Many of the techniques for fate mapping fade over time and cannot be.pdf

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Many of the techniques for fate mapping fade over time and cannot be used long term for tracing cell lineages. Describe a technique used for fate mapping that permanently marks cells and the advantages of this technique over other techniques (besides the longevity of its use). Solution Fate mapping is a method of understanding the embryonic origin of various tissues in the adult organisms by establishing the correspondence between individuals cells at one stage of development, and their progeny at later stages of development. When carried out at single-cell resolution, this process is termed cell-lineage tracing. Technique: Artificial markings: It is a method for labelling the only blastomere by which their fate can be traced:.

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explain why the jovian planet atmospheres are called primitive atmospheres while terrestrial planets are called evolved atmospheres. Solution Jupiter\'s atmosphere is composed mostly of hydrogen and helium, which are the same gases that dominate the Sun\'s and other stars\' composition. Likewise the early Earth atmosphere was most likely H2 and He. But the Earth\'s gravity is too weak to retain these gases for a long time. So contrary to the Jovian atmosphere the Earth\'s atmosphere changed and evolved. Hydrogen and helium were quickly replaced by an atmosphere produced by volcanic out-gassing. It contained gases like H2O, CO2, SO2, CO, S2, Cl2, N2, H2 and NH3 and CH4. But no free O2 as this is not found in volcanic gases. A lot more water came from outer space brought by comets bombarding the proto-Earth. As the Earth cooled, H2O rained out and the oceans formed. It didn\'t stop there, as oxygen started to buildup as soon as the Archean era, more than 2.5 billion years ago, by means of photochemical dissociation. That is breakup of water and ammonia molecules by ultraviolet. The O2 level rose to 1-2% of the current level. At these levels O3 (Ozone) can form to shield the Earth surface from ultraviolet light. Once the Earth was properly shielded, photosynthesis by bacteria and plants could start: CO2 + H2O + sunlight = organic compounds + O2, increasing the oxygen level to today\'s levels..

explain why the jovian planet atmospheres are called primitive atmos.pdf

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During digestion, in the intestinal lumen certain glycerols, FFA and bite salts from micelles which can then bond to ___ and ___ and promote their ___. fat soluble vitamins, cholesterol, absorption. water soluble vitamins, cholesterol, absorption. fat soluble vitamins, minerals, absorption. fat soluble vitamins, cholesterol, digestion. Solution The dietary lipids contains fat, triglycerides , phospholipids, cholesterol and fat soluble vitamins. They are absorbed back by the processes emulsification and diffusion. So the answer is Fat soluble vitamins and cholesterol- absorption.

During digestion, in the intestinal lumen certain glycerols, FFA an.pdf

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Describe the structures and functions associated with the water-vascular system echinoderms. Solution Echinodermata represents the phylum of animals in which spiny-skinned organisms are included. All the organisms of this phylum are aquatic in nature and follow either sedentary or mobile life- style. They have penta-radial symmetry in morphology and normally contain tube feets. The peculiar system in these organisms is the presence of a water-vascular canal. The features of these water vascular canals are enlisted as below: Various structures associated with water vascular system are discussed as below: Thus, the above given explanations decipher that the water vascular system of echinoderms is complex and has high physiological value. Madreporite: The madreporite represents the central tube along with which all the other tubules are associated. These water canals open directly on the surface of the organism and generate a water pore through which water can move to and fro. Stone canal: The stone canal is a tender tube associated with the madreporite. It has two openings, one of which opens towards the oral cavity and other towards the axial sinus. It runs longitudinally across the breadth of the organism\'s body. It is normally curved shaped but can be \"S\' shaped in some organisms. Axial sinus: The axial sinus and organ represent some accumulated connective tissues and are of germinal in nature. These axial sinuses directly communicate with the stone canal on their posterior end. Ring canal: The ring canal represents one of the most important organs of the echinoderm. It is a ring shaped structure located near the madreporite on the centre and have extensions of tubules which help in circulation of water and fluid. Since echinoderms have pentaradial symmetry, the ring canal often gives riset to atleast 5 extensions. Lateral vessels and the tube-feet: These are the organs of locomotion present just below the madreporite and lateral pentameric extensions of ring canal. Internally, the tube feet and lateral vessel communicate with the skeleton of water-vascular system..

Describe the structures and functions associated with the water-vascu.pdf

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Considering the atoms bonded and the shape of an H_2O molecule, is the H_2O molecule polar or non-polar? polar non-polar This is not the right answer. Do NOT click this answer! Solution A water molecule is polar because its O-H bonds are significantly polar and its bent geometry makes the distribution of those polar bonds asymmetrical.The side of the water molecule containing the more electronegative oxygen atom is partially negative and the side of the molecule containing the less electronegative hydrogen atoms is partially positive..

Considering the atoms bonded and the shape of an H_2O molecule, is th.pdf

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describe the differences and similarities between viruses, worms, and Trojan horses. Provide an example of at least one of these and describe that damage that can be done when a person or corporation is infected. Solution Lets me make it clear actually what these are first and then we move to the diffrences and similarities A computer virus connects itself to a program or record empowering it to spread starting with one PC then onto the next, leaving contamination as it travels. Like a human infection, a PC virus can extend in seriousness: some may bring about just somewhat irritating impacts while others can harm your hardware, programming or records. All viruses are joined to an executable record, which implies the infection may exist on your PC yet it really can\'t contaminate your PC unless you run or open the pernicious program. A worm is like an virus by outline and is thought to be a sub-class of an infection. Worms spread from PC to PC, yet not at all like an virus, it has the capacity to go with no human activity. A worm exploits document or data transport includes on your framework, which is the thing that permits it to travel unaided. A Trojan Horse is loaded with as much cunning as the fanciful Trojan Horse it was named after. The Trojan Horse, at first look will seem, by all accounts, to be valuable software yet will really do harm once introduced or keep running on your PC. Those on the less than desirable end of a Trojan Horse are normally deceived into opening them since they give off an impression of being getting true blue programming or documents from a honest to goodness source. Now I will Focus on the differences between Viruses, Worms and Torjan Horses On the basis of its mode of spread A virus attach itself to an executable file and then require a human to propogate, a worm does not require a human action to spread,torjan horse does not replicates itself and require human input to take into action. similarity is that they all are malicious programs that can cause damage to your computer. Examples:- VIRUSES WORMS Torjan.

describe the differences and similarities between viruses, worms, an.pdf

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A window has the rectangle surmounted by a semi circle the total per.pdf

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Can you identify how energy flows through an ecosystem? Can you identify how energy flows through an ecosystem? Solution Energy flow in ecosystem is basically in the form of heat by food chain. Ecosystem includes living (plants, animals, microrganismand non living things (sun, soil, climate, atmosphere etc) and the interactions between them. The food chain in our ecosystem comprises of herbivores, carnivores and omnivores and with every passage of energy from one organism to another there is a 70-80% loss of energy which is in the form of heat. So we can take an example where in green plants absorb sunlight, Co2 and water from surroundings and synthesise carbohydrates. These green plants are then eaten by herbivores like deer, cow, so the energy is tranferred from green plants to deer/ cow. Deer will be eaten by lion/ jackle, so again energy will be transferred from deer to lion/ jackle. Once lion/ jackle die, they will be eaten by eagle/ vulture (scavengers). Scavengers will then release nutrients in the environment and again the nutrients will be recycled. So the energy is never exhaused but recycled..

Can you identify how energy flows through an ecosystemCan you i.pdf

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You must use the full name, no abbreviations. Name the dynamic found around the cell. You will find this on the outside of a bacterium or a plant cell, but you will never find this n extracellular structure around an animal/mammal cell. The organelle is found in the cell, has a double membrane and is the location of DNA. This non-membrane organelle translates proteins. This membranous organelle has #4 associated with it. The membrane organelle receives material from #5 and modifies the lipids and the proteins. The digestive organelle of the cell is the The membrane organelle that produces ATP and other products is the: The network of proteins found through the cell, aiding in the placement and movement of the organelles is the: The specific proteins that comprise #9: The large \"long\" column of globular protein polymerized together: The microfilaments are composed of these 2 proteins. Name the membrane bound structure that moves material out of the cell. #8 is referred to as an endosymbiant. Why? Note the progression to form a protein. Solution 1. ANS: Cell membrane. 2. ANS: Cell wall. 3. ANS: Mitochondria (Circular DNA is present in this). 4. ANS: Ribosome\'s (The rRNA which present in the ribosomes can helps to translate mRNA into proteins). 5. ANS: Rough endoplasmic reticulum 6. ANS: Smooth endoplasmic reticulum 7. ANS: Lysosomes. 8. ANS: Mitochondria. 9. ANS: Cytoskeleton. 10- 12 ANS: G-Actin and F-Actin. Each microfilament is formed with hundreds of globular subunits G-Actin organized into a double stranded helix with a 36 nm repeat F-Actin. 13. ANS: Exocytosis (In this plasma membrane surrounding substances inside the cell and moves it outside the cell). 14. ANS: Mitochondria is a endosymbiont 1. Mitochondria had its own DNA (Which is naked and circular). 2. Mitochondria have ribosomes that are mostly similar to prokaryotes. 3. Mitochondria is only produced by pre-existing Mitochondria. 4. Mitochondria can able to synthesizes their own proteins. 15-18. ANS: Protein synthesis is a process in which cells build proteins from information in a DNA gene in a DNA gene in a two major steps. A. Transcription: In which a molecule of DNA is copied into a complimentary strand of RNA. B. Translation: In this the message in RNA is made into a protein..

You must use the full name, no abbreviations. Name the dynamic found.pdf

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You are burning the latest song you bought on ITunes to a disk. The song is 3 minutes, 23 seconds long and was sampled as a stereo signal (two sides, left and right) at 44.1 kHz using 16- bit encoding. How much space on the disk does this one track take up? If your CD is 640MBytes, how many songs of this size can you fit on one disk (assume 1 byte is 8-bits in length)? Remember, you don’t want fractional songs! How many songs could be saved on this same CD if the sampling rate was 128kHz using 32 bit encoding? Solution A) This stream of audio frames, as a whole, is then subjected to CIRC encoding, which segments and rearranges the data and expands it with parity bits in a way that allows occasional read errors to be detected and corrected. CIRC encoding also interleaves the audio frames throughout the disc over several consecutive frames so that the information will be more resistant to burst errors. Therefore, a physical frame on the disc will actually contain information from multiple logical audio frames. This process adds 64 bits of error correction data to each frame. After this, 8 bits of subcode or subchannel data are added to each of these encoded frames, which is used for control and addressing when playing the CD. CIRC encoding plus the subcode byte generate 33-bytes long frames, called \"channel-data\" frames. These frames are then modulated through eight-to-fourteen modulation (EFM), where each 8-bit word is replaced with a corresponding 14-bit word designed to reduce the number of transitions between 0 and 1. This reduces the density of physical pits on the disc and provides an additional degree of error tolerance. Three \"merging\" bits are added before each 14-bit word for disambiguation and synchronization. In total there are 33 × (14 + 3) = 561 bits. A 27-bit word (a 24-bit pattern plus 3 merging bits) is added to the beginning of each frame to assist with synchronization, so the reading device can locate frames easily. With this, a frame ends up containing 588 bits of \"channel data\" (which are decoded to only 192 bits music). The frames of channel data are finally written to disc physically in the form of pits and lands, with each pit or land representing a series of zeroes, and with the transition points—the edge of each pit—representing 1. A Red Book-compatible CD-R has pit-and-land-shaped spots on a layer of organic dye instead of actual pits and lands; a laser creates the spots by altering the reflective properties of the dye. The audio data stream in an audio CD is continuous, but has three parts. The main portion, which is further divided into playable audio tracks, is the program area. This section is preceded by a lead-in track and followed by a lead-out track. The lead-in and lead-out tracks encode only silent audio, but all three sections contain subcode data streams. The lead-in\'s subcode contains repeated copies of the disc\'s Table Of Contents (TOC), which provides an index of the start positions of the track.

You are burning the latest song you bought on ITunes to a disk. The .pdf

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Write a scientific research paper that investigates the properties of Digital (hardware) design from its inception till today. Be sure to pay special attention to the advent of Boolean algebra. Logic Gates and how they revolutionized today\'s technology. The paper must include the following sections: Introduction 1:5 pages Methodology 2 pages Project Analysis 2-3 pages Conclusion 1 page Annotated Bibliography The introduction should be no more than a page and a half long and should clearly state the objective(s) of your research paper. It should provide a brief history of the digital technology and the nature of your project. The methodology section should explain in great details to the layman, how the digital technology works from hardware and software point of view. The project analysis section should compare features of the digital technology with features from other technology (of your choice) from all aspects. Explain the advantages and disadvantages of both types of technologies. The conclusion should relate to the objective(s) stated in the introduction and summarizes the highlights of the paper. The annotated bibliography must conform to APA standards and you must remember to reference all your sources. Pay attention to the content, grammar and style. Solution DIGITAL TECHNOLOGY INTRODUCTION: Digital electronics or digital (electronic) circuits are electronics that handle digital signals – discrete bands of analog levels – rather than by continuous ranges (as used in analogue electronics). All levels within a band of values represent the same numeric value. Because of this discretization, relatively small changes to the analog signal levels due to manufacturing tolerance, signal attenuation or parasitic noise do not leave the discrete envelope, and as a result are ignored by signal state sensing circuitry. In most cases, the number of these states is two, and they are represented by two voltage bands: one near a reference value (typically termed as \"ground\" or zero volts), and the other a value near the supply voltage. These correspond to the \"false\" (\"0\") and \"true\" (\"1\") values of the Boolean domain respectively, named after its inventor, George Boole, yielding binary code. Digital techniques are useful because it is easier to get an electronic device to switch into one of a number of known states than to accurately reproduce a continuous range of values. Digital electronic circuits are usually made from large assemblies of logic gates, simple electronic representations of Boolean logic functions. METHODOLOGY: A digital circuit is a circuit where the signal must be one of two discrete levels. Each level is interpreted as one of two different states (for example, on/off, 0/1, true/false). Digital circuits use transistors to create logic gates in order to perform Boolean logic. This logic is the foundation of digital electronics and computer processing. Digital circuits are less susceptible to noise or degradation in qualit.

Write a scientific research paper that investigates the properties of.pdf

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// Write a function that multiplies each element in the array \"myArray\" // by the variable \"multiplyMe\". #include using namespace std; // TODO - Write your function prototype here int main() { const int SIZE = 10; int myArray [SIZE] = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50}; int multiplyMe = 5; // TODO - Add your function call here // print the array for(int i=0; i < SIZE; i++) { cout << myArray[i] << \" \"; } cout << endl; return 0; } // TODO - Write your function definition here Solution.

Write a function that multiplies each element in the array myA.pdf

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Which of the following is NOT a common feature of monosaccharides?A. similar or equal numbers of carbon and oxygen atomsB. a carbonyl (C=O) groupC. numerous hydroxyl (-OH) groupsD. All of the listed responses are correct. Solution The general formula for monosaccharides: (CH2O)n here n=3 or >3 On the basis of number of carbon atoms in the monosaccharides, these are classified as triose, tetrose, pentose, hexose. If n=3 then monosaccharides are called Trioses or if n=4 monosaccharides are known as Tetroses. Monosaccharides are also classified on the basis of carbonyl functional groups. If monosaccharide contains ketone group (>C=O) in the structure then these are classified as Ketose (example Dihydroxyacetone, Mannoheptulose), if monosaccharide contains aldehyde group (-CHO) then these are known as Aldoses (example Glucose, Glyceraldehyde). Monosaccharides contain numerous hydroxyl (-OH) groups. Triose, tetrose, pentose, and hexose contain 2, 3, 4 and 5 -OH groups respectively in their structure. Ans D will be correct answer..

Which of the following is NOT a common feature of monosaccharidesA..pdf

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Who are the owners of a cor[poration,and how is their ownership represented? Solution Shareholders are the real owners of a cooperation. they hold the share capital of the Company i.e., they bring the fund for the establishment and working of the Company. The shareholders own the corporation and are responsible for electing the directors also. Their ownership represented by the the number, value and class of shares they hold..

Who are the owners of a cor[poration,and how is their ownership repr.pdf

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Which of these is not a determinant of diseaserandom eventaging.pdf

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2A) You are the first to arrive at the scene of a house fire. You hear somebody in the house calling for help. Ignore the notion of helping people for the sake of helping people (a noble thing and highly recommended, but not immediately relevant here). In your own words, explain why it may be selectively advantageous for you to help, and risk serious injury or death in the process, if the person calling is a close relative. What if that person is a stranger? I am looking for a narrative explanation, not just a formula. 2B) Jane is an ant, an organism with a haplo-diploid mating system (females are diploid, males are haploid). In a parallel of J.B.S. Haldane’s statement, should Jane give up her life to save two of her sisters? Assume all ants can reproduce and will produce 2 offspring (neither is correct in reality, but assume it for your calculations). List all of your calculations, and your reasoning on why you plugged in specific numbers. Solution Answers: 2. (A) The concept of self-sacrificial behaviour was provided by William D. Hamilton. According to his arguements, natural selection favours genetic success, not reproductive success, and that individuals can pass copies of their genes on to future generations. Genes are passed from direct parentage and by assisting the reproduction of close relatives a concept referred to as kin selection. The conditions under which reproductive altruism evolves: r × B > C, where r = genetic relatedness of the altruist to the beneficiary, B = Benefit received by the recipient, C = Cost incurred/suffered by the donor. The coefficient of relatedness is different between everyone r =1/2 for parents/offspring r = 1/2 for siblings r = 1 for offspring and themselves r = 1/4 for half sibling Siblings are related by 1 to themselves but 1/2 to each other, while parents are equally related to all their offspring. So, its a fair deal in evolutionary terms if the person to be saved is a close relative like offsprings, siblings aorparents, but not for a stranger since r = 0.0 in a life and death situation. (B) Half of its genome comes from the father (haploid) and half from the mother (diploid). That means she carries all of her fathers genes and half of her mother’s genes. So does her sister, implying that they share of course the entire genome of their common father (i.e. already 50% of their genome), plus, on average, a quarter of their mother’s genome, yielding a coefficient of relatedness of 0.75. r for one sister = 0.75 So, r for two sisters = 0.75 + 0.75 = 1.5 Thus, the cost suffered by Jane is less than the benefit gained by the two sisters. So, Jane should give up her life to save her two sisters..

2A) You are the first to arrive at the scene of a house fire. You he.pdf

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Which of the following best describes link encryption? A. Data is encrypted at the point of origin and is decrypted at the point of destination. B. The message is decrypted and re-encrypted as it passes through each successive node using a key common to the two nodes. C. The KDC shares a user-unique key with each user. D. It requires a session key that the KDC shares between the originator and the final destination. Solution B. With link encryption, the message is decrypted and re-encrypted as it passes through each successive node using a key common to the two nodes. Answers A, C, and D are incorrect because they all describe end-to-end encryption.

Which of the following best describes link encryptionA. Data is e.pdf

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What structure surrounding the cytoplasm is composed mostly of a phospholipid bilayer?___ Which organelle is responsible for ATP production?___ Which organelle makes protein for use outside the cell?____ Which organelle in the cell produces lipids?___ The DNA that controls most of the function of the cell is found in what organelle?___ What cellular structure is responsible for ribosome production?___ What are the three parts of the cell cycle?___ Of the three parts of the cell cycle, in which one is DNA duplicated?___ In what part of the cell cycle do chromosomes first split apart?___ What part of the cell cycle involves the division of the cytoplasm?___ Describe the four phases of nuclear (mitotic) division and what occurs during those phases. Solution 4. Cell membrane 5. Mitochondria 6. rough endoplasmic reticulum 7. Smooth endoplasmic reticulum 8. Nucleus 9. Nucleolus 10. G1, G2, S and M phase 11. S phase.

What structure surrounding the cytoplasm is composed mostly of a phos.pdf

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What is happening in the starkeythiosulfate broth that results in a.pdf

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What is the relationship between molecular biology and cell biology? Solution Molecular biology is the study of biology at molecular level. Cell biology deals with the study of cell properties like the physiological properties, their structure, the organelles they contain. It also studies the interactions with their environment, their life cycle, division and death. Molecular and cell biology study the properties and function of a cell at molecular level..

What is the relationship between molecular biology and cell biology.pdf

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what are some examples of hazaedous waste remediation? Solution Hazardous Waste - Hazardous Waste is a Waste which is potentially dangerous or harmful to human health or the environment. Hazardous wastes can be liquids, solids, or contained gases. They can be the by-products of manufacturing processes, discarded used materials, or discarded unused commercial products, such as cleaning fluids (solvents) or pesticides. Wastes may be hazardous wastes if they exhibit any of the four characteristics of a hazardous waste - Ignitability, Corrosivity, Reactivity, and Toxicity. Hazardous Waste Remediation - Hazardous waste Remediation is the protection of human health and the environment by reducing risk. Primary approaches which can be used in site remediation to achieve acceptable levels of risk Hazardous waste site remediation usually addresses soils and groundwater . However, it can also include wastes, surface water, sediment , sludges, bedrock, buildings, and other man-made items. The hazardous constituents may be organic, inorganic and, occasionally, radioactive. They may be elemental ionic, dissolved, sorbed, liquid, gaseous, vaporous, solid, or any combination of these. Examples of Hazardous Waste Remediation - Incineration - Incineration is a waste treatment process that involves the combustion of organic substances contained in waste materials.Incineration of waste materials converts the waste into ash, flue gas and heat.Incineration with energy recovery is one of several waste-to-energy (WtE) technologies such as gasification, pyrolysis and anaerobic digestion. Bioremediation- Bioremediation is a process used to treat contaminated media, including water, soil and subsurface material, by altering environmental conditions to stimulate growth of microorganisms and degrade the target pollutants. Solidification / Stabilization- Solidification/stabilization (S/S) is a widely used treatment for the management/disposal of a broad range of contaminated media and wastes; particularly those contaminated with substances classified as hazardous. The treatment involves mixing a binding reagent into the contaminated media or waste. Thermal desorption -It is an environmental remediation technology that utilizes heat to increase the volatility of contaminants such that they can be removed (separated) from the solid matrix (typically soil, sludge or filter cake). Chemical Destructions -Chemical oxidation-reduction, acid-base neutralisation, precipitation, hydrolysis, ion exchange, thermal treatment methods, wet air oxidation photolysis.are various examples..

what are some examples of hazaedous waste remediationSolution.pdf

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