Maternal and parental chromosomes are separated in Anaphase I of mei.pdf

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Maternal and parental chromosomes are separated in Anaphase I of meiosis cell division. So sister chromatids are hold together in the Anaphase I. The Anaphase II is responsible for the separation of sister chromatids to form chromosomes. Sister chromatids are chromosomes of either maternal or paternal. Solution Maternal and parental chromosomes are separated in Anaphase I of meiosis cell division. So sister chromatids are hold together in the Anaphase I. The Anaphase II is responsible for the separation of sister chromatids to form chromosomes. Sister chromatids are chromosomes of either maternal or paternal..

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Particulars Amount $ Share Holders Funds Common Stock : Equity share capital as on 01.01.2013 500000 Issued During the Year 50000 550000 Retained Earnings : Previous Years 100000 Current Period Profit/ Loss: Revenue 150000 Expense (90000) 60000 160000 Less: Dividends declared -10000 Total 700000 Solution Particulars Amount $ Share Holders Funds Common Stock : Equity share capital as on 01.01.2013 500000 Issued During the Year 50000 550000 Retained Earnings : Previous Years 100000 Current Period Profit/ Loss: Revenue 150000 Expense (90000) 60000 160000 Less: Dividends declared -10000 Total 700000.

Particulars Amount $ Share Holders Funds Com.pdf

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The Nitro group is electron withdrawing, and therefore enhances the reactivity of the carbonyl group in the ester function towards nucleophiles, such as amines and hydrazine. The Nitro group ensures that any derived compounds will be crystalline solids. The Nitro group itself can be converted to a variety of other functional groups, such as hydroxylamine (RNHOH) and amine (RNH2) which can be used as building blocks for more complicated structures: anilines (derived from reducing NO2 to NH2) can be converted easily to quinolones with antibiotic properties generally. Solution The Nitro group is electron withdrawing, and therefore enhances the reactivity of the carbonyl group in the ester function towards nucleophiles, such as amines and hydrazine. The Nitro group ensures that any derived compounds will be crystalline solids. The Nitro group itself can be converted to a variety of other functional groups, such as hydroxylamine (RNHOH) and amine (RNH2) which can be used as building blocks for more complicated structures: anilines (derived from reducing NO2 to NH2) can be converted easily to quinolones with antibiotic properties generally..

The Nitro group is electron withdrawing, and ther.pdf

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The Systems Development Life Cycle Moderate and large firms with unique information needs often develop information systems in-house. That is to say that information technology (IT) professional within the firm design and program the systems. A greater number of smaller companies and large firms with relatively standardized information needs opt to purchase information systems from software vendors. Both approaches represent significant financial and operational risks. a model for reducing this risk through careful planning, execution, control, and documentation of key activities. The five phases of this model are: 1) Business Needs and Strategy Systems Strategy –Assess Strategic Information Needs –Develop a Strategic Systems Plan –Create an Action Plan 2) Project Initiation – Systems Analysis –Conceptualization of Alternative Designs –Systems Evaluation and Selection 3.) In-House Systems Development –Construct the System –Deliver the System 4). Commercial Packages –Trends in Commercial Packages –Choosing a Package 5) Maintenance and Support The participants in systems development can be classified into three broad groups: systems professionals, end users, and stakeholders. Systems professionals are systems analysts, systems designers, and programmers. These individuals actually build the system. They gather facts about problems with the current system, analyze these facts, and formulate a solution to solve the problems. The product of their efforts is a new system. End users are those for whom the system is built. Many users exist at all levels in an organization. These include managers, operations personnel, accountants, and internal auditors. In some organizations, it is difficult to find someone who is not a user. During systems development, systems professionals work with the primary users to obtain an understanding of the users’ problems and a clear statement of their needs. As defined in Chapter 1, stakeholders are individuals either within or outside the organization who have an interest in the system but are not end users. These include accountants, internal auditors, external auditors, and the internal steering committee that oversees systems development. Cost/Time Analysis: As stated before, the cost/time analysis is an attempt to calculate to what degree the project and system will meet the objectives. The SDLC must address two topics in its support of this area: the scope of the analysis and the algorithm for doing it. Cost/Benefit Scope The benefit side of the analysis should be expressed in quantitative terms wherever possible. Qualitative or intangible benefits usually are reflections of poorly analyzed tangible benefits. The SDLC should support the process of quantifying all benefits. On the costs side, the SDLC must address development costs, installation costs and ongoing operational costs. In doing these calculations it should differentiate between capital costs and expense costs. Cost/Benefit Algorithm The method of calcu.

The Systems Development Life Cycle Moderate and large firms with uni.pdf

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You are asked to determine DNA content of a liver tissue. Describe how you will porceed in finding amount of DNA in the sample Take a piece of live tissue and subject it to detergent solution i.e. SDS (sodiumdodecyl sulfate) so that the cell membrane of the cells will be removed out through micelle formation & then alcohol is added as DNA soluble in water. This is followed by the formation of aqueous phase with DNA, precipitate layer with protein and bottom organic layer. Now the purity of DNA sample is calculated to know the “total amount of DNA in the sample as ng DNA / µg dry weight”. There are various processes applicable to determine total DNA content using either cloning, qPCR, sequencing. However, initial purity of DNA can be calculated by measuring absorbance or OD260/OD280 ratio at 280nm using “spectrophotomemeter” (UV-Visible). The range of OD with absorption of UV light between 1.8 to 2.0 denotes the presence of DNA. Lower than this OD ratio, denotes the presence of proteins. The another method of examining the sample is through & this method is useful to deterimine total DNA concentration by use of the fluorochrome Hoechst 33258 & filters, & scanning fluorescence spectrophotometer at emission at 458 nm. Solution You are asked to determine DNA content of a liver tissue. Describe how you will porceed in finding amount of DNA in the sample Take a piece of live tissue and subject it to detergent solution i.e. SDS (sodiumdodecyl sulfate) so that the cell membrane of the cells will be removed out through micelle formation & then alcohol is added as DNA soluble in water. This is followed by the formation of aqueous phase with DNA, precipitate layer with protein and bottom organic layer. Now the purity of DNA sample is calculated to know the “total amount of DNA in the sample as ng DNA / µg dry weight”. There are various processes applicable to determine total DNA content using either cloning, qPCR, sequencing. However, initial purity of DNA can be calculated by measuring absorbance or OD260/OD280 ratio at 280nm using “spectrophotomemeter” (UV-Visible). The range of OD with absorption of UV light between 1.8 to 2.0 denotes the presence of DNA. Lower than this OD ratio, denotes the presence of proteins. The another method of examining the sample is through & this method is useful to deterimine total DNA concentration by use of the fluorochrome Hoechst 33258 & filters, & scanning fluorescence spectrophotometer at emission at 458 nm..

You are asked to determine DNA content of a liver tissue. Describe h.pdf

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when we draw a vertical line it will cross the graph twice(that is m.pdf

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We Know that Baking powder and Baking soda isused as a leave.pdf

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The fallowing program shows the simple transformation #define GLEW_STATIC #include #include #include #include #include #include #include const GLchar* vertexSource = \"#version 150 core\ \" \"in vec2 position;\" \"in vec3 color;\" \"in vec2 texcoord;\" \"out vec3 Color;\" \"out vec2 Texcoord;\" \"uniform mat4 trans;\" \"void main()\" \"{\" \" Color = color;\" \" Texcoord = texcoord;\" \" gl_Position = trans * vec4(position, 0.0, 1.0);\" \"}\"; const GLchar* fragmentSource = \"#version 150 core\ \" \"in vec3 Color;\" \"in vec2 Texcoord;\" \"out vec4 outColor;\" \"uniform sampler2D texKitten;\" \"uniform sampler2D texPuppy;\" \"void main()\" \"{\" \" outColor = mix(texture(texKitten, Texcoord), texture(texPuppy, Texcoord), 0.5);\" \"}\"; int main() { auto t_start = std::chrono::high_resolution_clock::now(); sf::ContextSettings settings; settings.depthBits = 24; settings.stencilBits = 8; sf::Window window(sf::VideoMode(800, 600, 32), \"OpenGL\", sf::Style::Titlebar | sf::Style::Close, settings); glewExperimental = GL_TRUE; glewInit(); GLuint vao; glGenVertexArrays(1, &vao); glBindVertexArray(vao); GLuint vbo; glGenBuffers(1, &vbo); GLfloat vertices[] = { -0.5f, 0.5f, 1.0f, 0.0f, 0.0f, 0.0f, 0.0f, 0.5f, 0.5f, 0.0f, 1.0f, 0.0f, 1.0f, 0.0f, 0.5f, -0.5f, 0.0f, 0.0f, 1.0f, 1.0f, 1.0f, -0.5f, -0.5f, 1.0f, 1.0f, 1.0f, 0.0f, 1.0f }; glBindBuffer(GL_ARRAY_BUFFER, vbo); glBufferData(GL_ARRAY_BUFFER, sizeof(vertices), vertices, GL_STATIC_DRAW); GLuint ebo; glGenBuffers(1, &ebo); GLuint elements[] = { 0, 1, 2, 2, 3, 0 }; glBindBuffer(GL_ELEMENT_ARRAY_BUFFER, ebo); glBufferData(GL_ELEMENT_ARRAY_BUFFER, sizeof(elements), elements, GL_STATIC_DRAW); GLuint vertexShader = glCreateShader(GL_VERTEX_SHADER); glShaderSource(vertexShader, 1, &vertexSource, NULL); glCompileShader(vertexShader); GLuint fragmentShader = glCreateShader(GL_FRAGMENT_SHADER); glShaderSource(fragmentShader, 1, &fragmentSource, NULL); glCompileShader(fragmentShader); GLuint shaderProgram = glCreateProgram(); glAttachShader(shaderProgram, vertexShader); glAttachShader(shaderProgram, fragmentShader); glBindFragDataLocation(shaderProgram, 0, \"outColor\"); glLinkProgram(shaderProgram); glUseProgram(shaderProgram); GLint posAttrib = glGetAttribLocation(shaderProgram, \"position\"); glEnableVertexAttribArray(posAttrib); glVertexAttribPointer(posAttrib, 2, GL_FLOAT, GL_FALSE, 7 * sizeof(GLfloat), 0); GLint colAttrib = glGetAttribLocation(shaderProgram, \"color\"); glEnableVertexAttribArray(colAttrib); glVertexAttribPointer(colAttrib, 3, GL_FLOAT, GL_FALSE, 7 * sizeof(GLfloat), (void*)(2 * sizeof(GLfloat))); GLint texAttrib = glGetAttribLocation(shaderProgram, \"texcoord\"); glEnableVertexAttribArray(texAttrib); glVertexAttribPointer(texAttrib, 2, GL_FLOAT, GL_FALSE, 7 * sizeof(GLfloat), (void*)(5 * sizeof(GLfloat))); GLuint textures[2]; glGenTextures(2, textures); int width, height; unsigned char* image; glActiveTexture(GL_TEXTURE0); glBindTexture(GL_TEXTURE_2D, textures[0]); image = SOIL_load_imag.

The fallowing program shows the simple transformation #define GLEW.pdf

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Two types of malware :- 1. Virus: A virus is a contagious program or code that attaches itself to another piece of software, and then reproduces itself when that software is run. Most often this is spread by sharing software or files between computers. Virus cannot be spread without a human action. So when you share infected files or send email with viruses as attachment in email, virus spread in another system also. If you have turned off your anti-virus protection your system is vulnerable to virus and other threats. 2. Trojan: This is the most dangerous Malware. Trojans are written with the purpose of discovering your financial information, taking over your computer’s system resources, and in larger systems creating a “denial-of-service attack ”. Denial-of-service attack: an attempt to make a machine or network resource unavailable to those attempting to reach it. Example: AOL, Yahoo or your business network becoming unavailable. 70% malware are trojan horse. When you open a file that have trojan horse or other kind of unscrupulous virus program then it replicate in your system and your system is also get infected. Solution Two types of malware :- 1. Virus: A virus is a contagious program or code that attaches itself to another piece of software, and then reproduces itself when that software is run. Most often this is spread by sharing software or files between computers. Virus cannot be spread without a human action. So when you share infected files or send email with viruses as attachment in email, virus spread in another system also. If you have turned off your anti-virus protection your system is vulnerable to virus and other threats. 2. Trojan: This is the most dangerous Malware. Trojans are written with the purpose of discovering your financial information, taking over your computer’s system resources, and in larger systems creating a “denial-of-service attack ”. Denial-of-service attack: an attempt to make a machine or network resource unavailable to those attempting to reach it. Example: AOL, Yahoo or your business network becoming unavailable. 70% malware are trojan horse. When you open a file that have trojan horse or other kind of unscrupulous virus program then it replicate in your system and your system is also get infected..

Two types of malware -1. Virus A virus is a contagious program o.pdf

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TlI -- Tl+ + I- Ksp = 8.9E-8 = [Tl+][I-] (eqn. 1) PbI2 .pdf

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The cells of the reproductive organs (Eg: ovaries ad testes) undergo meiosis. Meiosis is a type of nuclear division, which produces four haploid cells from a single diploid cell. Meiosis is divided into two phases, meiosis I (one diploid nucleus to two haploid nuclei) and meiosis II (two haploid nuclei to four haploid nuclei). Meiosis I and meiosis II are subdivided into four phases again namely, prophase, metaphase, anaphase and telophase. Thus, meiosis is divided into a total of eight phases. The phases of meiosis I are prohase-i, metaphase-I, anaphase-I and telophase-I. Similarly, the phases of meiosis-II are prohase-II, metaphase-II, anaphase-II and telophase-II. Prophase is the first phase of cell division in both mitosis and meiosis. During this phase, cohesion that is present throughout the length of the chromatids is removed except at the centromere. The chromatids become more visible during this phase. In meiosis, the substages of prophase-I include, leptonema, zygonema, pachynema, diplonema, diakinesis. During the metaphase, two chromatids are formed from each chromosome by longitudinal splitting. These chromatids form the chromosomes of daughter cells. During anaphase, centromeres are pulled and the chromatids become separated and move towards the opposite poles. These chromatids now are the chromosomes of the daughter cells, and are pulled towards the centrioles by means of spindle fibres. During “metaphase I and II” the chromosomes align at the midway between the two opposite spindle poles. In the stage of telophase, cytoplasmic division of the cell takes place (cytokinesis) during which each of the daughter cell receives one of the daughter nuclei. The interval between each division is called interphase. During meiosis, the homologous chromosomes align during the prophase. These chromosomes crossover at some point and rejoin to form recombinant chromatids, and move towards the opposite poles during anaphase. The chromosomal crossover ocurs during metaphase-I of the meiosis, In meiosis, Alignment of homologous chromosomes at Mid-Prophase –I is characteristic of meiosis. The two chromosomes sharing the same quality are called “homologous chromosomes.” These chromosomes crossover at some point and rejoin to form recombinant chromatids. Now, these chromosomes are separated at Anaphase-I and move to their opposite poles. During anaphase of mitosis, the sister chromatids get separated and move to the opposite poles. In contrast to mitosis, the “homologues chromosomes” (but not sister chromatids) move to the opposite poles in anaphase –I of meiosis. In meiosis, the separation of sister chromatids occur in anaphase –II. Solution The cells of the reproductive organs (Eg: ovaries ad testes) undergo meiosis. Meiosis is a type of nuclear division, which produces four haploid cells from a single diploid cell. Meiosis is divided into two phases, meiosis I (one diploid nucleus to two haploid nuclei) and meiosis II (two haploid nuclei to four h.

The cells of the reproductive organs (Eg ovaries ad testes) undergo.pdf

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Scientific evidences I would give in support of evolution: 1. Paleontological evidence: Fossils are those remains of plants and animals that could not be degraded. Some of them bear resemblances to present day animals and serve as connecting links, eg. Duck billed platypus is connecting link between reptiles and mammals. Some of them represent extinct organisms, eg. dinosaurs. Cross sections of the earth\'s crust from where the fossils have been isolated determines the geological time scale during which the organism existed. 2. Comparative anatomy/physiology: Homologous organs of animals such as forelimb bones of humans, tigers, whales and bats share similar type of arrangement and indicates common ancestry. Such type of evolution is called \"divergent\". Analogous organs are opposite to homologous organs, such as wings of butterfly and birds. Although they perform the same function, they are not anatomically similar. Thus, these are results of \"convergent \"evolution. 3. Similarities in genes and proteins among genetically different organisms also support for the existence of common ancestry. 4. Natural selection/Industrial melanism: The dominance of dark coloured moths in industrial areas compared to the white forms is an example of industrial melanism. This type of selection resulted in the dark coloured moths from becoming a prey of birds compared to the paler ones as the darker forms camouflage in the dark coloured soots. Another example of natural selection is antibiotic resistance in bacteria. The resistant bacteria can be isolated by plating the organisms in a LB agar plate containing the corresponding antibiotic. 5. Adaptive radiation: Darwin\'s finches in Galapagos island is an example of adaptive radiation. It was found that the different species of finches (with different beak structure) arose from an initial seed eating finch, enabling them to become vegetarian and insectivorous finches. Solution Scientific evidences I would give in support of evolution: 1. Paleontological evidence: Fossils are those remains of plants and animals that could not be degraded. Some of them bear resemblances to present day animals and serve as connecting links, eg. Duck billed platypus is connecting link between reptiles and mammals. Some of them represent extinct organisms, eg. dinosaurs. Cross sections of the earth\'s crust from where the fossils have been isolated determines the geological time scale during which the organism existed. 2. Comparative anatomy/physiology: Homologous organs of animals such as forelimb bones of humans, tigers, whales and bats share similar type of arrangement and indicates common ancestry. Such type of evolution is called \"divergent\". Analogous organs are opposite to homologous organs, such as wings of butterfly and birds. Although they perform the same function, they are not anatomically similar. Thus, these are results of \"convergent \"evolution. 3. Similarities in genes and proteins among genetically different o.

Scientific evidences I would give in support of evolution1. Paleo.pdf

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Package scopeis the range of visibility for aparticular element or construct contained in apackage.–“range of visibility” describe whether a packageitem (such as function) can be accessed fromoutside the package or only from other programunits within the same package.–Any elements declared in a package specificationare consideredpublic.–The elements are considered private if they can be called from other program units within the samepackage. Solution Package scopeis the range of visibility for aparticular element or construct contained in apackage.–“range of visibility” describe whether a packageitem (such as function) can be accessed fromoutside the package or only from other programunits within the same package.–Any elements declared in a package specificationare consideredpublic.–The elements are considered private if they can be called from other program units within the samepackage..

Package scopeis the range of visibility for aparticular element or c.pdf

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Point of zero charge related to the adsorption phenomenon I surface chemistry. Here it is the point if achievement of zero charge density on a surface. If pHzpc is 8.5 that means at this pH solid submerged in electrolyte will show zero electrical charge on the surface. P sorption refers to binding of phosphates in other forms to soil particles. At high pH, P availability is low because of reaction of phosphates to calcium which result in formation of apatite. At low pH, phosphate reacts with aluminum, iron etc and results in formation of various other minerals. Solution Point of zero charge related to the adsorption phenomenon I surface chemistry. Here it is the point if achievement of zero charge density on a surface. If pHzpc is 8.5 that means at this pH solid submerged in electrolyte will show zero electrical charge on the surface. P sorption refers to binding of phosphates in other forms to soil particles. At high pH, P availability is low because of reaction of phosphates to calcium which result in formation of apatite. At low pH, phosphate reacts with aluminum, iron etc and results in formation of various other minerals..

Point of zero charge related to the adsorption phenomenon I surface .pdf

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main.cpp #include iostream #include iomanip #include fs.pdf

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loan amount = 17500 present value of annuity = payment per period * [ 1 - (1+i)^-n ]/i => monthly payment *[1-(1+10.4%/12)^-39]/10.4%/12 = 17500 => monthly payment = 530.74 remaining balance = PV of remaining paymens = 530.74 * [1-(1+10.4%/12)^-28]/10.4%/12 = 13144.78 Solution loan amount = 17500 present value of annuity = payment per period * [ 1 - (1+i)^-n ]/i => monthly payment *[1-(1+10.4%/12)^-39]/10.4%/12 = 17500 => monthly payment = 530.74 remaining balance = PV of remaining paymens = 530.74 * [1-(1+10.4%/12)^-28]/10.4%/12 = 13144.78.

loan amount = 17500present value of annuity = payment per period .pdf

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import javax.swing.*; import java.awt.*; import java.awt.event.*; import java.text.*; public class convertTemperature extends JApplet implements ActionListener { JTextField inputField; JLabel label; JRadioButton celcius, fahrenheit; public void init(){ Container contain = getContentPane(); contain.setLayout (new FlowLayout()); inputField = new JTextField(\"\",10); contain.add(inputField); celcius= new JRadioButton (\"to Celcius\", true); contain.add(celcius); fahrenheit = new JRadioButton(\"to fahrenheit\", false); contain.add(fahrenheit); ButtonGroup selection = new ButtonGroup(); selection.add(celcius); selection.add(fahrenheit); JButton button1 = new JButton (\"Show Result\"); button1.addActionListener(this); contain.add(button1); label= new JLabel (\"Enter temperature and choose an option\"); contain.add(label); } public void actionPerformed(ActionEvent e) { DecimalFormat df = new DecimalFormat (\"#.##\"); double input = Double.parseDouble(inputField.getText()); double convertedTemp = 0.0; if (celcius.isSelected()) convertedTemp = ((5.0/9.0)*(input - 32.0)); if (fahrenheit.isSelected()) convertedTemp = ((9.0/5.0)*(input + 32.0)); label.setText(String.valueOf(df.format(convertedTemp))); } } Solution import javax.swing.*; import java.awt.*; import java.awt.event.*; import java.text.*; public class convertTemperature extends JApplet implements ActionListener { JTextField inputField; JLabel label; JRadioButton celcius, fahrenheit; public void init(){ Container contain = getContentPane(); contain.setLayout (new FlowLayout()); inputField = new JTextField(\"\",10); contain.add(inputField); celcius= new JRadioButton (\"to Celcius\", true); contain.add(celcius); fahrenheit = new JRadioButton(\"to fahrenheit\", false); contain.add(fahrenheit); ButtonGroup selection = new ButtonGroup(); selection.add(celcius); selection.add(fahrenheit); JButton button1 = new JButton (\"Show Result\"); button1.addActionListener(this); contain.add(button1); label= new JLabel (\"Enter temperature and choose an option\"); contain.add(label); } public void actionPerformed(ActionEvent e) { DecimalFormat df = new DecimalFormat (\"#.##\"); double input = Double.parseDouble(inputField.getText()); double convertedTemp = 0.0; if (celcius.isSelected()) convertedTemp = ((5.0/9.0)*(input - 32.0)); if (fahrenheit.isSelected()) convertedTemp = ((9.0/5.0)*(input + 32.0)); label.setText(String.valueOf(df.format(convertedTemp))); } }.

import javax.swing.; import java.awt.; import java.awt.event..pdf

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Hg is an atom , because an atom means that a smallest individual spece and it cannot be undivisible further. H+ is an ion, it is a spece was formed by the H atom after lost one electron from it. H2O is a molecule ,and it has obtained by the reaction of H2 atoms and O atoms. O2 and N2 , Ar and CO2 both are called as mixures becaues those were formed by the mixing from two individual species or molecules. NaCl is a compound , it has the fixed proportion of both Na and Cl in its compound. Solution Hg is an atom , because an atom means that a smallest individual spece and it cannot be undivisible further. H+ is an ion, it is a spece was formed by the H atom after lost one electron from it. H2O is a molecule ,and it has obtained by the reaction of H2 atoms and O atoms. O2 and N2 , Ar and CO2 both are called as mixures becaues those were formed by the mixing from two individual species or molecules. NaCl is a compound , it has the fixed proportion of both Na and Cl in its compound..

Hg is an atom , because an atom means that a smallest individual spe.pdf

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Hello! Please see the correction below! False: Hydrogen atoms are often the central atom of a lewis structure. (This doesn\'t make sense since hydrogen can only form 1 bond!) True: For a neutral molecule, the number of electrons in the Lewis structure is the sum of the valence electrons for the atoms. FALSE: Each atom of a Lewis structure must have eight electrons. (This is the \"octet rule.) Typically you would have each atom with eight electrons, but as your hint indicated, Hydrogen only gets 2 electrons. This exception would make this statement false. True: Electrons of covalent compounds may be shared between atoms. Sorry about that - Hope this helps! Please rate if it does : ) Solution Hello! Please see the correction below! False: Hydrogen atoms are often the central atom of a lewis structure. (This doesn\'t make sense since hydrogen can only form 1 bond!) True: For a neutral molecule, the number of electrons in the Lewis structure is the sum of the valence electrons for the atoms. FALSE: Each atom of a Lewis structure must have eight electrons. (This is the \"octet rule.) Typically you would have each atom with eight electrons, but as your hint indicated, Hydrogen only gets 2 electrons. This exception would make this statement false. True: Electrons of covalent compounds may be shared between atoms. Sorry about that - Hope this helps! Please rate if it does : ).

Hello! Please see the correction below! False Hydrogen atoms ar.pdf

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From LLMm has possible four gamates those are LM ,Lm,LM and Lm in meiosis 1.then it will be doubled in meiosis2. But when the secondary non disjunction occurs at M locus that means than gene is not segregated during meiosis 2. Then you get four gametes for L gene after meiosis 2 and in case of M gene you will have M and m allele which will not segregated in meiosis 2. So you will have Four gametes. (LM, Lm, LmM and L). Solution From LLMm has possible four gamates those are LM ,Lm,LM and Lm in meiosis 1.then it will be doubled in meiosis2. But when the secondary non disjunction occurs at M locus that means than gene is not segregated during meiosis 2. Then you get four gametes for L gene after meiosis 2 and in case of M gene you will have M and m allele which will not segregated in meiosis 2. So you will have Four gametes. (LM, Lm, LmM and L)..

From LLMm has possible four gamates those are LM ,Lm,LM and Lm in me.pdf

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D is correct. Formatting and code conversion are Presentation layer functions in the OSI model. The Application layer in TCP/IP maps to the top three layers of the OSI model. A, B, and C are incorrect. Solution D is correct. Formatting and code conversion are Presentation layer functions in the OSI model. The Application layer in TCP/IP maps to the top three layers of the OSI model. A, B, and C are incorrect..

D is correct. Formatting and code conversion are Presentation layer .pdf

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