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use properties of logarithms to expand the logarithmic expression as much as possilbe. Evaluate logarithmic expressions without using a calculator if possible. ln(e^5/5)= Solution Given : ln {(e^5)/5} To expand we use following log properties: ln(x*y/z) = lnx +lny - lnz So, ln {(e^5)/5} = ln(e^5) -ln(5) Now we use the property : ln(a^x) = x*lna So, ln {(e^5)/5} = ln(e^5) -ln(5) = 5ln(e) - ln(5) Now lne = lnee = lne/lne =1 So,ln {(e^5)/5} = ln(e^5) -ln(5) = 5ln(e) - ln(5) = 5 - ln5 (simplified form ).

$use properties of logarithms to expand the logarithmic expression as much as possilbe. Evaluate logarithmic expressions without using a calculator if possible. ln(e^5/5)= Solution Given : ln {(e^5)/5} To expand we use following log properties: ln(x*y/z) = lnx +lny - lnz So, ln {(e^5)/5} = ln(e^5) -ln(5) Now we use the property : ln(a^x) = x*lna So, ln {(e^5)/5} = ln(e^5) -ln(5) = 5ln(e) - ln(5) Now lne = lnee = lne/lne =1 So,ln {(e^5)/5} = ln(e^5) -ln(5) = 5ln(e) - ln(5) = 5 - ln5 (simplified form )$

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Describe the purpose for carrying out a DNAase Hypersensitivity assay. How do you analyze and interperet the DNAse Hypersensitivity assay? Solution DNase I hypersensitivity analysis is a powerful method to analyze chromatin structure and identify genomic regulatory elements. in this technique Euchromatin regions are associated with active form (open form so more accesible to various DNAase) and can be identified by their hypersensitivity to nuclease digestion. Hypersensitive sites can be identified by DNase hypersensitivity assays, in which intact, isolated nuclei are subjected to limited digestion with the enzyme DNase I. DNase I cleavage sites are identified by restriction digestion followed by Southern blotting using radiolabeled genomic fragments to probe a specific genomic region of interest. so in undigested lane you would be able to see only one band where as in digested one more than one band. now a days this technique combined with second genration sequencing for better understanding..

Describe the purpose for carrying out a DNAase Hypersensitivity assa.pdf

Describe the purpose for carrying out a DNAase Hypersensitivity assa.pdf

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compare and contrast the morphology of Cestodes, Trematodes, and Nematodes. Solution Feature Nematodes (roundworms) Cestodes (tapeworms) trematodes (flukes) Shape thread-like, cylindrical bodies; elongate, tapered at both ends Elongated, flat, ribbon like body Dorsoventrally flattened, oval; analogous to tree leaf Segmented No Yes with segments called proglottids No Symmetry bilaterally symmetrical bilaterally symmetrical bilaterally symmetrical length from microscopic to nearly 1 m long Few mm to serval meters Few cm to 9 cm Head end Suckers & hooks absent. Well-developed buccal cavity Suckers generally with hooks Suckers with no hooks Alimentary Canal Present; Complete with both mouth and an_s Absent Incomplete; no an_s Body Cavity Present (pseudocoeloms) Absent Absent Sexes Mostly dioecious hermaphroditic Hermaphroditic except Schisto Feature Nematodes (roundworms) Cestodes (tapeworms) trematodes (flukes) Shape thread-like, cylindrical bodies; elongate, tapered at both ends Elongated, flat, ribbon like body Dorsoventrally flattened, oval; analogous to tree leaf Segmented No Yes with segments called proglottids No Symmetry bilaterally symmetrical bilaterally symmetrical bilaterally symmetrical length from microscopic to nearly 1 m long Few mm to serval meters Few cm to 9 cm Head end Suckers & hooks absent. Well-developed buccal cavity Suckers generally with hooks Suckers with no hooks Alimentary Canal Present; Complete with both mouth and an_s Absent Incomplete; no an_s Body Cavity Present (pseudocoeloms) Absent Absent Sexes Mostly dioecious hermaphroditic Hermaphroditic except Schisto.

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compare and contrast the morphology of Cestodes, Trematodes, and Nem.pdf

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Carbon dioxide transport Drag each label to the appropriate location.pdf

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Carbon dioxide transport Drag each label to the appropriate location.pdf

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An important part of electrical engineering is PCB design. One important part of PCB design when using microcontrollers is determining which pin package to use. Develop a MATLAB function that requests input on whether to use an LQFP100, LQFP144, or LQFP176 pin package and outputs the corresponding pin package. The MATLAB function should receive an input of 1, 2, or 3 to determine the respective pin package. Think of it as a label maker with a dial input. It should be made sure that the full strings are returned and outputted to the console. Use a method other than the switch/case or if/else structures in order to develop this solution (i.e. build a matrix). Answer the following questions: 1) In this context, explain the terminology ‘map’. 2) What is a good way of storing such a map in MATLAB? 3) How would you change the function to use 0, 1, and 2 as inputs instead of 1, 2, and 3? 4) How does MATLAB differ from Python when using arrays / lists? Solution prompt = \'Enter values 1 for LQFP100 2 for LQFP144 and 3 for LQFP176 \'; c = {\'LQFP100\', \'LQFP144\', \'LQFP176\'}; x = input(prompt) c(1,x) A Map object is a data structure that allows you to retrieve values using a corresponding key. Keys can be real numbers or character vectors and provide more flexibility for data access than array indices, which must be positive integers. Values can be scalar or nonscalar arrays. Construction mapObj = containers.Map constructs an empty Map container mapObj. mapObj = containers.Map(keySet,valueSet) constructs a Map that contains one or more values and a unique key for each value. mapObj = containers.Map(keySet,valueSet,\'UniformValues\',isUniform) specifies whether all values must be uniform (either all scalars of the same data type, or all character vectors). Possible values for isUniform are logical true (1) or false (0). mapObj = containers.Map(\'KeyType\',kType,\'ValueType\',vType) constructs an empty Map object and sets the KeyType and ValueType properties. The order of the key type and value type argument pairs is not important, but both pairs are required. Input Arguments keySet 1-by-n array that specifies n unique keys for the map. All keys in a Map object are real numeric values or all keys are character vectors. If n > 1 and the keys are character vectors, keySet must be a cell array. The number of keys in keySet must equal the number of values in valueSet. valueSet 1-by-n array of any class that specifies n values for the map. The number of values in valueSet must equal the number of keys in keySet. \'UniformValues\' Parameter character vector to use with the isUniform argument. isUniform Logical value that specifies whether all values are uniform. If isUniform is true (1), all values must be scalars of the same data type, or all values must be character vectors. If isUniform is false (0), then containers.Map sets the ValueType to \'any\'. Default: true for empty Map objects, otherwise determined by the data types of values in valueSet. \'Key.

An important part of electrical engineering is PCB design. One impor.pdf

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Alternate Electron acceptors: Know what these electron acceptors will become after picking up electron. CO_2, SO_42-, Fe3+, NO_3-, NO_2-. Note these electron acceptors are not as efficient as oxygen. The final ATP yield will be lower if alternate electron acceptors are used instead of oxygen. Solution The alternate electron acceptor like CO2 ,SO42- ,FE3+,NO3-,NO2- are known to act as electron acceptors mainly under anaerobic conditions. In these organisms during electron transport chain instead of oxygen as the terminal electron acceptors the anaerobic bacteria use a wide variety of electron and play an important role in the processes of nitrification. As these molecules have a lesser redox potential than O2 they yield lesser ATP . Certain methylotropic bacteria use C1 compounds like methanol, formaldehyde ,formate and methylamines and completely oxidize them using the serine pathway. Acetogenesis is the process here in co2 and hydrogen are used to result in acetete . Similar to methanogenesis here also hydrogen is used as an electron donor while co2 is used a s an electron acceptor. De nitrification is the common process of fixing nitrogen by many nitrogenous bacteria . They use NO3-,NO2-, NO, N2 O to finally produce di nitrogen but some of the intermediate products that are formed like the N2 O contribute towards the increase n the green house gases and acid rain and global warming.many of the denitrifying bacteria also use other sources like Fe3 as electron acceptors. The de nitrification process is extensively used in sewage and waste water treatment. sulfate reduction SO42-Many gram negative bacteria uses sulphur as the terminal electron acceptor and produce hydrogen sulfide as the metabolic end product.These organisms use lactate or pyruvate as the electron donor. Some of the autotrophic sulfate reducing bacteria like Desulfotignum phosphitoxidans use phosphite HPO3- as electron donor while others like D.thiozymogenens are capable of sulphur disproprtination i.e spilliting of sulphur molecule and produce hydrogen sulpfide and sulfate. FE3+ or ferric ion acts as a terminal electron acceptor in many of the autotrophic and heterotrophic organisms this is similar tp normal electron transport chain but for the presence of ferric iron reductase as the final enzyme and the reaction results in nitrate .Some of the ferric reducing bacteria like Shewanella putrefaciens and Geobacter metallireducens can act upon toxic hydrocarbons like toluene they are used in decontaminating iron rich aquifers..

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A man with blood type A and a woman with blood type B have three children: a daughter with type B, and two sons, one with type AB and one with type O. What are the genotypes of the parents? Solution Answer: Human blood group is determined by multiple allele system, in which IA and IB are co-dominant to each other and dominant over IO. In the given problem, the possible genotype of father is IAIO and the possible genotype of the mother is, IBIO. This is because they had children of AB blood group and O blood group. IAIO* IBIO -à IAIO (1/4, blood group A), IAIB (1/4, blood group AB), IBIO (1/4, blood group B), IOIO (1/4, blood group O)..

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You have been exposed to each of the 8 microbes below. After a couple of days you develop the symptomes listed below. What microbe do you think has infected your (select all that apply). High fever up to 105 F Coughing out greenish, yellow, or bloody mucus Chills that make you shake Feeling like you can’t catch your breath, especially when you move around a lot Feeling very tired Low appetite Sharp or stabby chest pain (you might feel it more when you cough or take a deep breath) sweating a lot Fast breathing and heartbeat Lips and fingernails turning blue Question 3 options: Burkholderia mallei Yellow Fever virus Rickettsia rickettsii Epstein Barr Virus Rhizobium spp. Klebsiella pneumoniae Helicobacter pylori Aliivibrio fischeri Burkholderia mallei Yellow Fever virus Rickettsia rickettsii Epstein Barr Virus Rhizobium spp. Klebsiella pneumoniae Helicobacter pylori Aliivibrio fischeri Solution Yellow Fever virus: Yellow fever gets its name from two of its most obvious symptoms: fever and yellowing of the skin. The yellowing occurs because the disease causes liver damage, hepatitis. For some people, yellow fever has no initial symptoms, while for others, the first symptoms appear from three to six days after exposure to the virus from a mosquito bite. An infection with yellow fever typically has three phases. The first phase of symptoms can last for three to four days and then, for most people, disappears. The first phase is generally non- specific and cannot be distinguished from other viral infections. The initial symptoms of yellow fever are: Burkholderia pseudomallei: Burkholderia pseudomallei and Burkholderia mallei cause the diseases melioidosis and glanders, respectively. A well-studied aspect of pathogenesis by these closely-related bacteria is their ability to invade and multiply within eukaryotic cells. In contrast, the means by which B. pseudomallei and B. mallei adhere to cells are poorly defined..

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Which of the four levels of measurement is most appropriate for \"Weights of tennis players\"? ordinal ratio interval nominal Which of the four levels of measurement is most appropriate for \"Body temperatures of tennis players\"? nominal ordinal interval ratio Solution Weights are considered as RATIO Level of Measurement . In a Ratio scale , its zero position indicates the absence of the quantity being measured . Body Temperature are considered as INTERVAL Level of Measurement . Interval scales are numerical scales in which intervals have the same interpretation throughout as temperature difference between two values indicate the same magnitude of difference Note : If the Body Temperature is given in Kelvins we have to use RATIO scales.

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Which statement below regarding membrane structure and membrane transport is true? Increasing the proportion of unsaturated fatty acids of membrane phospholipids would be most likely to occur as environmental temperatures become warmer. Cyclization of membrane phospholipidfatty acids (like cyclopropane) would confer more rigid, straighter hydrocarbon chains Membrane lipids of by hyperthermophilicarchaebacteriaare of the ester-linked fatty acid type also found in non-archaebacteria and eukaryotemembranes, respectively. Cell wall-containing bacteria and archaea tend to possess a slightly hypotonic interior to create a positive turgor pressure thereby helping it to maintain cell shape and form. Group translocationis a transport mechanism that moves molecules via facilitated diffusion.A. Increasing the proportion of unsaturated fatty acids of membrane phospholipids would be most likely to occur as environmental temperatures become warmer.B. Cyclization of membrane phospholipidfatty acids (like cyclopropane) would confer more rigid, straighter hydrocarbon chainsC. Membrane lipids of by hyperthermophilicarchaebacteriaare of the ester-linked fatty acid type also found in non-archaebacteria and eukaryotemembranes, respectively.D. Cell wall-containing bacteria and archaea tend to possess a slightly hypotonic interior to create a positive turgor pressure thereby helping it to maintain cell shape and form.E. Group translocationis a transport mechanism that moves molecules via facilitated diffusion. Solution The changes in the side chains of fatty acids takes place via cyclization, it helps in rigidifying of the cell membranes that limits passive diffusion of small molecules. Therefore, the most appropriate option is B) Cyclization of membrane phospholipid fatty acids (like cyclopropane) would confer more rigid, straighter hydrocarbon chains.

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Why do cyanobacteria possess heterocysts? A- Heterocysts shield the organism\'s nitrogen fixation enzymes from oxygen B-Heterocysts serve as a primitive nucleus in these bacteria C- Bacterial photosynthetic pigments are located in heterocysts D-Heterocysts are reproductive structures similar to endospores in other bacteria Solution Heterocysts are nitrogen fixing cells formed during nitrogen starvation in some cyanobacteria, hence option A is answer..

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Which of the following is/are true regarding hex digits? A. Hex digits are made of 4 bits. B. Hex digits are made of 4 bytes. C. Hex can be expressed as 0–9 and A–G. D. Hex can be expressed as 0–9 and A–F. Solution A and D are correct. Hex digits are 4 bits in length and can be manipulated to display the alphanumeric characters 0–9 or A–F. B and C are incorrect. B is incorrect because hex digits are made of 4 bits, not 4 bytes. C is incorrect because hex digits can only represent characters up to F..

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which is true of the data shown in the histogram? I. The distribution is approximately symmetric. II. The mean and median are approximately equal. III. The median and IQR summarize the data better than the mean and standard deviation. A. I and II B. I and III C. III only D. I, II, and III E. I only Solution I. The distribution is approximately symmetric. II. The mean and median are approximately equal. We have 3 beccause, it symmetric by lookinng it for both right, left at hirarchy level Mean, Median are centered, and equal.

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what steps can be taken to prevent a core melt down when the turbine housing ruptures at a power plant what steps can be taken to prevent a core melt down when the turbine housing ruptures at a power plant what steps can be taken to prevent a core melt down when the turbine housing ruptures at a power plant Solution The prevention of a core melt down begins in the design stage of a nuclear power plant. Step 1 is the threat assessment for where the reactor is going to be operated. Seismic stability, tsunami, weather, and available water and power supplies are all evaluated and a probability of a destructive event during the plant\'s lifetime are all considered. Evey attempt is made to make these predictions as accurate as possible, but as Fukushima has shown, Mother Nature really is a mother, and can trump your models in a heartbeat. So the engineers take the data they get, give it the best look they can, add in some prudent \"fudge factors\" and design to that. Fukushima showed that the engineers who do this are a very pessimistic lot. The earthquake that tore up the region did not damage the plants. It was the tsunami, far larger than predicted, that did in the plants. There are two systems or loops, primary and secondary. The primary loop is the water that circulates through the core. The secondary loop is the steam/condensate loop the turns the turbine. The primary loop is the one we are concerned with here. The Step 2 is materials. Nothing goes into the build of the plant without rigorous (some have said insane) quality control checks. Every part of the primary loop, the base metal, the piping, the nuts and bolts, have their entire history tracked. And I mean the entire history, from where the metal came out of the ground to the last delivery on the parts shelf. Step 3 is the design. The design incorporates multiple, redundant methods to keep the core cooled. So the normal method of cooling the core is to circulate water by means of reactor coolant pumps. (each piping system, consisting of the pump and steam generator and piping is called a loop) But what if a pump failed, or there is a leak in the piping? The answer is (at a minimum) there must be a second loop, identical to the first, just in case. Some designs have 3 or four loops. This is the normal operating condition, where the steam plant is the heat sink. But what if the steam plant cannot function? A completely independent system, the Emergency Core Cooling system, is designed to take over the job if needed. But what if that fails? Then there is a design for injecting water directly into the core, using portable pumps (fire trucks) to cool the core. All these designs, plans, and procedures must be drawn, written up, reviewed and accepted by the builder, the customer, and the regulators, before the first part is put together Step 4 are the people who build, install and operate the plant. Every one of them are trained, tested, retrained, qualified and requalified, to insure they kn.

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Use the following code fragments for the next 3 questions. Assume that: . student. Empileves, and Betired are all child classes of Peraen Student, Employee, and Retired are all child classes of Person All four classes have different implementations of a method called getMoney Person p1 = new Person( ); int m1 =p1.getMoney(); // statement 1 Person p2 = new Student (. . .) ; int m2 = p2.getMoney ( ); // statement2 Person p3; if (m2 Solution Given :- Person p1 = new Person ( ); int m1 = p1.getMoney ( ); // statement1 Person p2 = new Student ( ); int m2 = p2.getMoney ( ); // statement2 Person p3; if(m2 < 100000) { p3 = new Employee( ); } else { p3 = new Retired( ); } int m3 = p3.getMoney ( ); // statement3 Question 50 :- statement1 :- Person p1 = new Person ( ); int m1 = p1.getMoney ( ); // statement1 The call to getMoney ( ) in statement1 will invoke the method defined in Person class. Explanation :- Non-private Non-static overridden method is executed from current object class. Question 51 :- statement2 :- Person p2 = new Student ( ); int m2 = p2.getMoney ( ); //statement2 The call to getMoney ( ) in statement2 will invoke the method defined in Student class. Explanation :- Non-private non-static overridden method is executed from current object class..

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Think about immigrant and refugee groups in your community. What is your impression of ways that elders in such families are involved or not involved in the lives of their children and grandchildren? What impact do a common language and shared cultural values appear to have on their interactions? To what extent have cultural differences created family conflicts and reduced the status of elders? Solution The elders who are refugess are not directly in contact with their children and grand children. They don\'t have any kind of family bond with them. Being far away from the families, they loose their bond with them and cannot led a nonrmal life with them. The elderly refugees will be make the most of healthcare and communal facilities in current country instead of returning to their nation state of birth, these services should arrange for augmented cultural and language variety amongst users. This may perhaps involve establishment of interpreters, precise outreach programmes, and socially modified services to come across elderly from varied language, spiritual, and social circumstances. The cultural differences creates the family conflicts when their comes the difference in the thougts of the family members and the elders. they find it difficult to adjust among themselves being diverse in the cultural aspects and the diversed linguistic problems. but the status of the elders should not be reduced as they are still the same elders..

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Think about immigrant and refugee groups in your community. What is .pdf

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This is for subject Microbiology Differentiate between physical and chemical methods used to control microbial growth; and give 5 examples of each type. Solution S.No. Physical methods Chemical methods 1. It is exclude microbes or reduce their number in solution or any non-living material which might come into contact with the individual It application of specific chemical agents which inhibits growth or kill microbe on surface of skin. 2. Filtration : is passing either a solution or glass device trap the microbe Chemical agent ar used on living tissue (antiseptics) and on inanimate (disinfectant) 3. Cold temperature PH of medium 4. Osmotic pressure Type of chemical agent 5. radiation Time of contact.

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This is for subject MicrobiologyDifferentiate between physical and.pdf

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The situation at Good Health Hospital has become a bit overwhelming ever since the outbreak of E. coli in Ward 10 on the second floor. It appears that there are six (6) cases of the disease. To better understand the situation, communication with the CDC has revealed that this particular bacterium can be found in many everyday items. The CDC recommends that investigation begins in the hospital kitchens, since E. coli can be transmitted via contaminated vegetables, as well as delicatessen meats such as salami and other cold cuts. The following table shows the number of cases identified so far: After meeting yesterday with chief administrator Joe Wellborn, it has been decided that the litigation issues with one (1) of the patients may need further information to determine if the patient was already symptomatic with the bacteria prior to admission. Research has indicated that hospitals operating in the Tampa Bay area have been known to have cases of E. coli contamination. This substantiates the need for further investigations by the County Health Department. Due to the severity of the cases, and the potential for further spread, it is recommended that the hospital seek alternate vending services since the present suppliers cannot verify sanitation protocols for their products. In conclusion, the study done at Good Health Hospital has indicated that the E. coli outbreak was in fact caused by spoil food from the cafeteria. As a social science researcher working for a regional hospital, you are placed in charge of contacting the DOH from your home state, as well as the CDC for statistical data on nosocomial diseases that frequently cause illness at your health care facility. Using information from both agencies, evaluate the data on common nosocomial diseases and compile a treatment contingency plan for dealing with the hospital infections. Identify a targeted audience within Good Health Hospital, and prepare an implementation plan based on your hypothetical meeting with the hospital health care administrator. Propose four (4) steps that will be useful in the final implementation plan.CaseAgeGender123M221F315M442F Solution Implementation Plan: 1.Improve communication between state and local health departments in US which should be taken care by CDC. 2.Enhance laboratory capacity 3.Identify the host, environment and behaviors that put people at increased risks for diseases which are infectious. 4.Provide training opportunities in US for infectious disease. 5.Increase the capacity and capability of research for new strains for infectious diseases..

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The situation at Good Health Hospital has become a bit overwhelming .pdf

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ARORACOCKERY2111

The law of large numbers states that as the number of observations .pdf

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ARORACOCKERY2111

The majority of new cancers are due to inherited DNA defects. True False Cancer incidence today is dramatically different compared to 1900. What is the single most important reason for the increase seen in cancer incidence rates today? Pesticide use Increase in number of days\' people work Increased environmental pollution Increase in life expectancy determine(s) how many times the cell can replicate and are, on average, shorter in older people. Chromatin Protein Telomeres Replication Factors Leukemia and lymphomas are cancers in which certain types of, and divide. red blood cells platelets phagocytes uncontrollably grow white blood cells Solution The majority of new cancers are due to inherited DNA defects: False Cancer incidence...rates today ? B. Increased environmental pollution. --------determineds...older people? C. Telomeres Leukemia and lymphomas...grow and devide. D. White blood cells.

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The majority of new cancers are due to inherited DNA defects. True Fa.pdf

The majority of new cancers are due to inherited DNA defects. True Fa.pdf

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