Some scientist and science fiction writers have envisioned life on ot.pdf

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Some scientist and science fiction writers have envisioned life on other Wanes based on silicon not carbon. Based on your understanding of the chemical nature of organic molecules why is silicon a logical substitute for carbon as basis of biomolecules? Solution although silicon based life is not well accepted theory of life, I will try to explain why silicon has been chosen as an alternative to carbon. initially, the search for carbon initiated to find life or aliens on any other planets, since carbon is the major component of biomolecules in all forms of life on Earth. However, it may not be true that life forms in other planets too would contain carbon as its major component. there might be some other component as well. then research began to find an alternative for carbon search. some scientists believe that there is a possibility of siliicon based life. on an extreme world like Titan, the atmosphere contains no oxygen, all the water had been frozen to solid. But silicon is not oxidized to form inert rock.Titan has liquid methane and ethane on its surface. methane acts a good solvent for silicon. silicon molecules mimics organisc chemistry on Earth. They are stable and may be a start of alien biochemistry..

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How are the epimysium and tendons related? A. The epimysium surrounds the tendon. B. Tendons surround the epimysium. C. Tendons and epimysium both surround the end of the muscle. D. The tendon is a continuation of the epimysium. Solution Answer: D. The tendon is a continuation of the epimysium. Explanation: At the ends of a muscle, the fibers of the Epimysium may form a cable-like tendon or a sheet- like aponeurosis..

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Gerard throws a discus distances of 10 meters, 14.5 meters, 14.8 meters, 16.7 meters, and 15.4 meters. On his last attempt he faults, which counts as 0 meters. What is his average throwing distance? A. 11.9 meters B. 12.8 meters C. 13.78 meters D. 14.28 meters Solution A. To find the average, add the numbers and divide by the number of numbers. In this problem add 10, 14.5, 14.8, 16.7, 15.4, and 0 to get 71.4. Divide this by 6. 71.4/6 = 11.9.

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For Printing and personal use ONLY. DO NOT Hand in this copy of the assignment. Only responses submitted to the corresponding D2L quiz will be graded. The D2L version is exactly like this document. Membrane proteins are proteins that are suspended within the lipid bilayer of a membrane. They are held in place because some R-groups of the protein interact with the phosphate heads of the phospholipids of the membrane, while other R-groups of the protein interact with the hydrocarbon tails of the phospholipids of the membrane. From the amino acids shown below, which one likely has an R-group that interacts with the phopholipid tails? If a cell has some salt inside it but lacks ion channels, and the cell is put into a beaker of pure water, what will happen? Solution Leucine (green color) is most likely to be part of the protein that interact with the hydrocarbon chain of phospholipid in the membrane (since it is hydrophobic amino acid).

For Printing and personal use ONLY. DO NOT Hand in this copy of the a.pdf

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Find and provide a link to a multi-year capital plan from any Illinois municipality. Discuss the presentation of the plan (i.e. - is there detail about specific projects, funding sources, etc) and how you think it could be improved. Solution The Link can\'t be provided so you can find it in the document centre of the village of the round lake beach and the multi year capital plan is for the years from 2015- 2024. The plan is presented in detail proving the names of the specific projects that would be undertaken by the village and all the ways in which they will be collecting the funds needed. The plan talks about the following specific projects: The funding is alot dependent upon the taxes from the people inhabiting the place and a very small amount has been estimated as allocations from outside. The plan is highly unrealistic as there may be instances where the people residing the place may not choose to pay so much or simply not buy the bonds they are expecting to issue..

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A geneticist is working with a new bacteriophage called phage Y3 that infects E. coli. He has isolated eight mutant phages that fail to produce plaques when grown on E. coli strain K. To determine whether these mutations occur at the same functional gene, he simultaneously infects E. coli K with paired combination of the mutants and looks to see whether plaques are formed. He obtains the following results. (A plus sign means that plaques were formed on E. coll K; a minus sign means that no plaques were formed on E. coli K.) To how many functional genes (cistrons) do these mutations belong? 5 1 8 2 7 3 Solution Answer: Correct answer: 3 Reason: The geneticist has conducted the transformation experiment in which “trans” portion of the cis–trans test to determine the complementation between the different phage mutants infected with E. coli K cells. Finally, these cells were looked whether the plaques are generated in the plates. It can be seen in the above results table, in which plus sign means that plaques were formed due to complimentaion in 3 mutants (5, 6 & 8) have meticulously specific defects in the same functional so that on E. coli K; a minus sign means that no plaques. Therefore, it can be concluded that complementation can happen when the mutations of the different phages come together in the sample as part of transformation.

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Assume that the four living species in the figure below evolved from their common ancestor without convergence or reversal. 1. Reconstruct their evolutionary relationships and label the character transitions. Show diagram. 2. Which evolved first: red eyes (living species 3) or fin stripes? Explain your answer. ***I have the answer but I don\'t know how we got it. I\'ve seen some incorrect/incomplete answers here and that wasn\'t helpful. Please explain you answer for BOTH parts and show diagram. If you\'re not sure don\'t answer! Thanks!*** Common ancestor Living species Solution a)the transition is as follows common ancestor->Panderichthys,->Tiktaalik with the fins->tetrapods with legs and wings- >amphibian b)The eyes came first because the fish first has a body with eyes and then it has fins to swim.The fins stripes increase with type into long wings and it can swim more and more.The eye is to see the direction to swim.

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Declining Balance The cost of the asset: $10,000.00 The salvage value: $800.00 The life of the asset: 4 years Calculate the Depreciation Expense to the Nearest One Dollar at End of Year 2. Solution Since No depreciation rate is given it is assumed theat depreciation is calclucated based on double decling balance method Hence depreciation Rate =1/4*2 =50% Depreciation Exxpense at year 2 is $3750Carrying Value at the beginningDepreciationCarrying value at the EndYear 11000025007500Year 2750037503750.

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A population of wild deer mice includes individuals with long or short hair. When two short haird indviduals breed, one-third of the pups have long hair. How do you explain the unusual phenotype ratio? Solution it can be autosomal inheritance where homozygous dominant is lethal. consider short hair to be dominant S, long hair to be recessive s so cross between two short haired can be Ss X Ss S s S SS Ss s Ss ss 1:2:1 ratio for SS:Ss:ss but we considered homozygous dominant to be lethal, so SS are dead, therefore Ss:ss is 2:1 and 1/3rd population have long hair..

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Who is it important that the lymphatic system operate separately from the cardiovascular system? Explain Describe the relationship between the lymphatic vessels and blood vessels. Who is it important that the lymphatic system operate separately from the cardiovascular system? Explain Describe the relationship between the lymphatic vessels and blood vessels. Describe the relationship between the lymphatic vessels and blood vessels. Solution It is important that the lymphatic system operate separately from the cardiovascular system becuase lymphatic system is a closed circulatory system that consists of lymph capillaries, lymph vessels, and lymph nodes different to blood vessels. The lymph present in capillaries flows into lymphatic vessels, which are presented deep or superficial but blood vessels are deep innervating various parts of body to maintain pheripheral resistance & mean arterial blood pressure. Blood flows through three types of blood vessels (arteries, capillaries and veins) distributed throughout the body. Capillaries are the fine endings of blood vessels. These capillaries are responsible for the exchange of respiratory gases, nutrients and waste products. The molecules move across the capillaries through a process called “diffusion.” These lymph vessels combine to form lymphatic trunks that empty into the large collecting vessels of lymphatic system called “thoracic duct and right lymphatic duct.” The lymphoid tissue is mostly composed of a reticular connective tissue (a type of loose connective tissue). The fibres of the reticular network contain macrophages. A number of lymphocytes from the postcapillary venule walls occupy this network and leave to the body again. This ensures the lymphocytes to identify the infected sites at a faster rate. Lymphatic system is a closed circulatory system that consists of lymph capillaries, lymph vessels, and lymph nodes. Lymph is a tissue fluid, which is formed by the diffusion of substances from blood capillaries into tissue spaces. The lymph present in capillaries flows into lymphatic vessels, which are presented deep or superficial. These lymph vessels combine to form lymphatic trunks that empty into the large collecting vessels of lymphatic system called “thoracic duct and right lymphatic duct.” It is important that the lymphatic system operate separately from the cardiovascular system due to the follwoing reasons and exerts differentail effects of innate immune system. The lymph is generated to produce defensive mechanisms via innate immunity include physical barriers, phagocytes, immune surveillance,interferons, complement system, inflammation and fever. 1. Fluid recovery: Filtration of the fluid into the tissue spaces from the blood capillaries will occur continuously. 85% of it will be absorbed by the blood capillaries. About 15% will be remained unabsorbed. This unabsorbed portion of the fluid is about 2-4 litres of water and 1/4th to 1/2th of plasma protein. If this water protein an.

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Provide a full explanation to the below question. 1. Summarize the 802.11 standard and describe the various flavors of the standard, its architecture and how the standard is contributing to easing of congestion in cellular networks. Solution IEEE 802.11 is a standard that defines the physical and MAC layers of Wireless Local Area Network (WLAN). Under this standard, the Mobile terminals (MTs) can communicate with a Access Point (AP) in two modes. First mode is the Infrastructure Mode, in which the MTs communicate with the APs which forward their data to the WLAN. Second mode is the Adhoc mode, in which the MTs communicate directly with each other without the use of a AP. The IEEE 802.11 has a very robust Medium Access Control (MAC) mechanism that helps in alleviating te congestion in the LANs. The inherent working of MAC protocol employed by 802.11 standard is same as that of Carrier Sense Multiple Access/ Collision Aviodance (CSMA/CA). However, in IEEE 802.11 the protocol is implemented in two different ways. First way is called as the Distributed Coordination Function based Wireless MAC (DCFWMAC), and the second way is Point Coordination Function based Wireless MAC (PCFWMAC). In the DCFWMAC, every node tries to access the medium based on some fixed duration of time defined in the standard. Normally there are three such time intervals defined based on which a node attempts to access the channel and transmit its packets. The first interval is the Short Inter frame Spacing (SIFS). It is the smallest duration of time between two frames and gives the node higher priority for sending its data. Such a interval is employed only when the node has correctly sent its data through the channel and the receiver node needs to send a Acknowlegement (ACK) after waiting for SIFS period of time. The second most priority interval is PCF Inter Frame Spacing (PIFS) used during Polling mechanism by the AP. Its duration is between SIFS and DCF Inter frame Spacing (DIFS). The DIFS is the longest time duration and hence of least priority, in which a AP has to wait between two successive channel accesses for the given duration. Once the channel is sensed idle, the AP waits for DIFS period of time to transmit data through the channel. If the channel is sensed busy, then it backs off for a period of time based on the minimum and maximum value of a contention window. During the back off period, if the node senses the channel as busy, then it freezes the backoff counter and starts the backoff(with the remaining time left) once the channel is idle. In this manner, the nodes which have waited longer get higher priority over others which accessing the channel. Once the node sends the data after accessing the channel, the receiver waits for SIFS duration of time to finally send the acknowledgement to the sender. In order to tackle the hidden terminal problem, the DCFWMAC employs an additional mechanism called as the RTS-CTS mechanism, in which any node that wants to transmit.

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Write a program to generate the entire calendar for one year. The program must get two values from the user: (1) the year and (2) the day of the week for January 1st of that year. The year, which should be positive, is needed to check for and handle leap years1. The day of the week for January 1st is needed so that you know where to start the calendar. The user should enter 0 for Sunday, 1 for Monday, ... or 6 for Saturday. As always, you need to validate the user\'s input. To actually print the calendar, you must use a single method that prints out the calendar for one month and then call this function 12 times from main(), once for each month in the year. To check for a leap year you will need to write another method that takes the year as a parameter and returns true if it’s a leap year, or false otherwise. Stubs (i.e. method signatures without any code) for both of these methods have been provided for you. The calendar should be printed in the form below. In this example, January starts on a Saturday (day 6). Note that February starts on a Tuesday in this example because January ended on a Monday. January 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 2829 30 31 February 1 2 3 4 5 6 7 8 9 10 11 12 ... The format he gave us is as follows: public class PA6a { /** * Error to output if year is not positive */ static final String E_YEAR = \"The year must be positive!\"; /** * Error to output if the day is not between 0 and 6 */ static final String E_DAY = \"The day of January 1st must be between 0 and 6!\"; /** * Determines if an input is a leap year * * @param year year in question * @ereturn true if a leap year */ public static boolean isLeapYear(int year) { return false; // TODO: replace with your code } /** * Outputs a month to the console * * @param month title * @param startDay 0=Sunday ... 6=Saturday * @param numDays number of days in the month * @return day of the week of the last day of the month */ public static int printMonth(String month, int startDay, int numDays) { return 0; // TODO: replace with your code } /** * Program execution point: * input year, day of the week (0-6) of january 1 * output calendar for that year * * @param args command-line arguments (ignored) */ public static void main(String[] args) { // TODO: write your code here } I am confused especially with the print month method, help is greatly appreciated. Solution package org.students; import java.util.Scanner; import com.sun.org.apache.bcel.internal.generic.IUSHR; public class Calander { //declaring static variables static final String E_YEAR = \"The year must be positive!\"; static final String E_DAY = \"The day of January 1st must be between 0 and 6!\"; public static void main(String[] args) { //Declaring variables int getFirstDay; int year; Scanner scanner = new Scanner(System.in); //This while loop continue to execute until user enters valid year while (true) { //Getting the year entered by the user System.out.print(\"Enter the year:.

Write a program to generate the entire calendar for one year. The pr.pdf

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Hyenas are diploid organisms, and their gametes contain 20 chromosomes. Fill in the blank in each sentences using the terms below. Each term may be used more than once or not at all. When the somatic cells of a hyena are in interphase, the ploidy level is denoted as The ploidy level of the hyena\'s gametes is denoted as The 2n number of a hyena is 40 20 1n 80 2n Solution When the somatic cells of hyena in interphase = diploid [2n] because they are not divided and tend to divide by preparing enzymes and multiplying DNA. The ploidy level of hyena is haploid or 1n because they are developed through meiosis process. The 2n number of chromosomes of hyena is 40 because gametes are haploid so n=20 [as given] and the 2n = 2x 20 = 40.

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Why doesn\'t hemoglobin have any intermediate conformations between R-state and T-state? Solution Answer: X-ray crystallogarphy has determined two major conformations of hemoglobin- the T state which is stabilized in the absence of oxygen and the R state which is relatively more stable in the presence of oxygen. The pair of identical dimers of Hemoglobin are linked by several ion pairs at the interface that stabilize it in its deoxy-state. Many of these interactions are disrupted upon binding to oxygen and new ones are formed. There are intermediate conformations between T and R states, but the intermediate conformations have intermediate binding affinities. Also, these transition states are thermodynamically unfavourable and short lived. These states have been utilized in studying the allosteric pathways of hemoglobin. The binding of the ligand destabilizes the T state and makes it more likely to switch to the high-affinity oxygenated (oxy)-R state..

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4. What is gyanandromorphy? How does this happen? Solution Answer : Gynandromorphy is a phenamenan in which there is an occurence of both male and female charectorstics in a single individual (part of the body expresses as male charectors, where as the other part express female charectors. Individuals with this phenotype are called gynandromorphs. It results due to the loss of an X chromosome in a particular cell during development through mitosis.For instance, a cell with XY through mitosis produces XXYY upon duplication which in turn divides to form XY cells. But in gynandromorphs, this cell divides into an X cell and an XYY cells. Due to this the organism develops with both male and female charectors..

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Which of the following requires the mitochondria to create contact points with the ER? mt fission mt fusion mt biogenesis of more mitochondria autophagy of mitochondria (Mitophagy) apoptosis Solution Answer= The mitochondria requires the following to create a contact with ER is mt fission - mitochondrial fission Explaination= The mitochondria & RE are tubular organelles with a characteristic network structure that facilitates the formation of interorganeller connections The mito.& ER joins together at multiple contact sites to form specific domains called as mitochondria ER associated membranes.

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why nitrogen fixation is so important Solution Nitrogen fixation It is characteristic and manufactured, is basic for all types of life since nitrogen is required to biosynthesize fundamental building pieces of plants, creatures and other living things, e.g., nucleotides for DNA and RNA, the coenzyme nicotinamide adenine dinucleotide for its part in digestion system (exchanging electrons amongst atoms), and amino acids for proteins. Along these lines, as a major aspect of the nitrogen cycle, it is fundamental for horticulture and the fabricate of compost. It is additionally, in a roundabout way, important to the make of every concoction intensify that contain nitrogen, which incorporates explosives, most pharmaceuticals, colors, and so on. Nitrogen fixation : it happens normally in the dirt by nitrogen settling microscopic organisms (for instance, Azotobacter partnered with a few plants, particularly vegetables). Some nitrogen- settling microbes have cozy associations with plants, alluded to as harmonious nitrogen obsession. Looser connections between nitrogen-settling microorganisms and plants are regularly alluded to as cooperative or non-advantageous, as found in nitrogen obsession happening on rice roots. It additionally happens actually noticeable all around by method for lightning. All natural nitrogen fixation is finished by method for metalloenzymes called nitrogenases. These catalysts contain press, frequently with a moment metal, generally molybdenum yet here and there vanadium. Microorganisms that can settle nitrogen are prokaryotes (both microscopic organisms and archaea, conveyed all through their particular Domains) called diazotrophs. Some higher plants, and a few creatures (termites), have shaped affiliations (advantageous interaction) with diazotrophs..

why nitrogen fixation is so importantSolutionNitrogen fixation.pdf

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