I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancyBalanc.pdf

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I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancy Balanced MolecularEquation is : CuCl2 +H2SiO4 ---> CuSiO4 (s) +2HCl Total ionic Equation is : Cu2+ +2 Cl - + 2H+ +SO42- ---> CuSiO4 +2H+ + 2Cl- Spectator ions are : Cl - &H+ Spectator ions are : Cl - &H+ Net ionic Equation is : Cu2+ + SO42- --->CuSiO4 Solution I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancy Balanced MolecularEquation is : CuCl2 +H2SiO4 ---> CuSiO4 (s) +2HCl Total ionic Equation is : Cu2+ +2 Cl - + 2H+ +SO42- ---> CuSiO4 +2H+ + 2Cl- Spectator ions are : Cl - &H+ Spectator ions are : Cl - &H+ Net ionic Equation is : Cu2+ + SO42- --->CuSiO4.

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Static arrays are structures whose size is fixed at compile time and therefore cannot be extended or reduced to fit the data set. A dynamic array can be extended by doubling the size but there is overhead associated with the operation of copying old data and freeing the memory associated with the old data structure. One potential problem of using arrays for storing data is that arrays require a contiguous block of memory which may not be available, if the requested contiguous block is too large. However the advantages of using arrays are that each element in the array can be accessed very efficiently using an index. However, for applications that can be better managed without using contiguous memory we define a concept called “linked lists”. A linked list is a collection of objects linked together by references from one object to another object. By convention these objects are named as nodes. So the basic linked list is collection of nodes where each node contains one or more data fields AND a reference to the next node. The last node points to a NULL reference to indicate the end of the list. Types of Linked Lists Linked lists are widely used in many applications because of the flexibility it provides. Unlike arrays that are dynamically assigned, linked lists do not require memory from a contiguous block. This makes it very appealing to store data in a linked list, when the data set is large or device (eg: PDA) has limited memory. One of the disadvantages of linked lists is that they are not random accessed like arrays. To find information in a linked list one must start from the head of the list and traverse the list sequentially until it finds (or not find) the node. Another advantage of linked lists over arrays is that when a node is inserted or deleted, there is no need to “adjust” the array. There are few different types of linked lists. A singly linked list as described above provides access to the list from the head node. Traversal is allowed only one way and there is no going back. A doubly linked list is a list that has two references, one to the next node and another to previous node. Doubly linked list also starts from head node, but provide access both ways. That is one can traverse forward or backward from any node. A multilinked list (see figures 1 & 2) is a more general linked list with multiple links from nodes. For examples, we can define a Node that has two references, age pointer and a name pointer. With this structure it is possible to maintain a single list, where if we follow the name pointer we can traverse the list in alphabetical order of names and if we traverse the age pointer, we can traverse the list sorted by ages. This type of node organization may be useful for maintaining a customer list in a bank where same list can be traversed in any order (name, age, or any other criteria) based on the need. Designing the Node of a Linked List Linked list is a collection of linked nodes. A node is a struct with at least a.

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Interphase. This is the phase where the cell prepares for the next round of division. It has three phases further, Gap1, synthesis, and gap 2. During Gap1, the cell acquired nutrients and increases in size. The checkpoints ensure that the cell is normal and all the things necessary for DNA synthesis are prepared. During synthesis phase, DNA replicates. During the Gap 2 phase, the cell grow further and the checkpoints ensure that the cell is correctly prepared for mitosis phase. The cell spends most of its time in G1 phase, followed by S phase. Remaining phases comprise only very short period of the cell cycle. Solution Interphase. This is the phase where the cell prepares for the next round of division. It has three phases further, Gap1, synthesis, and gap 2. During Gap1, the cell acquired nutrients and increases in size. The checkpoints ensure that the cell is normal and all the things necessary for DNA synthesis are prepared. During synthesis phase, DNA replicates. During the Gap 2 phase, the cell grow further and the checkpoints ensure that the cell is correctly prepared for mitosis phase. The cell spends most of its time in G1 phase, followed by S phase. Remaining phases comprise only very short period of the cell cycle..

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In the above conversation it is belonging to stereotypes. Stereotypes are a form of generalization. When we generalize, we group or classify people, places, or things according to the traits they have in common. For example, we may say most Masai tribesmen are unusually tall or Scandinavians are usually fair-skinned. If our observations are careless or too limited, the generalization may be faulty, as when someone says, \"Hollywood hasn\'t produced any quality movies in the past fifteen years.\" But stereotypes are more serious than mere faulty generalizations. They are fixed, unbending generalizations about people, places, or things. When a stereotype is challenged, the person who holds it is unlikely to modify or discard it, because it is based on a distortion of perception. As Walter Lippmann explains, when we stereotype,\" we do not so much see this man and that sunset; rather we notice that the thing is man or sunset, and then see chiefly what our mind is full of on those subjects.\"75 The most common kinds of stereotyping are ethnic and religious. Jews are shrewd and cunning, clannish, have a financial genius matched only by their greed. Italians are hot-tempered, coarse, and sensual. The Irish, like the Poles, are big and stupid; in addition, they brawl, lust after heavy liquor and light conversation. Blacks are primitive and slowwitted. (Often each of these stereotypes includes a virtue or two – Jews are good family members, Italians artistic, Poles brave, the Irish devout, blacks athletic.) Beyond these stereotypes are numerous other, less common ones; Swedish women, foreign film directors, Southern senators, physical education instructors, fundamentalist clergymen, agnostics, atheists, democrats, republicans, Mexicans, scientist, prostitutes, politicians, English teachers, psychiatrists, construction workers, black militants, college dropouts, homosexuals, and society matrons. There are stereotypes of institutions as well: marriage, the church, government, the military, the Founding Fathers, Western culture, the Judeo-Christian tradition. (A full list would include even God and mother.) FACTS DON\'T MATTER It is pleasant to assume that when the facts are known, stereotypes disappear. However, that is seldom the case. The late Harvard psychologist Gordon Allport, in The nature of Prejudice, pointed out that \"it is possible for a stereotype to grow in defiance of all evidence…\"76 People who stereotype don\'t dust accept the facts that are offered to them. They measure those facts against what they already \"know.\" That is, they measure them against the stereotype itself. Instead of seeing that the stereotype is false and therefore dismissing it, they reject the unfamiliar facts. People who think in terms of stereotypes tend to be selective in their perceptions. They reject conditions that challenge their preformed judgment and retain those that reinforce it. Thus a person can notice the Jewish employer promoting another Jew but ign.

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if one can understand a few things it is easier to solve these kind of questions. first comes to I which codes for repressor. there are 3 types - I+ Wild type and is normal and make Z and Y products in the presence of lactose only. I- is a recessive mutation will make products of Z and Y in the presence and the absence of lactose. Is is a dominant mutation will stop both the products Z and Y. if you have atlease one Is in the question neither the products will from no matter whether the lactose is presence or not. then comes to Operator (O) ,O+ is a wild type and can produce both the products of Z and Y. in the presence of lactose.and Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose. next comes to Z+ represents production of -galactosidase and Z- represents no production of - galactosidase. 1.Ans. I+P-O+Z+Y+ In this haploid genotype the transcription of operon genes totally absent. Transcription of structural genes is not occured. because here P- indicates the mutation in the promoter site, the RNA polymerase unable to bind to the Promoter and initiate transcription of structural genes. so these cells cannot able to grow on media which contain lactose as sole carbon source. the Strain is lac- 2.Ans.I+P+OcZ+Y- here Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose means constitutively. but here only the product beta galctosidase produced constitutively and the permease is not produced because Permease gene is is mutated (Y-). these cells also cannot able to grow on media which contain lactose as sole carbon source.because the absence of functional permease enzyme so the Strain is lac- 3.Ans. I-P+OcZ+Y+ here I- is a recessive mutation will make products of Z and Y in the presence and the absence of lactose and Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose. so this haploid genotype transcribes operon genes constitutively. and both the products of Z and Y expressed continuously, so these cells able to grow on the media which contain lactose as sole carbon source. the strain is lac+ b. partially diploid cap-I+P+O+Z-Y+/Cap+I-P+O+Z+Y- these cells can able to grow on lactose medium because these cells have presence of wild promoter which can able to active the transcription of structural genes. so these cells can able to grow on the media which contain lactose as sole carbon source. ii. Transcription of beta-galactosidase and permease is Inducible though diploid cell have mutated inducer region which can able to activate constitutive expression but Wild type (I+) is dominant over I- . so it is inducible operon. Solution if one can understand a few things it is easier to solve these kind of questions. first comes to I which codes for repressor. there are 3 types - I+ Wild type and is normal and make Z and Y products in the presence of lactose only. I- is a recessive mutation .

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class Mesh{ public: /*Mesh data*? vectorvertices; vector indices; vectortextures; ?*functions*? Mesh(vectorvertices,vectorindices,vector textures); void draw(shader shader); private: /* Render data*/ gluint vao,vbo,ebo; /* functions */ void setupmesh(); }; Solution class Mesh{ public: /*Mesh data*? vectorvertices; vector indices; vectortextures; ?*functions*? Mesh(vectorvertices,vectorindices,vector textures); void draw(shader shader); private: /* Render data*/ gluint vao,vbo,ebo; /* functions */ void setupmesh(); };.

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Bacteria was used in transformation experiment, to prove Dna was the genetic material and replicated semi conservatively, used in gene expression and regulation, preferred host for cloning genes, and in manipulation of DNA Solution Bacteria was used in transformation experiment, to prove Dna was the genetic material and replicated semi conservatively, used in gene expression and regulation, preferred host for cloning genes, and in manipulation of DNA.

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