In repressive operon if RNA polymerase cannot bind to the promoter, .pdf

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In repressive operon if RNA polymerase cannot bind to the promoter, genes will not be transcribed. And this will hamper further transcription of the genes. Solution In repressive operon if RNA polymerase cannot bind to the promoter, genes will not be transcribed. And this will hamper further transcription of the genes..

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The 3 main important things required for migration or transition of data from old system to new system are as follows: 1) Conversion Plan 2) Testing Plan 3) User Training and Documentation let me discuss in brief about this 3 plans. 1) Conversion Plan - This plan name itself suggets migration or transfer. Here we use different softwares and data conversion softwares that are required to change or migrate the data from old system to the new system. While transfering data there may be lose of data or hacking of data. So, Inorder to avoid this kind of problems we use different softwares that are required for transfer of data and also for encrypting the data inorder to make its size little small for a flexible transfer of data. For this transfer of data from one system to another system requires a sync between the system which is possible through the installed softwares. 2) Testing is mainly the crucial part of this three scenarios. testing consists of 3 categories which include a) unit testing b) integration testing c) Acceptance testing a) Unit testing : Inorder to test the transfered data from old system to new system testing is required.Unit testing is used to find the bugs or defects in the transfered data. b) Integration Testing: When individual software modules are merged and tested as a group than it is known as integration testing. Integration testing is sets between Unit Testing and System Testing. Integration testing is executed to establish whether the components interact with each other consort to the specification or not. Integration testing in large refers to joining all the components resulting in the complete system. It is further performed by the developer or the software Tester or by both. Example- checking that a Payroll system interacts as required with the Human Resource system. c) Acceptance Testing : After the system test has corrected all or most defects, the system will be delivered to the user or customer for acceptance testing. Acceptance testing is basically done by the user or customer although other stakeholders may be involved as well. The goal of acceptance testing is to establish confidence in the system. Acceptance testing is most often focused on a validation type testing. Acceptance testing may occur at more than just a single level, for example: A Commercial Off the shelf (COTS) software product may be acceptance tested when it is installed or integrated. Acceptance testing of the usability of the component may be done during component testing. Acceptance testing of a new functional enhancement may come before system testing. 3) User Training and Documentation : This is the most important part of this whole process. a docment should be maintained with this changes and what all are done for future reference. and User training is the mandatory part of this for the basic knowledge of using the system. Solution The 3 main important things required for migration or transition of data from old system to new system .

The 3 main important things required for migration or transition of .pdf

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Solution : public class Figure2 { public static void main(String args[]) { int i,j,k; for(i=1;i<=12;i++) { System.out.println(); for(j=12;j>=i;j--) System.out.print(\" \"); if(i==1) System.out.print(\"+-----------------------------+\"); else if(i<=11) { System.out.print(\"/\"); for(j=12;j>i+1;j--) System.out.print(\" \"); for(k=i-1;k>=1;k--) System.out.print(\"__/\"); } else if(i==12) System.out.print(\"+-----------------------------+\"); for(k=i-2;k>=1;k--) System.out.print(\"/\"); } System.out.println(); for(i=1;i<=5;i++) { System.out.print(\"| Building Java Programs |\"); for(j=5;j>=i;j--) System.out.print(\"//\"); System.out.println(); } System.out.print(\"+-----------------------------+\"); } }.

Solution public class Figure2 { public static void main(Str.pdf

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Ques-1: Answer: Thinking of a mumps infection how do the symptoms involve the GI tract yet lead to airborne transmission of the disease? why isn\'t disease transmitted by fecal oral contamination instead Mumps virus is a negative-sense RNA virus and it can infect parotid glands and gastrointestinal tract from mouth to respiratory organs. This agent can cause viral gastroenteritis and contagious through respiratory droplets from person to person. Coughing and sneezing include the respiratory droplets to transmit a fresh host though air bone transmission (the droplets aerosolize) of the disease. This infectious resides inside the cells of respiratory and sputum droplets coming out of mouth. It has no vector to transmit from person to person through fecal-oral contamination but it can transmit if a person to touch oral another viral infected person. Fecal contamination potentially requires a vector to transmit. Solution Ques-1: Answer: Thinking of a mumps infection how do the symptoms involve the GI tract yet lead to airborne transmission of the disease? why isn\'t disease transmitted by fecal oral contamination instead Mumps virus is a negative-sense RNA virus and it can infect parotid glands and gastrointestinal tract from mouth to respiratory organs. This agent can cause viral gastroenteritis and contagious through respiratory droplets from person to person. Coughing and sneezing include the respiratory droplets to transmit a fresh host though air bone transmission (the droplets aerosolize) of the disease. This infectious resides inside the cells of respiratory and sputum droplets coming out of mouth. It has no vector to transmit from person to person through fecal-oral contamination but it can transmit if a person to touch oral another viral infected person. Fecal contamination potentially requires a vector to transmit..

Ques-1AnswerThinking of a mumps infection how do the symptoms .pdf

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POCl3Phosphorus chloride oxideSolutionPOCl3Phosphorus .pdf

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Option 4 is correctSince Si2 is a molecule.Its bonding is Si -.pdf

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m=1Solutionm=1.pdf

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What we\'re looking for here are salts in which the anion will not accept protons to form a stable molecular species. That is, a compound will dissolve in dilute HCl if the reaction: X- + H+ ------> HX occurs to some extent. But this is just an acid association reaction (the reverse of the more common acid dissociation reaction). Thus, what we want are acids that do not fully ionize, or in other words, salts of weak acids. The strong acids are: HNO3, HClO4, HI, HBr, HCl, and H2SO4. So, salts of NO3-, ClO4-, I-, Br-, Cl-, and HSO4- will not dissolve with dilute HCl, others will (at least to some degree). Looking at the salkts you have listed, that means that: AgNO2, Hg2SO4, AlPO4, NiCO3, Cu2S, and CaF2 will dissolve (whole or in part) in dilute HCl. Solution What we\'re looking for here are salts in which the anion will not accept protons to form a stable molecular species. That is, a compound will dissolve in dilute HCl if the reaction: X- + H+ ------> HX occurs to some extent. But this is just an acid association reaction (the reverse of the more common acid dissociation reaction). Thus, what we want are acids that do not fully ionize, or in other words, salts of weak acids. The strong acids are: HNO3, HClO4, HI, HBr, HCl, and H2SO4. So, salts of NO3-, ClO4-, I-, Br-, Cl-, and HSO4- will not dissolve with dilute HCl, others will (at least to some degree). Looking at the salkts you have listed, that means that: AgNO2, Hg2SO4, AlPO4, NiCO3, Cu2S, and CaF2 will dissolve (whole or in part) in dilute HCl..

What were looking for here are salts in which the anion will not a.pdf

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Step-1: Calculation of Avilable assets and liabilities: Cash 75000 Inventory 140000 Building 155000 370000 Accounts payable 205000 Note Payable 105000 Secured by Inventory Liabilities 35000 With priority Bond Payable 200000 Secured by Building 545000 Step-2: Determination od payment plan: Stage of Payment in case of bankruptcy 1 Liabilities with Priority 2 Secured Debt 3 Unsecured debt Step-3: Payment of debt with priority and secured debt Paid from Debt Cash Balance Debt Closing Cash Liabilities With Priority 35000 35000 0 40000 Paid from realisation of Inventory Closing Amount of Inventory Notes Payable 105000 105000 0 35000 Paid from realisation of Building Closing Amount of Building Bonds Payable 200000 155000 45000 0 Step-4:Calculation of remaining unsecured debt and Assets Unsecured Liabilities: Bonds Payable 45000 Accounts Payable 205000 Total 250000 Remaining Assets: Cash 40000 Realisation from Inventory 35000 Step-5: Calculation of proprtionate distribution: Proportionate distribution: Bonds Payable 75000*45/250 13500 Accounts Payable 75000*205/250 61500 75000 Step-6: Calculation of amount received by bond holders Payment to Bonds Holder: Secured part 155000 Unsecured Part 13500 Total 168500 Thus, holders of bond can expect tp receive $ 168,500 Solution Step-1: Calculation of Avilable assets and liabilities: Cash 75000 Inventory 140000 Building 155000 370000 Accounts payable 205000 Note Payable 105000 Secured by Inventory Liabilities 35000 With priority Bond Payable 200000 Secured by Building 545000 Step-2: Determination od payment plan: Stage of Payment in case of bankruptcy 1 Liabilities with Priority 2 Secured Debt 3 Unsecured debt Step-3: Payment of debt with priority and secured debt Paid from Debt Cash Balance Debt Closing Cash Liabilities With Priority 35000 35000 0 40000 Paid from realisation of Inventory Closing Amount of Inventory Notes Payable 105000 105000 0 35000 Paid from realisation of Building Closing Amount of Building Bonds Payable 200000 155000 45000 0 Step-4:Calculation of remaining unsecured debt and Assets Unsecured Liabilities: Bonds Payable 45000 Accounts Payable 205000 Total 250000 Remaining Assets: Cash 40000 Realisation from Inventory 35000 Step-5: Calculation of proprtionate distribution: Proportionate distribution: Bonds Payable 75000*45/250 13500 Accounts Payable 75000*205/250 61500 75000 Step-6: Calculation of amount received by bond holders Payment to Bonds Holder: Secured part 155000 Unsecured Part 13500 Total 168500 Thus, holders of bond can expect tp receive $ 168,500.

Step-1 Calculation of Avilable assets and liabilities Cash 750.pdf

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Here the required number of ways is 128SolutionHere the requir.pdf

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From the incorrectpicture shown, simply move the ethyl group inbetwe.pdf

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All of the aboveSolutionAll of the above.pdf

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A. The ATH command is the Hayes-compatible command to hang up an existing call in progress. The Hayes command set was introduced by Hayes Communications many years ago and is the de facto standard for modem communications and protocol. Solution A. The ATH command is the Hayes-compatible command to hang up an existing call in progress. The Hayes command set was introduced by Hayes Communications many years ago and is the de facto standard for modem communications and protocol..

A. The ATH command is the Hayes-compatible command to hangup an ex.pdf

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a) Lysosomal proteins Lysosomal proteins as they enter the cis Golgi are phosphorylated on the specific mannose sugars attached to them leading to the formation of mannose-6- phosphates on the lysosomal proteins. The mannose-6-phosphate is identified by a receptor protein in the trans Golgi which binds to the lysosomal proteins. The transport vesicles bud from the trans Golgi network in order to transport the lysosomal proteins to late endosomes which later form lysosomes. b) Membrane proteins Membrane proteins are retrieved from the trans Golgi by endocytosis and then transported to the correct domain by transcytosis. The cytoplasmic sorting signals guide membrane proteins selectively to the basolateral plasma membrane. c) ER resident proteins ER resident proteins are transported from the cis end of the Golgi complex back to the rough endoplasmic reticulum with the help of a protein known as COPI. COPI is a protein complex that coats the vesicles transporting ER resident proteins. Solution a) Lysosomal proteins Lysosomal proteins as they enter the cis Golgi are phosphorylated on the specific mannose sugars attached to them leading to the formation of mannose-6- phosphates on the lysosomal proteins. The mannose-6-phosphate is identified by a receptor protein in the trans Golgi which binds to the lysosomal proteins. The transport vesicles bud from the trans Golgi network in order to transport the lysosomal proteins to late endosomes which later form lysosomes. b) Membrane proteins Membrane proteins are retrieved from the trans Golgi by endocytosis and then transported to the correct domain by transcytosis. The cytoplasmic sorting signals guide membrane proteins selectively to the basolateral plasma membrane. c) ER resident proteins ER resident proteins are transported from the cis end of the Golgi complex back to the rough endoplasmic reticulum with the help of a protein known as COPI. COPI is a protein complex that coats the vesicles transporting ER resident proteins..

a) Lysosomal proteinsLysosomal proteins as they enter the cis Golg.pdf

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A language can be seen as a system suitable for expression of certain ideas, facts and concepts. For formalizing the notion of a language one must cover all the varieties of languages such as natural (human) languages and programming languages. Let us look at some common features across the languages. One may broadly see that a language is a collection of sentences; a sentence is a sequence of words; and a word is a combination of syllables. If one considers a language that has a script, then it can be observed that a word is a sequence of symbols of its underlying alphabet. It is observed that a formal learning of a language has the following three steps. 1. Learning its alphabet - the symbols that are used in the language. 2. Its words - as various sequences of symbols of its alphabet. 3. Formation of sentences - sequence of various words that follow certain rules of the language. In this learning, step 3 is the most difficult part. Let us postpone to discuss construction of sentences and concentrate on steps 1 and 2. For the time being instead of completely ignoring about sentences one may look at the common features of a word and a sentence to agree upon both are just sequence of some symbols of the underlying alphabet. For example, the English sentence \"The English articles - a, an and the - are categorized into two types: indefinite and definite.\" may be treated as a sequence of symbols from the Roman alphabet along with enough punctuation marks such as comma, full-stop, colon and further one more special symbol, namely blank-space which is used to separate two words. Thus, abstractly, a sentence or a word may be interchangeably used 4 for a sequence of symbols from an alphabet. With this discussion we start with the basic definitions of alphabets and strings and then we introduce the notion of language formally. Further, in this chapter, we introduce some of the operations on languages and discuss algebraic properties of languages with respect to those operations. We end the chapter with an introduction to finite representation of languages via regular expressions. Solution A language can be seen as a system suitable for expression of certain ideas, facts and concepts. For formalizing the notion of a language one must cover all the varieties of languages such as natural (human) languages and programming languages. Let us look at some common features across the languages. One may broadly see that a language is a collection of sentences; a sentence is a sequence of words; and a word is a combination of syllables. If one considers a language that has a script, then it can be observed that a word is a sequence of symbols of its underlying alphabet. It is observed that a formal learning of a language has the following three steps. 1. Learning its alphabet - the symbols that are used in the language. 2. Its words - as various sequences of symbols of its alphabet. 3. Formation of sentences - sequence of various words that follow certain rul.

A language can be seen as a system suitable for expression of certai.pdf

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1. RbI Rubidium iodide2. Cs2SO4 Cesium sulfate3. Al2S3 Alumin.pdf

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1) Yes. Hazardous wastes can be liquids, solids, contained gases, or sludges. 2) No. Under RCRA, the term \"solid waste\" means any waste, whether it is a solid, semisolid, or liquid. It is also regulated by RCRA. 3) RCRA Subtitle D focuses on the the primary planning, regulating, implementing, and enforcement entities for the management of nonhazardous solid waste, It regulates management of non hazardous waste. Solution 1) Yes. Hazardous wastes can be liquids, solids, contained gases, or sludges. 2) No. Under RCRA, the term \"solid waste\" means any waste, whether it is a solid, semisolid, or liquid. It is also regulated by RCRA. 3) RCRA Subtitle D focuses on the the primary planning, regulating, implementing, and enforcement entities for the management of nonhazardous solid waste, It regulates management of non hazardous waste..

1) Yes. Hazardous wastes can be liquids, solids, contained gases, or.pdf

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