Videos
Which of the following substances can limit plant growth in nature?
- a. sunlight
- b. water
- c. carbon dioxide
- d. fixed nitrogen
- e. all of the above
Introduction: Nature comprises of several components that can either proliferate or limit the growth of the plant. Plant growth can be limited due to several specific factors.
Answer to Problem 1TY
Correct answer: Plant requires an optimum amount of sunlight, water, fixed nitrogen, and carbon dioxide for the growth and development. If any of these factor is not optimum, it can limit the growth of the plant. Hence, the correct answer is option e.
Explanation of Solution
Reason for the correct answer:
Plants require sunlight, carbon dioxide, and water to carry out the process of photosynthesis. Photosynthesis is important for the growth and development of plants. Nitrogen-fixing bacteria help in the conversion of nitrogen into a form usable for the plants.
Option e. is given as “all of the above”.
Plant growth can be limited due to sunlight, water, carbon dioxide, and nitrogen. All of these play a crucial role in the growth and development of the plants. Hence, the correct answer is option e.
Reasons for the incorrect answer:
Option a. is given as, “sunlight”.
Plants are also called producers and it utilize the sunlight for the process of photosynthesis. Plants convert light energy into chemical energy for efficient photosynthesis. However, sunlight as a sole factor cannot help in the growth and development of the plants. Hence option a. is incorrect.
Option b. is given as, “water”.
Plants also require water along with sunlight for the process of photosynthesis. Water is utilized by the plants for photosynthesis as well as for the growth and development of the plants. However, water as a sole factor cannot help in the growth and development of the plants. Hence, option b. is incorrect.
Option c. is given as, “carbon dioxide”.
Carbon dioxide released by human beings play a crucial role in the process of photosynthesis. Carbon dioxide along with sunlight and water helps in the production of glucose which is used as an energy source by the plants. However, carbon dioxide as the sole factor cannot help in the growth and development of the plants. Hence, option c. is incorrect.
Option d. is given as, “fixed nitrogen”.
Plants cannot use atmospheric nitrogen for the process of growth and development. The nitrogen is fixed by nitrogen-fixing bacteria to make it usable for the cellular processes in plants. However, nitrogen as a sole factor cannot limit the growth of the plant. Hence, option d. is incorrect.
Hence, the options a., b., c., and d. are incorrect.
A combination of sunlight, carbon dioxide, water and fixed nitrogen can limit the growth and development of the plant.
Want to see more full solutions like this?
Chapter 38 Solutions
Biology
Additional Science Textbook Solutions
Biological Science (6th Edition)
Genetics: From Genes to Genomes
Physical Science
Campbell Essential Biology (7th Edition)
HUMAN ANATOMY
Chemistry: A Molecular Approach (4th Edition)
- You expect to find which of the following in the Microtubule Organizing Center (MTOC)...(mark all that apply) A. Gamma tubulin B. XMAP215 C. Centrioles D. Kinesin-13arrow_forwardThe actin-nucleating protein formin has flexible “arms” containing binding sites that help recruit subunits in order to enhance microfilament polymerization. What protein binds these sites? A.Thymosin B.Profilin C.Cofilin D.Actin E.Tropomodulinarrow_forwardWhile investigating an unidentified motor protein, you discover that it has two heads that bind to actin. Based on this information, you could confidently determine that it is NOT... (mark all that apply) A. A myosin I motor B. A dynein motor C. A myosin VI motor D. A kinesin motorarrow_forward
- You isolate the plasma membrane of cells and find that . . . A. it contains regions with different lipid compositions B. it has different lipid types on the outer and cytosolic leaflets of the membrane C. neither are possible D. A and B both occurarrow_forwardYou are studying the mobility of a transmembrane protein that contains extracellular domains, one transmembrane domain, and a large cytosolic domain. Under normal conditions, this protein is confined to a particular region of the membrane due to the cortical actin cytoskeletal network. Which of the following changes is most likely to increase mobility of this protein beyond the normal restricted region of the membrane? A. Increased temperature B. Protease cleavage of the extracellular domain of the protein C. Binding to a free-floating extracellular ligand, such as a hormone D. Protease cleavage of the cytosolic domain of the protein E. Aggregation of the protein with other transmembrane proteinsarrow_forwardTopic: Benthic invertebrates as an indicator species for climate change, mapping changes in ecosystems (Historical Analysis & GIS) What objects or events has the team chosen to analyze? How does your team wish to delineate the domain or scale in which these objects or events operate? How does that limited domain facilitate a more feasible research project? What is your understanding of their relationships to other objects and events? Are you excluding other things from consideration which may influence the phenomena you seek to understand? Examples of such exclusions might include certain air-born pollutants; a general class of water bodies near Ottawa, or measurements recorded at other months of the year; interview participants from other organizations that are involved in the development of your central topic or issue. In what ways do your research questions follow as the most appropriate and/or most practical questions (given the circumstances) to pursue to better understand…arrow_forward
- The Esp gene encodes a protein that alters the structure of the insulin receptor on osteoblasts and interferes with the binding of insulin to the receptor. A researcher created a group of osteoblasts with an Esp mutation that prevented the production of a functional Esp product (mutant). The researcher then exposed the mutant strain and a normal strain that expresses Esp to glucose and compared the levels of insulin in the blood near the osteoblasts (Figure 2). Which of the following claims is most consistent with the data shown in Figure 2 ? A Esp expression is necessary to prevent the overproduction of insulin. B Esp protein does not regulate blood-sarrow_forwardPredict the per capita rate of change (r) for a population of ruil trees in the presence of the novel symbiont when the soil moisture is 29%. The formula I am given is y= -0.00012x^2 + 0.0088x -0.1372. Do I use this formula and plug in 29 for each x variable?arrow_forwardPlease answer the following chart so I can understand how to do it.arrow_forward
- Digoxin: Intravenous Bolus - Two Compartment Model Drug Digoxin Route: IV Bolus Dose: 0.750 mg Plasma Concentration Time Profile Beta Alpha Time (hrs) Conc (ng/ml) LN (ng/ml) LN (ng/ml) LN 0.00 #NUM! #NUM! #NUM! 0.10 12.290 2.509 #NUM! #NUMI 0.60 6.975 1.942 #NUM! #NUMI 1.00 4.649 1.537 #NUM! #NUMI 2.00 2.201 0.789 #NUM! #NUM! 3.00 1.536 0.429 #NUM! #NUM! 4.00 1.342 0.294 #NUM! #NUM! 5.00 1.273 0.241 #NUM! #NUMI 6.00 1.238 0.213 #NUM! #NUM! 7.00 1.212 0.192 #NUM! #NUM! 8.00 1.188 0.172 #NUMI #NUM! 9.00 1.165 0.153 #NUM! #NUMI 10.00 1.143 0.134 #NUMI #NUM! 11.00 1.122 0.115 #NUM! 12.00 1.101 0.096 #NUMI 13.00 1.080 0.077 #NUMI 16.00 1.020 0.020 #NUMI 24.00 0.876 -0.132 #NUMI Pharmacokinetic Parameters Parameter Value Alpha B Beta Units ng/ml hr-1 ng/ml hr-1 CO ng/ml H.C AUC ng x hr/ml Vc Vbeta Vss C L/hr TK (alpha) hr TX (beta) days 5+ F3 F4 F5 0+ F6 F7 % 6 95 14 #3 29 & t F8 F9 FW EWarrow_forwardLinuron, a derivative of urea, is used as an herbicide. Linuron serum levels were measured in 4Kg rabbits following a bolus IV injection of 10mg/kg. Time (minutes) Serum Linuron Levels (ug/ml) following IV dose 10 15.48 20 8.60 30 5.90 45 3.78 60 2.42 90 1.49 120 0.93 180 0.60 240 0.41 300 0.29 360 0.22 Analyze this data and perform the necessary calculations to determine the following pharmacokinetic parameters from the IV data: (5 points per parameter, 24 parameters/variables ■ 120 points possible). You do NOT need to submit graphs or data tables. Give the terminal regression line equation and R or R² value: Give the x axis (name and units, if any) of the terminal line: Give the y axis (name and units, if any) of the terminal line: Give the residual regression line equation and R or R² value: Give the x axis (name and units, if any) of the residual line: Give the y axis (name and units, if any) of the residual line:arrow_forward3. In the tomato, red fruit (O+) is dominant over orange fruit (0), and yellow flowers (W+) are dominant over white flowers (w). A cross was made between true-breeding plants with red fruit and yellow flowers, and plants with orange fruit and white flowers. The F₁ plants were then crossed to plants with orange fruit and white flowers, which produced the following results: a. b. 333 red fruit, yellow flowers 64 red fruit, white flowers 58 orange fruit, yellow flowers 350 orange fruit, white flowers Conduct a chi-square analysis to demonstrate that these two genes DO NOT assort independently. Make sure to interpret the P value obtained from your chi-square test. Calculate and provide the map distance (in map units) between the two genes.arrow_forward
Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning
Concepts of BiologyBiologyISBN:9781938168116Author:Samantha Fowler, Rebecca Roush, James WisePublisher:OpenStax College
Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage LearningEssentials of Pharmacology for Health ProfessionsNursingISBN:9781305441620Author:WOODROWPublisher:CengageBasic Clinical Lab Competencies for Respiratory C...NursingISBN:9781285244662Author:WhitePublisher:Cengage