | [
"Popište variabilitu a dědičnost.",
"Vysvětlete rozdílné přežívání a rozmnožování.",
"Uveďte příklad odolnosti vůči antibiotikům."
] | biology | Answer: Variabilita, dědičnost a rozdílná zdatnost umožňují selekci.
Steps:
1. Popište variabilitu a dědičnost.
2. Vysvětlete rozdílné přežívání a rozmnožování.
3. Uveďte příklad odolnosti vůči antibiotikům. | evolution |
Only about 10 percent of energy at one trophic level becomes biomass at the next, because most is lost as heat through respiration, movement, and incomplete digestion, which limits food chain length. | null | bio_0076 | concept | en | What is the 10 percent rule of energy transfer between trophic levels? | [
"Track energy loss to respiration and heat.",
"Explain 90 percent loss per transfer.",
"Link to short food chains and pyramids."
] | biology | Answer: About 10 percent of energy moves up each level.
Steps:
1. Track energy loss to respiration and heat.
2. Explain 90 percent loss per transfer.
3. Link to short food chains and pyramids. | ecology |
Rising blood osmolarity triggers ADH release from the posterior pituitary, which inserts aquaporins in kidney collecting ducts so more water is reabsorbed, concentrating urine and diluting blood back toward normal. | null | bio_0077 | concept | en | Explain how antidiuretic hormone (ADH) regulates water balance. | [
"Detect high osmolarity via hypothalamic receptors.",
"Release ADH and insert aquaporins.",
"Reabsorb water until osmolarity normalizes."
] | biology | Answer: ADH saves water by adding kidney aquaporins when blood is concentrated.
Steps:
1. Detect high osmolarity via hypothalamic receptors.
2. Release ADH and insert aquaporins.
3. Reabsorb water until osmolarity normalizes. | physiology |
DNA is a double-stranded deoxyribose helix with thymine that stores hereditary information, while RNA is usually single-stranded ribose with uracil that carries messages (mRNA), builds ribosomes (rRNA), and delivers amino acids (tRNA). | null | bio_0078 | concept | en | Compare DNA and RNA in structure and function. | [
"Contrast sugars, strands, and bases.",
"Contrast storage versus working roles.",
"Name mRNA, tRNA, and rRNA."
] | biology | Answer: DNA stores information in a double helix; RNA works in single strands.
Steps:
1. Contrast sugars, strands, and bases.
2. Contrast storage versus working roles.
3. Name mRNA, tRNA, and rRNA. | biochemistry |
In prophase chromosomes condense and the spindle forms, in metaphase they align at the equator, in anaphase sister chromatids separate to opposite poles, and in telophase nuclear envelopes reform around two identical sets. | null | bio_0079 | concept | en | Describe the four stages of mitosis and the key event of each. | [
"Trace condensation and spindle formation.",
"Trace alignment then separation.",
"End with two nuclei and cytokinesis."
] | biology | Answer: Prophase condenses, metaphase aligns, anaphase separates, telophase reforms.
Steps:
1. Trace condensation and spindle formation.
2. Trace alignment then separation.
3. End with two nuclei and cytokinesis. | cell_biology |
Při kodominanci se u heterozygota plně projeví obě alely zároveň; člověk s genotypem IAIB má na červených krvinkách antigeny A i B, tedy krevní skupinu AB, zatímco alela i je recesivní vůči oběma. | null | bio_0080 | concept | cs | Vysvětlete kodominanci na příkladu krevních skupin systému AB0. | [
"Definujte současný projev obou alel.",
"Přiřaďte genotyp IAIB k antigenům A i B.",
"Odlište od neúplné dominance."
] | biology | Answer: Genotyp IAIB nese antigeny A i B zároveň.
Steps:
1. Definujte současný projev obou alel.
2. Přiřaďte genotyp IAIB k antigenům A i B.
3. Odlište od neúplné dominance. | genetics |
(C) Mutation; only changes in DNA sequence create novel alleles, while other mechanisms reshuffle or remove existing variation. | {
"A": "Genetic drift",
"B": "Natural selection",
"C": "Mutation",
"D": "Nonrandom mating"
} | bio_0081 | multiple_choice | en | Which mechanism introduces entirely new alleles into a population's gene pool? | [
"Recall drift, selection, and mating act on existing alleles.",
"Identify DNA change as the novelty source.",
"Choose mutation."
] | biology | Answer: (C) Mutation creates new alleles.
Steps:
1. Recall drift, selection, and mating act on existing alleles.
2. Identify DNA change as the novelty source.
3. Choose mutation.
Answer: (C) | evolution |
(D) Commensalism; for example, barnacles gain transport on whales that are largely unaffected. | {
"A": "Mutualism",
"B": "Parasitism",
"C": "Competition",
"D": "Commensalism"
} | bio_0082 | multiple_choice | en | Which species interaction benefits one species while the other is neither helped nor harmed? | [
"Classify options by plus-minus outcomes.",
"Eliminate plus-plus, plus-minus, and minus-minus.",
"Choose commensalism."
] | biology | Answer: (D) Commensalism helps one partner only.
Steps:
1. Classify options by plus-minus outcomes.
2. Eliminate plus-plus, plus-minus, and minus-minus.
3. Choose commensalism.
Answer: (D) | ecology |
(A) Red blood cells; they lack nuclei and are packed with hemoglobin that binds oxygen in the lungs. | {
"A": "Red blood cells",
"B": "Platelets",
"C": "B lymphocytes",
"D": "Neutrophils"
} | bio_0083 | multiple_choice | en | Which blood cells are specialized for oxygen transport? | [
"Eliminate clotting and immune cells.",
"Recall hemoglobin in erythrocytes.",
"Choose red blood cells."
] | biology | Answer: (A) Red blood cells carry oxygen.
Steps:
1. Eliminate clotting and immune cells.
2. Recall hemoglobin in erythrocytes.
3. Choose red blood cells.
Answer: (A) | physiology |
(C) Mitochondrial matrix; pyruvate is oxidized there to CO2, feeding NADH and FADH2 to the electron transport chain. | {
"A": "Cytosol",
"B": "Nucleus",
"C": "Mitochondrial matrix",
"D": "Thylakoid lumen"
} | bio_0084 | multiple_choice | en | Where does the citric acid (Krebs) cycle occur in eukaryotes? | [
"Separate glycolysis in cytosol from aerobic steps.",
"Eliminate nucleus and chloroplast.",
"Choose mitochondrial matrix."
] | biology | Answer: (C) The Krebs cycle runs in the mitochondrial matrix.
Steps:
1. Separate glycolysis in cytosol from aerobic steps.
2. Eliminate nucleus and chloroplast.
3. Choose mitochondrial matrix.
Answer: (C) | biochemistry |
(B) Ribozom; spojuje aminokyseliny do polypeptidů podle informace z mRNA. | {
"A": "Lyzozom",
"B": "Ribozom",
"C": "Vakuola",
"D": "Peroxizom"
} | bio_0085 | multiple_choice | cs | Která organela je hlavním místem syntézy bílkovin? | [
"Vylučte trávicí, zásobní a detoxikační organely.",
"Přiřaďte translaci ribozomu.",
"Zvolte ribozom."
] | biology | Answer: (B) Ribozom syntetizuje bílkoviny.
Steps:
1. Vylučte trávicí, zásobní a detoxikační organely.
2. Přiřaďte translaci ribozomu.
3. Zvolte ribozom.
Answer: (B) | cell_biology |
(D) 50%; each son receives Y from his father and either XH or Xh from his mother with equal probability. | {
"A": "0%",
"B": "25%",
"C": "100%",
"D": "50%"
} | bio_0086 | multiple_choice | en | Hemophilia is X-linked recessive. A heterozygous woman (XHXh) and a normal man (XHY) have sons. What proportion of their sons is expected to have hemophilia? | [
"Write cross XHXh x XHY.",
"List sons XHY and XhY in equal ratio.",
"Choose 50 percent affected."
] | biology | Answer: (D) 50% of sons are expected to be affected.
Steps:
1. Write cross XHXh x XHY.
2. List sons XHY and XhY in equal ratio.
3. Choose 50 percent affected.
Answer: (D) | genetics |
About 17190 years old; 12.5 percent equals three half-lives of decay. | null | bio_0087 | worked_example | en | Carbon-14 has a half-life of 5730 years. A bone contains 12.5 percent of its original carbon-14. Estimate its age. | [
"Express 12.5 percent as 1/8 = (1/2) cubed.",
"Count 3 half-lives.",
"Multiply 3 x 5730 = 17190.",
"Recompute: 50, 25, then 12.5 percent after 1, 2, 3 half-lives."
] | biology | Answer: About 17190 years.
Steps:
1. Express 12.5 percent as 1/8 = (1/2) cubed.
2. Count 3 half-lives.
3. Multiply 3 x 5730 = 17190.
4. Recompute: 50, 25, then 12.5 percent after 1, 2, 3 half-lives. | evolution |
About 165 cells; exponential growth multiplies the starting population by e to the 0.5. | null | bio_0088 | worked_example | en | A bacterial colony starts with 100 cells and grows exponentially with per-capita rate r = 0.1 per hour. Estimate size after 5 hours using N = N0 times e to the power rt. | [
"Compute rt = 0.1 x 5 = 0.5.",
"Evaluate e^0.5 = about 1.6487.",
"Multiply 100 x 1.6487 = 164.87.",
"Recompute: ln(164.87/100) = 0.5, divided by 5 gives r = 0.1."
] | biology | Answer: About 165 cells after 5 hours.
Steps:
1. Compute rt = 0.1 x 5 = 0.5.
2. Evaluate e^0.5 = about 1.6487.
3. Multiply 100 x 1.6487 = 164.87.
4. Recompute: ln(164.87/100) = 0.5, divided by 5 gives r = 0.1. | ecology |
Vital capacity is 4600 mL and total lung capacity is 5800 mL; vital capacity excludes residual volume while total capacity includes it. | null | bio_0089 | worked_example | en | Given inspiratory reserve 3000 mL, tidal volume 500 mL, expiratory reserve 1100 mL, and residual volume 1200 mL, calculate vital capacity and total lung capacity. | [
"VC = 3000+500+1100 = 4600 mL.",
"TLC = 4600+1200 = 5800 mL.",
"Recompute: 5800-1200=4600 and 4600-500-1100=3000."
] | biology | Answer: VC=4600 mL, TLC=5800 mL.
Steps:
1. VC = 3000+500+1100 = 4600 mL.
2. TLC = 4600+1200 = 5800 mL.
3. Recompute: 5800-1200=4600 and 4600-500-1100=3000. | physiology |
Je potřeba 20 ml zásobního roztoku; ředění zachovává látkové množství rozpuštěné látky. | null | bio_0090 | worked_example | cs | Ze zásobního roztoku o koncentraci 5 mol na litr je třeba připravit 500 ml roztoku o koncentraci 0,2 mol na litr. Jaký objem zásobního roztoku je potřeba? | [
"Zapište V1 = C2V2 / C1.",
"Dosaďte 0,2 x 500 / 5.",
"Vypočítejte 100 / 5 = 20 ml.",
"Přepočítejte: 5 x 20 = 100 a 0,2 x 500 = 100."
] | biology | Answer: 20 ml zásobního roztoku.
Steps:
1. Zapište V1 = C2V2 / C1.
2. Dosaďte 0,2 x 500 / 5.
3. Vypočítejte 100 / 5 = 20 ml.
4. Přepočítejte: 5 x 20 = 100 a 0,2 x 500 = 100. | biochemistry |
About 25 cells; the fraction of the cycle spent in mitosis predicts the fraction of cells observed there. | null | bio_0091 | worked_example | en | A cell cycle lasts 24 hours and mitosis occupies 1 hour. In a sample of 600 unsynchronized cells, how many are expected to be in mitosis? | [
"Fraction in mitosis = 1/24.",
"Multiply 600 x 1/24 = 25.",
"Recompute: 25/600 = 1/24 and 25 x 24 = 600."
] | biology | Answer: About 25 cells in mitosis.
Steps:
1. Fraction in mitosis = 1/24.
2. Multiply 600 x 1/24 = 25.
3. Recompute: 25/600 = 1/24 and 25 x 24 = 600. | cell_biology |
Males have only one X chromosome, so a single recessive allele is fully expressed, while females need two copies because a dominant normal allele on the second X masks it. | null | bio_0092 | concept | en | Explain why X-linked recessive traits affect males more often than females. | [
"Recall male XY versus female XX.",
"Explain hemizygous expression in males.",
"Give hemophilia or color blindness as example."
] | biology | Answer: Males express any X-linked allele; females need two copies.
Steps:
1. Recall male XY versus female XX.
2. Explain hemizygous expression in males.
3. Give hemophilia or color blindness as example. | genetics |
Adaptive radiation is rapid diversification of one ancestor into many species filling different niches; finches on the Galapagos evolved varied beaks for seeds, insects, and cactus from a common ancestor. | null | bio_0093 | concept | en | What is adaptive radiation? Illustrate with Darwin's finches. | [
"Define rapid niche-filling diversification.",
"Describe finch beak variation by diet.",
"Link isolation and selection to speciation."
] | biology | Answer: One ancestor diversifies fast into many niches, as in Galapagos finches.
Steps:
1. Define rapid niche-filling diversification.
2. Describe finch beak variation by diet.
3. Link isolation and selection to speciation. | evolution |
A keystone species holds the community together beyond its abundance, like sea otters; an indicator species reveals environmental health, like lichens for air quality; a flagship species rallies conservation support, like pandas. | null | bio_0094 | concept | en | Distinguish keystone, indicator, and flagship species. | [
"Define disproportionate ecological role.",
"Define sensitivity revealing conditions.",
"Define charisma mobilizing protection."
] | biology | Answer: Keystones structure, indicators reveal, flagships rally.
Steps:
1. Define disproportionate ecological role.
2. Define sensitivity revealing conditions.
3. Define charisma mobilizing protection. | ecology |
Srdeční rytmus udává sinoatriální uzel, který vysílá vzruchy přes síně k atrioventrikulárnímu uzlu a Hisovým raménkům do komor; sympatikus činnost zrychluje a parasympatikus zpomaluje podle potřeb těla. | null | bio_0095 | concept | cs | Vysvětlete, jak je řízena srdeční činnost. | [
"Popište SA uzel jako udavatele rytmu.",
"Sledujte šíření vzruchu do komor.",
"Vysvětlete vliv sympatiku a parasympatiku."
] | biology | Answer: SA uzel udává rytmus, nervy ho zrychlují či zpomalují.
Steps:
1. Popište SA uzel jako udavatele rytmu.
2. Sledujte šíření vzruchu do komor.
3. Vysvětlete vliv sympatiku a parasympatiku. | physiology |
Lock-and-key treats the active site as a rigid match for the substrate, while induced fit holds that the enzyme reshapes slightly around the substrate for a tighter fit, which better explains broad specificity and catalysis. | null | bio_0096 | concept | en | Contrast the lock-and-key and induced-fit models of enzyme-substrate binding. | [
"Describe rigid complementarity.",
"Describe conformational adjustment.",
"Note induced fit fits kinetic evidence better."
] | biology | Answer: Lock-and-key is rigid; induced fit reshapes around substrate.
Steps:
1. Describe rigid complementarity.
2. Describe conformational adjustment.
3. Note induced fit fits kinetic evidence better. | biochemistry |
Osmosis moves water toward higher solute concentration; animal cells burst in hypotonic and shrivel in hypertonic solutions, while plant cells become turgid or plasmolyzed because the wall resists bursting. | null | bio_0097 | concept | en | Explain osmosis and predict effects of hypotonic, hypertonic, and isotonic solutions on animal and plant cells. | [
"Define water movement down its gradient.",
"Trace animal cell swelling or shrinking.",
"Contrast with walled turgor and plasmolysis."
] | biology | Answer: Water follows solutes; walls decide bursting versus turgor.
Steps:
1. Define water movement down its gradient.
2. Trace animal cell swelling or shrinking.
3. Contrast with walled turgor and plasmolysis. | cell_biology |
DNA is transcribed to RNA in the nucleus and mRNA is translated to protein by ribosomes in the cytoplasm, with reverse transcription in some viruses as the famous exception. | null | bio_0098 | concept | en | State the central dogma of molecular biology and where each step occurs in eukaryotes. | [
"Map transcription to the nucleus.",
"Map translation to cytoplasmic ribosomes.",
"Note retroviral reverse transcription."
] | biology | Answer: DNA to RNA in nucleus, RNA to protein in cytoplasm.
Steps:
1. Map transcription to the nucleus.
2. Map translation to cytoplasmic ribosomes.
3. Note retroviral reverse transcription. | genetics |
Homologous structures like vertebrate forelimbs share bone patterns from common ancestry despite different uses, and early embryos of fish, chickens, and humans look strikingly similar with gill slits and tails before diverging. | null | bio_0099 | concept | en | Summarize two anatomical lines of evidence for evolution: homologous structures and embryology. | [
"Describe shared limb bone plans.",
"Describe similar early embryos.",
"Infer common descent."
] | biology | Answer: Shared limbs and similar embryos point to common descent.
Steps:
1. Describe shared limb bone plans.
2. Describe similar early embryos.
3. Infer common descent. | evolution |
Potravní řetězec je lineární sled producent, konzument a rozkladač, potravní síť propojuje více řetězců do reálné spleti vztahů a trofické úrovně označují postavení organismu od primárních producentů po vrcholové predátory. | null | bio_0100 | concept | cs | Vysvětlete potravní řetězec, potravní síť a trofické úrovně. | [
"Definujte lineární řetězec.",
"Vysvětlete síť jako propojení řetězců.",
"Seřaďte trofické úrovně."
] | biology | Answer: Řetězec je linie, síť je splet, úrovně značí postavení.
Steps:
1. Definujte lineární řetězec.
2. Vysvětlete síť jako propojení řetězců.
3. Seřaďte trofické úrovně. | ecology |
End of preview. Expand in Data Studio
Dataset eval_09_subjects publikovany z projektu KucLab LLM.
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