Videos
Introduction: Plant hormones are chemical molecules produced by plants in extremely low concentration for regulating the growth and development. There are five major hormones involved in the growth process. They are auxin, gibberellin, ethylene, abscisic acid, and cytokinin. Each of these hormones has its own functions at specific locations. The presence of these hormones in a definite amount is very essential for the normal growth and development of the plant.
Answer to Problem 1TYU
Correct answer: In the signal transduction process for the hormone auxin, the molecule ubiquitin tags certain proteins for destruction.
Hence, the correct answer is option (c).
Explanation of Solution
Reason for the correct answer:
General mechanism of action of auxin hormone:
Auxins are a group of related hormones responsible for a wide range of consequences on the growth and development of plants. Within a certain concentration, Auxin stimulates cell elongation in stems and coleoptiles. Auxin inhibits elongation growth by increasing the cell wall extensibility, according to the acid growth hypothesis. The main effects of auxin are to promote cell elongation according to the concentration.
Many plant hormones bind to the receptors which trigger the enzymatic reaction and results in the changes for cell growth and development of the plants. Both external and internal signal triggers the auxin hormone synthesis. The cytosol or the nucleus of the cell contains one receptor with three-dimensional shape (TIR1 receptor) that binds to auxin molecule. As the auxin binds to its receptor, ubiquitin molecule attaches to the repressor molecule and inhibits the auxin response genes. Thereafter, the ubiquitinylated protein is targeted and degraded into peptide fragments in a proteasome. This causes transcription of auxin response gene which acidifies the cell wall of target cells. The acidified target cell walls become more plastic which enables it to expand due to increased force of the cell’s turgor pressure. Thus, the action of auxin hormone causes cell expansion without cell division.
Option (c) is given as “tags certain proteins for destruction”.
The small regulatory protein called ubiquitin attaches to other proteins and are used to tag certain proteins for destruction. This process of tagging is referred to as ubiquitination and it is a post-translational modification. Ubiquitin-mediated proteolysis normally takes place during auxin signaling in plants.
Hence, the correct answer is option (c).
Reasons for the incorrect answers:
Option (a) is given as “absorbs blue light”.
Phototropins are the proteins that absorb blue light and they help to control the photosynthetic efficacy of plants, not auxins.
Hence, option (a) is incorrect.
Option (b) is given as “becomes phosphorylated”.
Phototropins are the blue-light receptors that control the photosynthetic efficacy of plants located at the shoot tips. These proteins become phosphorylated in response to blue light.
Hence, option (b) is incorrect.
Option (d) is given as “interacts antagonistically with gibberellins”.
Gibberellin is a growth-regulating hormone that enhances the elongation of stems and leaves and in the development of fruits. During auxin signaling in plants, the ubiquitin molecule does not antagonistically interact with gibberellins. Instead, ubiquitin-mediated proteolysis takes place in the signal transduction process in auxin.
Hence, option (d) is incorrect.
Option (e) is given as “binds to a receptor in the plant cell’s plasma membrane”.
During auxin signaling in plants, the ubiquitin molecule does not bind to a receptor in the plasma membrane of the plant cell. Instead, ubiquitin-mediated proteolysis takes place in the signal transduction process in auxin.
Hence, option (e) is incorrect.
Hence, options (a), (b), (d), and (e) are incorrect.
As the auxin binds to its receptor, ubiquitin molecule attaches to the repressor molecule in order to target it for destruction.
Want to see more full solutions like this?
Chapter 38 Solutions
EBK BIOLOGY
- Molecular Biology Please help with question. Thank you in advance. Discuss, compare and contrast the structure of promoters inprokaryotes and eukaryotes.arrow_forwardMolecular Biology Please help with question. Thank you You are studying the expression of the lac operon. You have isolated mutants as described below. In the absence of glucose, explain/describe what would happen, for each mutant, to the expression of the lac operon when you add lactose AND what would happen when the bacteria has used up all of the lactose (if the mutant is able to use lactose).1. Mutations in the lac repressor gene that would prevent the binding of lactose2. Mutations in the lac repressor gene that would prevent release of lactose once lactose hadbound3. Normally the lac repressor gene is located next to (a few hundred base pairs) and upstreamfrom the lac operon. Mutations in the lac repressor gene that move the lac repressor gene 100,000base pairs downstream.4. Mutations in the lac operator that would prevent binding of lac repressorarrow_forwardYou have returned to college to become a phylogeneticist. One of the first things you wish to do is determine how mammals, birds, and reptiles are related. Like any good scientist, you need to consider all available data objectively and without a preconceived “correct” answer. In pursuit of that, you should produce a phylogenetic tree based only on morphological features that show birds and mammals are more closely related. You will then produce a totally different tree, also using morphological features, that shows birds and reptiles are more closely related. Do not forget to include all three groups in both your trees. Based solely off the trees you produce, which relationship would you consider the more likely and why? Once you have answered that question, provide a brief summary of the “modern” understanding of the relationship between these three groups.arrow_forward
- true or false, the reason geckos can walk on walls is hydrogen bonding between their foot pads and the moisture on the wall.arrow_forwardBiology laboratory problem Please help. thank you You have 20 ul of DNA solution and 6X DNA loading buffer solution. You have to mix your DNA solution and DNA loading buffer before load DNA in an agarose gel. The concentration of the DNA loading buffer must be 1X in the DNA and DNA-loading buffer mixture after you mix them. For that, I will add _____ ul of 6X loading buffer to the 20 ul DNA solution.arrow_forwardBiology lab problem To make 20 ul of 5 mM MgCl2 solution using 50 mM MgCl2 stock solution and distilled water, I will mix ________ ul of 50 mM MgCl2 solution and ________ ul of distilled water. Please help . Thank youarrow_forward
- Biology Please help. Thank you. Biology laboratory question You need 50 ml of 1% (w/v) agarose gel. Agarose is a powder. How would you make it? You can ignore the volume of agarose powder. Don't forget the unit.TBE buffer is used to make an agarose gel, not distilled water. I will add _______ of agarose powder into 50 ml of distilled water (final 50 ml).arrow_forwardAn urgent care center experienced the average patient admissions shown in the Table below during the weeks from the first week of December through the second week of April. Week Average Daily Admissions 1-Dec 11 2-Dec 14 3-Dec 17 4-Dec 15 1-Jan 12 2-Jan 11 3-Jan 9 4-Jan 9 1-Feb 12 2-Feb 8 3-Feb 13 4-Feb 11 1-Mar 15 2-Mar 17 3-Mar 14 4-Mar 19 5-Mar 13 1-Apr 17 2-Apr 13 Forecast admissions for the periods from the first week of December through the second week of April. Compare the forecast admissions to the actual admissions; What do you conclude?arrow_forwardAnalyze the effectiveness of the a drug treatment program based on the needs of 18-65 year olds who are in need of treatment by critically describing 4 things in the program is doing effectively and 4 things the program needs some improvement.arrow_forward
- I have the first half finished... just need the bottom half.arrow_forward13. Practice Calculations: 3 colonies were suspended in the following dilution series and then a viable plate count and microscope count was performed. Calculate IDF's, TDF's and then calculate the CFU/mL in each tube by both methods. Finally calculate the cells in 1 colony by both methods. Show all of your calculations in the space provided on the following pages. 3 colonies 56 cells 10 μL 10 μL 100 μL 500 με m OS A B D 5.0 mL 990 με 990 με 900 με 500 μL EN 2 100 με 100 μL 118 colonies 12 coloniesarrow_forwardDescribe and give a specific example of how successionary stage is related to species diversity?arrow_forward
Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning
Biology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStax
Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning
Biology: The Unity and Diversity of Life (MindTap...BiologyISBN:9781337408332Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa StarrPublisher:Cengage Learning