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BIO Bird Migration. Canada geese migrate essentially along

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 39E Chapter 3

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 39E

BIO Bird Migration.? Canada geese migrate essentially along a north–south direction for well over a thousand kilo-meters in some cases, traveling at speeds up to about 100 km/h. If one goose is flying at 100 km/h relative to the air but a 40-km/h wind is blowing from west to east, (a) at what angle relative to the north–south direction should this bird head to travel directly southward relative to the ground? (b) How long will it take the goose to cover a ground distance of 500 km from north to south? (?Note?: Even on cloudy nights, many birds can navigate by using the earth’s magnetic field to fix the north–south direction.)

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Copyright: © Hannah Kennedy, Kent State University 1 Chapter 10 Lecture Notes 1. Bacterial Cell Division a. Binary fission = a closing process bacteria use in which bacterium divide into 2 identical bacterium b. Daughter cells = 2 cells produced as a result of cell division c. Origin of replication = the site at which DNA begins to be replicated so binary fission can occur d. Septum = a new membrane that forms between the 2 daughter cells i. Facilitated by FtsZ 2. Eukaryotic Chromosomes a. Each cell of the human body contains 23 pairs of chromosomes i. Aka 46 chromosomes ii. Haploid = n = 1 complete set of chromosomes 1. Ex = sperm and egg cells iii. Diploid = 2n = 2 copies of each chromosomes 1. Ex = all cells that aren’t sperm and egg cells iv. Homologue = each pair of chromosomes b. Chromatin = DNA + protein i. Genes are localized to specific chromosomes 1. Each chromosome contains 1 DNA molecule compacted to fit into the nucleus c. Histone proteins = positively­charged proteins in which DNA wraps around for consolidation 3. The cell cycle = the process in which cellular replication occurs with periods of growth followed by division; 24 hours a. The cell cycle Stage of cell cycle What is happening Interphase/M phase G 1= gap phase 1 ­ primary growth phase, cell is Interphase becoming bigger and expanding everything ­ making sure theres enough cytoplasm, organelles, etc ­ the longest S = synthesis ­ cells make a copy of the genome Interphase (chromosomes replicate to form identical copy) ­ DNA replication nd G 2= gap phase 2 ­ 2 growth phase that prepares Interphase the cell for division ­ microtubules reorganize ­ cell gets bigger and is doubled up ­ chromosomes condense and become even tighter Mitoses ­ sister chromatids are pulled M phase apart during 5 subdivided states ­ prophase Copyright: © Hannah Kennedy, Kent State University 2 ­ prometaphase ­ metaphase ­ anaphase ­ telophase Cytokinesis ­ cytoplasm divides to form 2 M phase daughter cells G0 ­ cells exit the cell cycle and go N/A into a state of rest in which they don’t divide ­ can re­enter the cell cycle later b. Interphase i. Cohesins = proteins that hold chromosome copies together along their lengths ii. Sister chromatic = the identical copy of each chromosome iii. Kinetochore = disc­like structures of proteins that help join sister chromatids 1. Microtubules attach here during mitosis 2. Centromere = the point at which the kinetochore is anchored a. Where the sister chromatids constrict c. Mitosis i. Spindle = component involved in mitosis that will separate sister chromatids; one spindle is on each side of the cell with microtubules attached Mitosis stage What is happening Prophase ­ chromosomes continue to condense ­ cytoskeleton is disassembled ­ nuclear envelope breaks down ­ golgi and ER disperse Prometaphase ­ microtubules attach to kinetochore ­ chromosomes begin to move to middle of the cell Metaphase ­ chromosomes line up in middle of the cell ­ chromosomes begin to experience tension Anaphase ­ cohesin proteins are broken down ­ sister chromatids are pulled to opposite sides of the cell ­ tubulin subunits are continuously removed to shorten the length of the microtubules Telophase ­ chromosomes decondense ­ nucleus re­forms ­ ER re­forms ­ golgi re­forms Cytokinesis ­ belt of actin contracts and pinches the cell into 2 daughter cells 4. Control of the cell cycle (2 factors we focus on) a. Cyclin­dependent kinases = CDKs = influence protein activity through phosphorylation b. Cyclins = bind to and activate CDKs c. There are several stopping point as the cell progresses through the cell cycle to check and make sure that the phases are completed appropriates Copyright: © Hannah Kennedy, Kent State University 3 Name of checkpoint What it is based on What is happening What prevents a cell from making it through this checkpoint G 1S checkpoint ­ External factors ­ increase in G1 cyclin ­ damage to DNA (growth factors) triggers the activity of ­ starvation conditions of ­ deciding whether or enzymes needed for the cell not to replicate DNA replication ­ lack of growth factors G 2M checkpoint ­ successful DNA ­ damaged DNA can ­ damage to DNA replication inactivate CDKs ­ cell decides if it will enter mitosis Spindle checkpoint ­ cell makes sure that ­ anaphase­promoting ­ if chromosomes aren’t the spindles have complex sets a series of aligned properly attached properly events to break down cohesin if they are attached properly 5. Growth factors = proteins which bind to membrane receptors and initiate pathways that result in cell growth and division a. Cells have specific receptors for the growth factors i. Growth factors can only trigger certain cells b. These trigger the production of cyclins so that the cell cycle may begin c. Without these the cells leave the cell cycle and enter G 0 6. Cancer a. Characteristics of a cancer cell is the ability to ignore cell cycle checkpoints and zip through them b. P53 gene = involved in the G /S and G /S checkpoints that codes for a protein that checks 1 2 for proper DNA replication; triggers enzymes to repair the DNA if it is damaged i. Apoptosis = cell death that occurs if the DNA cannot be repaired 1. P53 is missing/faulty in cancerous cells Copyright ©: Hannah Kennedy, Kent State University 1 Chapter 8 1. Photosynthesis = a light­dependent process that occurs in plants (leaves), bacteria, and algae (3 steps) 6CO 2 12H O 2 light 6 C12 6 + 6H2O + 6O 2 a. Capture energy from sunlight b. Make ATP i. Reduce NADP to NADPH c. Use ATP, NADPH, and CO to mak2 organic molecules 2. Chloroplasts – inner and outer membrane a. Thylakoid disks = flattened stacks of chloroplasts b. Grana = composed of stacks of thylakoid disks c. Chlorophyll = pigment contained within the thylakoid membrane that captures light energy and makes ATP d. Stroma = the inside of chloroplast that contains enzymes 3. Photons and pigments a. Photon = a particle of light that contains energy of which is transferred by chloroplasts to electrons to excite and raise them to a higher energy level i. Energy is inversely proportional to wavelength (i.e. long wavelengths have low energy and short wavelengths have high energy) b. Pigments = absorb light energy in the visible range (i.e. the color that you so is the one that hasn’t been absorbed) i. Contained in chlorophyll and carotenoids 1. Chlorophyll = pigments that absorb violet­blue and red light a. Electrons in the central ring are excited and then channeled through the hydrocarbon tail 2. Carotenoids = pigments that absorb light that chlorophyll can’t a. Important for the neutralization of free radicals 4. Light­dependent reactions (4 steps in the thylakoids) a. Primary photo­event i. Photon of light is captured b. Charge separation i. Reaction center transfers electron to acceptor c. Electron transport i. Electron carriers shuttle electrons and transport proton to create a gradient d. Chemiosmosis i. Protons flow down gradient to create ATP 5. Photosystems = components that include pigments and proteins that trap light energy as excited electrons—photosystem I and photosystem II (2 components) a. Antenna complex = web of hundreds of chlorophyll molecules held together within thylakoid membrane by a protein matrix that captures photons from the sun and channels them to chlorophylls i. Energy passes from 1 pigment to another until it reaches the reaction center b. Reaction center = component of a photosystem that eventually receives the energy and the excited electron i. Once the chlorophyll in the reaction center is excited it can be oxidized (because it is now a strong electron donor) Copyright ©: Hannah Kennedy, Kent State University 2 1. This electrons is passed to a chain of electron acceptors (analogous to the electron transport chain) and light energy is converted into chemical energy a. Electron acceptor is Quinone and the donor is water 6. Noncyclic photophosphorylation (represented as the Z diagram) a. Photosystem II oxidizes water to replace the electrons it transfers to photosystem I i. To make 1 oxygen molecule, 4 photons of light are absorbed and 2 molecules of water are oxidized ­ b. Cytochrome/b f com6lex = proton pump that links the photosystems together i. Receives a pair of excited electrons from Quinone from photosystem II ii. 1 electron pair causes 1 proton to be pumped from the stroma into the thylakoid space c. Photosystem I transfers electrons to NADP to make NADPH i. NADP reductase = catalyzes this transfer 7. Chemiosmosis = the proton gradient used to generate ATP a. ATP synthase is used to make the gradient and to pump H protons; contained in the thylakoid i. As protons move through the channel of the enzyme, ATP is made 8. Calvin cycle = cycle through which carbon fixation occurs with the goal to produce sugars a. During the process: i. Ribulose 1,5­biphosphate (RuBP) = molecule produced by photosynthetic cells that is made by joining 2 intermediates from glycolysis 1. 6 molecules of CO a2e bound to 6 molecules of RuBP via rubisco a. Makes 12 molecules of 3­phophoglycerate (PGA) b. 6 turns of this cycles makes 1 glucose molecules

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Chapter 3, Problem 39E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The answer to “BIO Bird Migration.? Canada geese migrate essentially along a north–south direction for well over a thousand kilo-meters in some cases, traveling at speeds up to about 100 km/h. If one goose is flying at 100 km/h relative to the air but a 40-km/h wind is blowing from west to east, (a) at what angle relative to the north–south direction should this bird head to travel directly southward relative to the ground? (b) How long will it take the goose to cover a ground distance of 500 km from north to south? (?Note?: Even on cloudy nights, many birds can navigate by using the earth’s magnetic field to fix the north–south direction.)” is broken down into a number of easy to follow steps, and 111 words. Since the solution to 39E from 3 chapter was answered, more than 320 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 39E from chapter: 3 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This textbook survival guide was created for the textbook: University Physics, edition: 13. This full solution covers the following key subjects: North, South, relative, direction, Ground. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. University Physics was written by and is associated to the ISBN: 9780321675460.

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BIO Bird Migration. Canada geese migrate essentially along