Table of contents
- 1. Introduction to Biology2h 40m
- 2. Chemistry3h 40m
- 3. Water1h 26m
- 4. Biomolecules2h 23m
- 5. Cell Components2h 26m
- 6. The Membrane2h 31m
- 7. Energy and Metabolism2h 0m
- 8. Respiration2h 40m
- 9. Photosynthesis2h 49m
- 10. Cell Signaling59m
- 11. Cell Division2h 47m
- 12. Meiosis2h 0m
- 13. Mendelian Genetics4h 41m
- Introduction to Mendel's Experiments7m
- Genotype vs. Phenotype17m
- Punnett Squares13m
- Mendel's Experiments26m
- Mendel's Laws18m
- Monohybrid Crosses16m
- Test Crosses14m
- Dihybrid Crosses20m
- Punnett Square Probability26m
- Incomplete Dominance vs. Codominance20m
- Epistasis7m
- Non-Mendelian Genetics12m
- Pedigrees6m
- Autosomal Inheritance21m
- Sex-Linked Inheritance43m
- X-Inactivation9m
- 14. DNA Synthesis2h 27m
- 15. Gene Expression3h 20m
- 16. Regulation of Expression3h 31m
- Introduction to Regulation of Gene Expression13m
- Prokaryotic Gene Regulation via Operons27m
- The Lac Operon21m
- Glucose's Impact on Lac Operon25m
- The Trp Operon20m
- Review of the Lac Operon & Trp Operon11m
- Introduction to Eukaryotic Gene Regulation9m
- Eukaryotic Chromatin Modifications16m
- Eukaryotic Transcriptional Control22m
- Eukaryotic Post-Transcriptional Regulation28m
- Eukaryotic Post-Translational Regulation13m
- 17. Viruses37m
- 18. Biotechnology2h 58m
- 19. Genomics17m
- 20. Development1h 5m
- 21. Evolution3h 1m
- 22. Evolution of Populations3h 52m
- 23. Speciation1h 37m
- 24. History of Life on Earth2h 6m
- 25. Phylogeny2h 31m
- 26. Prokaryotes4h 59m
- 27. Protists1h 12m
- 28. Plants1h 22m
- 29. Fungi36m
- 30. Overview of Animals34m
- 31. Invertebrates1h 2m
- 32. Vertebrates50m
- 33. Plant Anatomy1h 3m
- 34. Vascular Plant Transport2m
- 35. Soil37m
- 36. Plant Reproduction47m
- 37. Plant Sensation and Response1h 9m
- 38. Animal Form and Function1h 19m
- 39. Digestive System10m
- 40. Circulatory System1h 57m
- 41. Immune System1h 12m
- 42. Osmoregulation and Excretion50m
- 43. Endocrine System4m
- 44. Animal Reproduction2m
- 45. Nervous System55m
- 46. Sensory Systems46m
- 47. Muscle Systems23m
- 48. Ecology3h 11m
- Introduction to Ecology20m
- Biogeography14m
- Earth's Climate Patterns50m
- Introduction to Terrestrial Biomes10m
- Terrestrial Biomes: Near Equator13m
- Terrestrial Biomes: Temperate Regions10m
- Terrestrial Biomes: Northern Regions15m
- Introduction to Aquatic Biomes27m
- Freshwater Aquatic Biomes14m
- Marine Aquatic Biomes13m
- 49. Animal Behavior28m
- 50. Population Ecology3h 41m
- Introduction to Population Ecology28m
- Population Sampling Methods23m
- Life History12m
- Population Demography17m
- Factors Limiting Population Growth14m
- Introduction to Population Growth Models22m
- Linear Population Growth6m
- Exponential Population Growth29m
- Logistic Population Growth32m
- r/K Selection10m
- The Human Population22m
- 51. Community Ecology2h 46m
- Introduction to Community Ecology2m
- Introduction to Community Interactions9m
- Community Interactions: Competition (-/-)38m
- Community Interactions: Exploitation (+/-)23m
- Community Interactions: Mutualism (+/+) & Commensalism (+/0)9m
- Community Structure35m
- Community Dynamics26m
- Geographic Impact on Communities21m
- 52. Ecosystems2h 36m
- 53. Conservation Biology24m
51. Community Ecology
Community Structure
1:30 minutes
Problem 12
Textbook Question
Textbook QuestionA team of 102 scientists spent a year surveying a small area of the San Lorenzo rain forest in Panama to count the number of species of arthropods living there. After collecting 129,494 specimens—using nets, traps, shovels, tree-climbing harnesses, helium balloons, and other creative gear—it took the team eight years to sort and identify the arthropods. Rather than measuring the entire 6000-hectare (ha) forest, the researchers sampled arthropod diversity by intensively collecting as many arthropods as they could in 12 plots that measured 20 m×20 m square. If 1 ha=10,000 m2,, how many hectares of forest did they sample in all? a. 20 m×20 m×12=4800 ha b. 4800 m2×10,000 m2/1 ha=48,000,000 ha c. 20 m×20 m=400 ha d. 4800 m^2 x 1ha/10,000 m^2 = 0.48 ha
Verified step by step guidance
1
Step 1: Understand the problem. The problem is asking for the total area of the forest that the scientists sampled in hectares. They sampled 12 plots, each measuring 20 m by 20 m.
Step 2: Calculate the total area in square meters that the scientists sampled. This is done by multiplying the area of one plot (20 m x 20 m = 400 m^2) by the number of plots (12). So, 400 m^2 x 12 = 4800 m^2.
Step 3: Convert the total area from square meters to hectares. We know that 1 hectare is equal to 10,000 square meters. So, we divide the total area in square meters by the number of square meters in one hectare. So, 4800 m^2 / 10,000 m^2/ha = 0.48 ha.
Step 4: Check the options given in the problem. The correct answer is the one that matches the result we calculated.
Step 5: The correct answer is option d. 4800 m^2 x 1ha/10,000 m^2 = 0.48 ha.
Recommended similar problem, with video answer:
Verified Solution
This video solution was recommended by our tutors as helpful for the problem above
Video duration:
1mPlay a video:
Was this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Area Calculation
To determine the total area sampled by the researchers, one must calculate the area of a single plot and then multiply it by the number of plots. Each plot measures 20 m by 20 m, which gives an area of 400 m². By multiplying this area by the 12 plots surveyed, the total sampled area can be found.
Recommended video:
Guided course
02:10
Mark & Recapture: Calculations & Assumptions
Unit Conversion
Understanding unit conversion is essential for translating the area from square meters to hectares. Since 1 hectare equals 10,000 square meters, converting the total area sampled from square meters to hectares involves dividing the area in square meters by 10,000. This step is crucial for expressing the results in the appropriate unit of measurement.
Recommended video:
Guided course
05:05
Atoms- Smallest Unit of Matter
Sampling Methodology
The sampling methodology used by the scientists is important for understanding the context of the study. By selecting specific plots within the larger forest area, the researchers aimed to estimate the diversity of arthropods without surveying the entire forest. This approach allows for efficient data collection while still providing insights into the overall biodiversity of the ecosystem.
Recommended video:
Guided course
02:31
Sampling Stationary Organisms: Transects & Quadrats
Watch next
Master Community Structure with a bite sized video explanation from Jason Amores Sumpter
Start learningRelated Videos
Related Practice