Help me out my elementary surveying class.Write a one-page essay a.pdf

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Help me out my elementary surveying class. Write a one-page essay about the following map projections: UTM: Universal Transverse Mercator & SPCS: State Plane Coordinate System You may try to write introduction, cons & pros, conclusion Solution UNIVERSAL TRANSVERSE MERACATOR (UTM) INTRODUCTION The Universal Transverse Mercator (UTM) conformal projection uses a 2-dimensional Cartesian coordinate system to give locations on the surface of the Earth. Like the traditional method of latitude and longitude, it is a horizontal position representation, i.e. it is used to identify locations on the Earth independently of vertical position. However, it differs from that method in several respects. The UTM system is not a single map projection. The system instead divides the Earth into sixty zones, each being a six-degree band of longitude, and uses a secant transverse Mercator projection in each zone. Mercator projection is very well known, and is the most common way to represent the globe in a 2D map. The good side of the projection is that it will keep the shape of the continents with minimal distortion, but the area will not, and it gets worse as you get closer to the poles. ADVANTAGES DIS-ADVANTAGES STATE PLANE CO-ORDINATE SYSTEM INTRODUCTION The State Plane Coordinate System (SPS or SPCS) is a set of 124 geographic zones or coordinate systems designed for specific regions of the United States. Each state contains one or more state plane zones, the boundaries of which usually follow county lines. The system is widely used for geographic data by state and local governments. Its popularity is due to at least two factors:- DIS-ADVANTAGES The main problem with the state plane coordinate system is that each zone uses a different coordinate system. This is not a major problem as long as one\'s needs are within the boundaries of a given state plane zone, as is the case with most county and city governments. However, the need to transform spatial data from one coordinate system to another can be burdensome..

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of a given state plane zone, as is the case with most county and city governments. However, the
need to transform spatial data from one coordinate system to another can be burdensome.

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Give background information concerning EMT and include what is known about EMT in normal development as well as in neoplasms. What are strategies that could be used to overcome drug resistance in cancer cells by targeting EMT (minimum four paragraphs). Solution Answer: Cancer stem cell (CSC) hypothesis: This hypothesis elucidating that the CSC are the subpopulation and able to exert higher resistance associated with explicit capability to self-renew finally to differentiate into different offspring of cancer cells to generate a tumor. This hypothesis is based on the cellular mechanisms exclusively developed inside the CSC to exert resistance such pathways are mainly through Wnt signaling, Notch signaling. There are other targets to regulate proliferation and differentiation through ABC transporter system and receptor tyrosine kinases. These targets are the future stem cells anti-cancer therapies in regulating the proliferation & differentiation of cancer cells as CSCs have a potential relationship with epithelial-mesenchymal transition (EMT). Metastasis has many steps. The main cancer cell morphological features that support invasion, intravasation, and extravasation are epithelial-mesenchymal transition via cell -adhesion molecules. Invasion is mainly due to presence of cell adhesion molecules such as integrins on the cancer cells finally promote the movement of cancer cells to another region. This process is going to connect adjacent cells to tumor cells finally undergo epithelial-mesenchymal transition (EMT) result in formation of \"continuously dividing neoplasms\" with higher drug resistence to cure compared to normal cells Strategies that could be used to overcome drug resistance in cancer cells by targeting EMT: \"A combination therapy of neutraceuticals & chemotherapeutic agents\" are most promising strategies to overcome drug resistance in cancer cells by targeting EMTs, cancer stem cells (CSC) & Modulation of microRNAs do promote effective strategy to overcome drug resistance in cancer cells Oligonucletodie deliver: Another strategy could be used to overcome drug resistance in cancer cells by targeting EMT. In this method, synthesized microRNAs are going to make in the form of \"nanoparticle or microsomal\" formulation finally used to target microRNAs in cancer cells result in \"complete destruction of pre-mRNA\" finally there will no translation mechanism to synthesize \"cell survival proteins\" to proliferate.

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X-Linked Recessive Write three rules to keep in mind when counseling someone about the likelihood of inheriting an X-linked recessive condition: About parents\' genotype. About parents\' phenotype. About the probability of the offspring showing the trait. If instead of having Hemophilia C (autosomal recessive), Lisa Keller\'s brother had Hemophilia A (X-linked recessive), what would be the probability that she was a carrier? 25% 50% 67% 100% More information is required to answer this question. Solution About genotype of parents- If the mother is a carrier i.e. carrying the recessive allele on one of her X chromosome and the father do not have the recessive allele in his genotype, then none of the daughters will suffer from the disease (since they will inherit only one recessive allele from the mother) but sons will have a 50% probability of suffering from the disease as they will inherit the recessive allele from the mother and the Y chromosome from the father, although there is a 50% probability of inheriting the wild type allele from the mother. Now if the mother is suffering from the disease i.e. has the recessive allele on both the X chromosome while the father is healthy having no allele for the disease, the daughters will always be a carrier but never suffer from the disease as they eill inherit the healthy wild type allele from the father as one X chromosome comes from the father and the other X chromosome from the mother. The sons will however all suffer from the disease as they inherit the Y chromosome from the father and the X chromosome having the recessive allele causing the disease will be inherited from the mother. Now suppose the mother is a carrier i.e. carries the recessive allele only on one X chromosome but the father suffers from the disease making his genotype XrY (Xr denoting recessive allele carrying X chromosome), then, 50%of their son will duffer from the disease while 50% of the daughters will also have a chance of suffering from the disease. If both the parents do not have a single recessive X linked disease gene on any of their X chromosome then none of the children will suffer from the disease. About phenotype of parents- If the mother is a carrier i.e. have the diseased allele on only one X Chromosome, then the phenotype of mother will be normal. But the father can never be a carrier for a X linked recessive disease as males have only one X chromosome. If the diseased allele in present on that X chromosome in males it will be expressed as there is no dominant healthy wild type allele to dominate the diseased gene. Thus if the father has the recessive diseased allele on X chromosome, the father\'s phenotype will be expresding the disease while if the father do not have the allele he will be healthy. On the other hand for the mother to express the disease the recessive allele has to be present on bith her X chromosome. Probability of offsprings showing the trait- If mother is a carrier and father is normal, then 5.

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Write a class (BasketballTeam) encapsulating the concept of a team of basketball players. This method should have a single instance variable which is an array of basketball player objects. In addition to that class, you will need to design and a code a Playerclass to encapsulate the concept of a basketball player, assuming a basketball player has the following attributes: a name, a position, number of shots taken, and number of shots made. In your Player class you should write the following methods: A constructor taking a String for the name, a String for the position, an int for the number of shots taken, and an int for the number of shots made. Accessor, mutator, toString, and equals methods. A method returning the individual shooting percentage. Avg = Shots Made / Shots Taken. Hint: This method should return a double!!! In your BasketballTeam class you should write the following methods: A constructor taking an array of Player objects as its only parameter and assigning that array to the array data member of the class, its only instance variable. Accessor, mutator, toString, and equals methods. A method checking if all positions are different, returning true if they are, false if they are not. A method returning the shooting percentage of the team. Team Shooting Percentage = Total Shots Made / Total Shots Taken Hint: This should return a double!!! Also, this is not the average of the the averages. A method checking that we have a center (that is, the name of the position) on the team. If we do not have any return false; otherwise, it returns true; A method returning void that sorts the array of Player objects in ascending order using the number of shots taken as the sorting key. A method returning the array of Player objects sorted in ascending alphabetical order by name. Hint: Use compareTo method of the String class. A method returning the name of the player with the most shots made. A method returning the name of the player with the highest shooting average. In your BasketballTeamClient class, when you test all your methods, you can hard-code five basketball Player objects. This client should call all 12 methods including the constructor for full credit. When writing the Player class you should write a PlayerClient class that tests all its methods before integrating into BasketballTeam class. Solution public class Player { //Attributes private String name; private String position; private int shotsTaken; private int shotsMade; /** * Constructor * @param name * @param position * @param shotsTaken * @param shotsMade */ public Player(String name, String position, int shotsTaken, int shotsMade) { this.name = name; this.position = position; this.shotsTaken = shotsTaken; this.shotsMade = shotsMade; } /** * @return the name */ public String getName() { return name; } /** * @param name the name to set */ public void setName(String name) { this.name = name; } /** * @return the position */ public String getPosition() { return position; } /** * @p.

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Which of the following would you expect to happen in a plant that does not have enough water (multiple answers are correct) Without water, the plant cannot make the sugar it needs to grow the stem begins to dry out growth would be stunted because it is difficult to move nutrients around Without water, the plant would become limp. Solution Answer: All the choices given are correct! If Plant that does not have enough water; (a) Without water, plants cannot make the sugar it needs to grow. Plants make their own food by the process of photosynthesis.The amount of water, sunlight and carbon dioxide directly influences the process of food production for plant growth. Thus the process of photosynthesis slows down and even stops when water is absent, which causes internal food supplies to diminish. (b) the stem begins to dry out Scarcity of water leads to loss of turgor pressure. Stomata on the stem surface close as a result of less water (for water conservation) and as a result stem dries out. (c) the growth would be stunted because it is difficult to move nutrients around. As water moves above the plant system it carries vital minerals and nutrients to different areas of the plant. The stomas on the leaf and stem surface release waste products (oxygen) into the environment and takes CO2 in. This process of transpiration maintains the turgor pressure and also cools the plant in sunlight. In scarcity of water, transpiration process slows and, the plant show stunded growth due to lack of nutrients supply. (d) Without water the plant would become limp Turgor pressure holds the plant cells and keeps them erect. This pressure is lost without water and plants begin to collapse (wilting!) or give a limp appearance..

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What adaptive benefit do these abilities give wolbachia? In other words, why does manipulating the host reproduction help its survival survival as parasite? What adaptive benefit do these abilities give wolbachia? In other words, why does manipulating the host reproduction help its survival survival as parasite? What adaptive benefit do these abilities give wolbachia? In other words, why does manipulating the host reproduction help its survival survival as parasite? Solution Manipulating the host reproduction help its survival as parasite.The improvement of complex lifestyles cycles can growth parasite health. Some parasites have, via host manipulation, further improved their reproductive success. Host manipulation in a few parasites, where they effectively can change the behavior or morphology of their intermediate host in a way that increases the parasite\'s trophic transmission It is a method of parasite transmission wherein the parasite manipulates the behavior of its host to growth transmission success to better trophic levels.

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Think about autumn (Fall season) in places like New England (Pennsylvania, New York, Vermont, New Hampshire), where the leaves change color with the seasons. During the fall, which pigments are the first to be broken down in the leaves? How does that explain the fall colors we see? Why does the green color disappears during autumn (Fall season)? Please fully and clearly explain why the green color disappeares in trees located in New England during the autumn. Solution A number of factors play crucial role in changing the color of leaves in winter. Some are listed below: Similar is the reason with the New England trees..

Think about autumn (Fall season) in places like New England (Pennsyl.pdf

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The orange is which type of fruit? Simple - Legume Simple - Drupe Simple - Berry Multiple Aggregate The peas is which type of fruit? Simple - Legume Simple - Drupe Simple - Berry Multiple Aggregate The pistachio is which type of fruit? Simple - Legume Simple - Drupe Simple - Berry Multiple Aggregate Solution Please find the answers below and reasons in parentheses: Answer 22: Chocie 3 (Botanically, orange fruit is considered as a multiple aggregate of berry and the juicy part or pulp is considered as succulent testa of the ovary. The peel of orange is considered as ovarian wall.) Answer 23: Choice 1 (A pea pod is considered as a simple legume by botanical nature. It can open across both walls which eliminates the seeds inside the pod) Answer 24: Choice2 (Pistachios belong to family of cashew or family Anacardiaceae botanically. The fruits of this family are simple drupes in nature. These fruits contain the ovarian wall as the hard covering of the fruit and the ovary itself modifies into edible seed).

The orange is which type of fruit Simple - Legume Simple - Drupe .pdf

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The _________ is the protective chamber that houses the ovule and Later matures to become the fruit. A. Petal B. Ovary C. Carpel D. Stigma E. Sepals. What adaptation enhances bird\'s flying ability? A. hollow bones B. sternum. C. air sacs D. homeothermy E. all of the above. What plant characteristics are adaptations to land? A. internal fertilizations B. vascular tissues. C. seeds with waterproof coats. D. waxy leaves. E. all of the above. Which of which is characteristic of alternation of generation in land plants? A. haploid sporophytes make haploid spores. B. gametophytes produce spores that develop into gametes C. sporophytes and gametophytes are typically similar in appearance. D. meiosis in sporophytes produces haploid spores. E. either the gametophyte or the sporophyte is unicellular. The haploid stage in a plant life in a plant life cycle is called A. embryo B. zygote C. gametophyte D. sporophyte E. conspicuous Animals in which the blastopo Solution 31. ovary is the protective chamber that houses the ovule and later matures the fruit 32. tha hollow bones, air sacs , the sternum and homeothermy helps the birds to fly. the hollow bones helps in reducing the body weight which helps in flying. the air sacs are responsible for extracting more oxygen which hepls them to frly high. the sternum bones are present in the wings which are strong and flexible for the movement of the wings 33. all of the above are the plants characteristics for the plant to adapt land 34. 35. the haploid stage of plant life in plant life cycle is known as gametophyte. 36.in protostomes the blastopore becomes the mouth.

The _________ is the protective chamber that houses the ovule and Lat.pdf

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