Anatomy Question Please answer them all.A. Jane is blood type A a.pdf

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Anatomy Question: Please answer them all. A. Jane is blood type A and her husband is blood type B. Jane is puzzled because their daughter is type O. Explain how the daughter inherited a blood type that neither of their parents has. B. If Jane and her husband were both type AB, could they have a type O daughter? Explain your answer. C. An archeologist discovers an unopened tomb in Egypt. Inside she finds the mummies of two adults and two children. Inscriptions identify the tow adults and the Chief Scribe and his wife. The inscriptions describe how their son and his best friend drowned in the Nile River when their boat overturned. The mummies are so well preserved that lab technicians can type the blood of the four individuals. Both adults are type B. Child 1 is type A. Child 2 is type O. Which child is not the son? How do you know? Solution A). Human blood group is determined by multiple allele system, in which IA and IB are codominant to each other and dominant over IO. In the given problem, the both the mother is type “A” and her husband\'s blood group is, B. But they had a daughter with O blood group, so her genotype is IOIO. This indicates that both the parents are heterozygous for the blood group. AO* BO --> AB (1/4), AO (1/4), BO (1/4), OO (1/4) Thus, 1/4th of theur children will be AB. B). The genotype of AB blood group is, AB. So, if both the parents are AB, the blood group of their offspring will be, AB*AB ---> AA (1/4), AB (1/2), BB (1/4) Thus, the children with \"O\" group blood are zero..

Anatomy Question: Please answer them all.
A. Jane is blood type A and her husband is blood type B. Jane is puzzled because their daughter
is type O. Explain how the daughter inherited a blood type that neither of their parents has.
B. If Jane and her husband were both type AB, could they have a type O daughter? Explain your
answer.
C. An archeologist discovers an unopened tomb in Egypt. Inside she finds the mummies of two
adults and two children. Inscriptions identify the tow adults and the Chief Scribe and his wife.
The inscriptions describe how their son and his best friend drowned in the Nile River when their
boat overturned. The mummies are so well preserved that lab technicians can type the blood of
the four individuals. Both adults are type B. Child 1 is type A. Child 2 is type O. Which child is
not the son? How do you know?
Solution
A). Human blood group is determined by multiple allele system, in which IA and IB are
codominant to each other and dominant over IO. In the given problem, the both the mother is
type “A” and her husband's blood group is, B. But they had a daughter with O blood group, so
her genotype is IOIO. This indicates that both the parents are heterozygous for the blood group.
AO* BO --> AB (1/4), AO (1/4), BO (1/4), OO (1/4)
Thus, 1/4th of theur children will be AB.
B). The genotype of AB blood group is, AB. So, if both the parents are AB, the blood group of
their offspring will be,
AB*AB ---> AA (1/4), AB (1/2), BB (1/4)
Thus, the children with "O" group blood are zero.

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Page 1“Why Is He Different from Both Parents?” by Liang and Rice NATIONAL CENTER FOR CASE STUDY TEACHING IN SCIENCE by Jun Liang, Science Department, Borough of Manhattan Community College / City University of New York William J. Rice, Simons Electron Microscopy Center, New York Structural Biology Center Why Is He Different from Both Parents? Part I – Human ABO Blood Type “But I already know all this stuff! I’ll do well on the test tomorrow, I promise,” begged Kevin. He really wanted to go out that night with his friend, but his mother had other ideas. “No, absolutely not. You’re going to stay home and study; tomorrow’s test is very important,” his mother replied. Kevin ended up staying home that night, reluctantly. The next day Kevin met his best friend Peter. “How’s it going?” asked Peter. “Not so good. We had a test today. I guess it went all right,” said Kevin. “How’s your stomach doing? Is it still sore?” Peter asked. “Oh, it’s better now. You know, there’s something weird going on. Last time I was in the hospital, neither of my parents could donate blood to me. The doc said they aren’t a match for my blood type.” “They don’t match? What do you mean?” Peter asked. “Well, the doc said I’m type O positive, my mom is type A positive, and my dad is type B negative. We’re all different, and I don’t get it. They’re my parents—I mean, I should inherit my blood type from at least one of them, if not both, right?” “That’s a good point,” Peter agreed. “Well, I’ve got a very cool science teacher, Ms. B. She would know better than us. Let’s ask her. She has a class tomorrow. We can meet her afterward. What do you say?” The next day, Kevein and Peter met Ms. B. and explained Kevin’s puzzle about his family’s blood types. “Ah, good question,” replied Ms. B. “Let me explain how blood types work so that it will help you to understand why your parents couldn’t donate blood to you and how you can all have different blood types in one and the same family.” “The ABO blood group system is one of the most important systems that hospitals use to identify individuals for blood transfusion. ABO stands for the type of carbohydrate that sits on the surface of red blood cells; it’s this carbohydrate that allows cells to recognize themselves. They are also called surface markers. Positive (+) refers to the presence of Rhesus (Rh) factor. Some individuals lack Rh factor, and they are negative (−). Rh factor is a predominant protein surface marker on red blood cells.” “Interesting. What is carbohydrate exactly? The stuff we eat, like bread and sugar?” Peter asked. “Yes. Sugar is a carbohydrate and bread is made of lots of different carbohydrates. Red blood cells may have carbohydrate A on the cell surface, which results in type A blood; but if they have carbohydrate B, then they become type B blood. Some red blood cells have both carbohydrate A and B, which makes them have type AB blood. When neither carbohydrate A nor B is prese ...

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Page 1Why Is He Different from Both Parents” by Liang and Ri.docx

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1 BIOL 102: Lab 9 Simulated ABO and Rh Blood Typing Objectives: After completing this laboratory assignment, students will be able to: • explain the biology of blood typing systems ABO and Rh • explain the genetics of blood types • determine the blood types of several patients Introduction: Before Karl Landsteiner discovered the ABO human blood groups in 1901, it was thought that all blood was the same. This misunderstanding led to fatal blood transfusions. Later, in 1940, Landsteiner was part of a team who discovered another blood group, the Rh blood group system. There are many blood group systems known today, but the ABO and the Rh blood groups are the most important ones used for blood transfusions. The designation Rh is derived from the Rhesus monkey in which the existence of the Rh blood group was discovered. Although all blood is made of the same basic elements, not all blood is alike. In fact, there are eight different common blood types, which are determined by the presence or absence of certain antigens – substances that can trigger an immune response if they are foreign to the body – on the surface of the red blood cells (RBCs also known as erythrocytes). ABO System: The antigens on RBCs are agglutinating antigens or agglutinogens. They have been designated as A and B. Antibodies against antigens A and B begin to build up in the blood plasma shortly after birth. A person normally produces antibodies (agglutinins) against those antigens that are not present on his/her erythrocytes but does not produce antibodies against those antigens that are present on his/her erythrocytes. • A person who is blood type A will have A antigens on the surface of her/his RBCs and will have antibodies against B antigens (anti-B antibodies). See picture below. • A person with blood type B will have B antigens on the surface of her/his RBCs and will have antibodies against antigen A (anti-A antibodies). • A person with blood type O will have neither A nor B antigens on the surface of her/his RBCs and has BOTH anti-A and anti-B antibodies. • A person with blood type AB will have both A and B antigens on the surface of her/his RBCs and has neither anti-A nor anti-B antibodies. The individual’s blood type is based on the antigens (not the antibodies) he/she has. The four blood groups are known as types A, B, AB, and O. Blood type O, characterized by an absence of A and B agglutinogens, is the most common in the United States (45% of the population). Type A is the next in frequency, found in 39% of the population. The incidences of types B and AB are 12% and 4%, respectively. 2 Table 1: The ABO System Blood Type Antigens on RBCs Antibodies in the Blood Can GIVE Blood to Groups: Can RECEIVE Blood from Groups: A A Anti-B A, AB O, A B B Anti-A B, AB O, B AB A and B Neither anti-A nor anti-B AB O, A, B, AB O Neither A nor B Both anti-A.

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3. A Peace Corps physician is stationed in a small area of the Carpathian Mountains. A few of her patients have hemolytic ancmia- -inemia that results from fragile red blood cell membranes that rupture easily. These patieats have a mutation in the gene A K l (adenylate kinase). A K l is on chromosome 9 and is linked to the gene encoding the ABO blood group (locus h ). Here are the genotypes and phenotypes for different blood types (we are not considering the Bombay phenotype): Both I A and I l alleles are dominant over i . The I A and I B alleles are co-dominant. Here is a pedigree with affected individuals indicated in black. The two alleles at the AKl locus are A (dominant, wild type) and a (recessive, mutant). The blood types for this family are also noted on this pedigree. A. What are the two possible ways to write the genotype of individual I-3 at the A K I and I loci with phase information? Remember to use horizontal parallel lines or a " p " 6 indicate the orientation of the pairs of alleles on different bomologs (phase). B. What is the genotype of individual 1.4 at the AKI and I loci? Is there any ambiguity as to the phase of I 4 ? C. What is the genotype of individual II-I at the A K l and / loci? Which alleles of each gene are carried on the chromosome he inherited from his father, and which alleles are carried on the chromosome he inberited from his mother? D. Individuals If-1 and I1-2 just had a baby boy who has blood type B and hemolytic anemia. (Put him on the pedigree above.) Explain how this could happen by drawing a diagram of meiosis in individual II-1. Include drawings of boch chromosome 9 and the sex chromosomes ( X and Y ) . .

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Gregor Mendel used pea plants to study heredity in a series of experiments. Mendel worked by carefully observing and recording traits in successive generations of plants. Knowledge about DNA and chromosomes came later. This lab will apply genetic laws to human inheritance using Punnett squares. Recall that DNA is wound tightly into chromosomes. Cells with only one set of chromosomes, such as sex cells, are haploid . When two haploid cells fuse during fertilization, a diploid zygote with two full sets of chromosomes is formed. Most cells of a mature individual are diploid. Homologous chromosomes have the same genes, but they might have different versions ( alleles ) of those genes. Diploid cells have two alleles for each gene. These alleles might be identical (gene A) or different (gene B). Each gene’s locus is its location on a chromosome. Human traits come through dominant or recessive inheritance. For example, the cystic fibrosis traits carried by a dominant allele are always expressed, even if the recessive gene is present (FF or Ff). The recessive is only expressed when two copies of the recessive allele are present (ff). Mother: Healthy carrier F f Father: Healthy carrier F FF Healthy non-carrier Ff Healthy carrier f Ff Healthy carrier FF Affected Human gender is carried on the X and Y chromosomes. Females are XX and males are XY. Heredity traits such as color blindness, which is the inability to distinguish among some colors, are carried on the X chromosome (X c ). The presence of one normal X C will allow normal vision. In this next set of exercises, you will determine the genotypes of the parents by considering the inheritance patterns of traits in their children. The following is a table of the phenotypes of the family members: Phenotype Alleles Parents Mother Not color blind Freckles Type B blood X c X C Ff I B i Father Color blind No freckles Type A blood X c Y Ff I A i Children Abby Color blind Freckles X c X c Ff or ff Brady Not color blind No freckles X C Y ff Carly Not color blind No freckles X c X C ff Dennis Color blind Freckles X c Y Ff or ff Exercise 1: Color Blindness Using the alleles XC (not color blind) and Xc (color blind), distribute the gametes from each parent to the outside of the Punnett square. Drag and drop the child with the correct phenotype to the box within the Punnett square that has the corresponding genotype that would occur from the fusion of egg and sperm as indicated by your placement of the gametes. Exercise 2: Freckles Freckles are groups of cells on the skin that produce the pigment melanin, often in response to exposure to ultraviolet (UV) light. The gene for freckles is inherited in a dominant/recessive pattern. A person carrying even a single copy of the dominant allele (F) will have freckles. A person who is homozygous recessive (ff) will have no freckles. Using the alleles F (freckles) and f (no freckles), distribute the gametes from each parent to the outside of the Punne.

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The ABO and MN blood groups are given below for four sets of parents (1 to 4) and four children (a to d). Recall that the ABO blood group has three alleles, I^A, I^B, and i. The MN blood group has two codominant alleles, M and N. When all the genetic data is considered together and by the process of elimination you should be able to match each of the four children with the correct set of parents. Solution Beginning from the first set of parents, let us try to determine the genotype of all sets. Parents 1: The father can be IBIB or IBi and the mother can be IAIAor IAi. For the MN group, The father is NN and the mother is MN. So, their children can of A, B, AB or O blood type and of either MN or N blood group. Parents 2: The father can be IAIA or IAi and the mother can be IAIAor IAi. For the MN group, The father is NN and the mother is also NN. So, their children can of A or O blood type and of N blood group. Parents 3: The father can be IAIAor IAi and the mother is IAIB. For the MN group, both the parents are of MN type. So, their children can of A, B, or AB blood type and of either M, N or MN blood group. Parents 4: The father is ii and the mother is IAIAor IAi. For the MN group, both the parents are of MM type. So, their children can of A or O blood type and of M blood group. Amongst the given set of children only one has the N type blood group. So this child (a) definitely belongs to the parents who can have children with only N blood group i.e. Parents 2. For the remaining 3 cases (b, c and d), we want parents with possibility of giving children of M type blood for children (c) and (d). Parents 1 cannot give an offspring with M type blood group, so it has to be matched with child (b) where it can easily give AB and MN blood types. For child (d), Only parents 4 are a match because that is the only set of parents from the remaining parents which can have a child with O blood group. That leaves parents 3 to give birth to child (c) with A blood group and M blood type and it is possible. So, 1- b, 2- a, 3-c, 4-d.

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Write a class called Student that extends the provided Person class. Ensure that your student class is a concrete class. The foo method for a person should display (to standard output) the person\'s age and name in a nice way (e.g, \"Nigel\'s age is 14\"). When a person makes a noise it is their name repeated N times where N is the person\'s age. public abstract class Person implements Noisy Person Class public String name; 3 ublic int age; s public abstract void foo); public interface Noisy 2 String noise) 1 public interface Noisy{ Noisy Interface 1 public interface Comparable 2 i Comparable Interface int compareTo(Type other); 1 public class Date 2 int day, month, year; Date Class public Date(int d, int m, int y){ dayd; month - m; year-y; Solution public class Student extends Person{ Student(String name, int age){ this.name = name; // initialising the student properties this.age = age; } public void foo(){ if(isNoisy()){ //check if person makes noise for(int i=0 ; i< age; i++ ){ System.out.print(\" \"+name+\" \"); } }else{ System.out.println(name+\"\\\'s age is \"+age); } } } // Note: Information about the Noisy class is required..

Write a class called Student that extends the provided Person class..pdf

Write a class called Student that extends the provided Person class..pdf

Write a class called Student that extends the provided Person class..pdf

why NADH and FADH2 are a type of energy currency Solution Energy used by the body for its many activities ultimately comes from the chemical energy in our food. The chemical energy in our food is converted to reducing agents NADH and FADH2 . These reducing agents are then used to make ATP. ATP stores chemical energy ,so that it is available to the body in a readily accessible form. The energy needed by the body is coming from NADH and FADH2, that why they are type of energy currency..

why NADH and FADH2 are a type of energy currencySolutionEnergy.pdf

why NADH and FADH2 are a type of energy currencySolutionEnergy.pdf

why NADH and FADH2 are a type of energy currencySolutionEnergy.pdf

Who is D.L Shell? Solution Hi friend, Donald L. Shell (March 1, 1924 – November 2, 2015) was an American computer scientist who designed the Shell sort sorting algorithm. He acquired his Ph.D. in Mathematics from the University of Cincinnati in 1959, and published the shell sort algorithm in the Communications of the ACM in July that same year. YOu can read on: https://en.wikipedia.org/wiki/Donald_Shell.

Who is D.L ShellSolutionHi friend,Donald L. Shell (March 1,.pdf

Who is D.L ShellSolutionHi friend,Donald L. Shell (March 1,.pdf

Who is D.L ShellSolutionHi friend,Donald L. Shell (March 1,.pdf

Which of the following correctly states the functions of the rough endoplasmic reticulum ? lipid metabolism, cholesterol synthesis, and synthesis of the lipid components of lipoproteins absorption, synthesis, and transport of fats. storage of calcium ions synthesis of the cell\'s membranes Solution As per the given options rough endoplasmic reticulum functions are :- 2) Absorbtion, synthesis and transport of fats & 4) Synthesis of cell membranes Generally, the eucaryotic cell contains two types of endoplasmic reticulum i) Rough Endoplasmic Reticulum. Rough Endoplasmic Reticulum why named as rough means it contains ribosomes on it so, the surface of the reticulum looks rough. The main function of the rough ER is protein synthesis due to the ribosomes. Rough ER through the membrane contact sites contact with the other organells allowing the transfer of lipids and other small molecules. The intestinal cells, pancreatic cells, which involves in the absorption process contains more rough ER why because the proteins synthesized by the Rough ER involved in the intracellular transport of resynthesized lipids.So the rough ER indirectly helps in the absorption process. ii)Smooth Endoplasmic Reticulum:- Due to the absence of ribosomes it looks smooth so named as smooth ER. The main role is lipid metabolism, and synthesis of carbohydrates and steroid harmones and storage of calcium ions So, in the given question the first option(Lipid metabolism, cholesterol synthesis, synthesis of lipid components of lipo proteins) indicates smooth Endoplasmic Reticulum functions 2nd option Absorbtion, synthesis and transport of fats indicates Rough Endoplasmic Reticulum functions 3rd Option Indicates(Storage of Calcium ions) smooth Endoplasmic Reticulum functions 4th Option is synthesis of cell membrane, For this function the main role of rough Endoplasmic reticulum is during the lipid transport the replenishment of new membranes is done by the morphological change of rough endoplasmic reticulum involved..

Which of the following correctly states the functions of the roug.pdf

Which of the following correctly states the functions of the roug.pdf

Which of the following correctly states the functions of the roug.pdf

What is isomerism? How many kinds are there? Define each kind and explain with examples? What is isomerism? How many kinds are there? Define each kind and explain with examples? Solution The process of isomerism involves the isomers .These are basically defined as the molecules which have the same molecular formula but they usually possess a different arrangement of the atoms in space. They are of two types: 1.Constitutional isomers :They are defined as the compounds in which the various atoms are connected or bonded in a different manner .They are also called as structural isomers 2.Stereoisomers:On the other hand ,they are the compounds which differ on the basis of orientation in space.They are of two types a)Enantiomers :They are the compounds which are usually a mirror image of one another and thehese mirror images are also non-superimposable b)Diastereomers :They are isomers which are not mirror images of one another.Moreover these compounds are non-superimposable. They are of two types: cis and trans ->Cis-isomers are the isomers which usually contain atoms which are oriented on the same side of a bond. ->On the other hand the Trans-isomers are the compounds that have atoms oriented on opposite sides of a bond.

What is isomerism How many kinds are there Define each kind and.pdf

What is isomerism How many kinds are there Define each kind and.pdf

What is isomerism How many kinds are there Define each kind and.pdf

What is the definition of the words, and state the difference between them if there is: 1- D-Flip Flop 2- Register 3- Equality 4- Counter 5- Adder 6- Multiplexer What is the definition of the words, and state the difference between them if there is: 1- D-Flip Flop 2- Register 3- Equality 4- Counter 5- Adder 6- Multiplexer What is the definition of the words, and state the difference between them if there is: 1- D-Flip Flop 2- Register 3- Equality 4- Counter 5- Adder 6- Multiplexer Solution 1. The D flip-flop tracks the input, making transitions with match those of the input D. The D stands for \"data\"; this flip-flop stores the value that is on the data line. It can be thought of as a basic memory cell. The Q output always takes on the state of the D input at the moment of a rising clock edge. (or falling edge if the clock input is active low). 2. A register is a group of binary cells suitable for holding binary information. A group of cascaded flip flops used to store related bits of information is known as a register. 4. A counter is a device which stores (and sometimes displays) the number of times a particular event or process has occurred, often in relationship to a clock signal. Each pulse applied to the clock input increments or decrements the number in the counter. 5. An adder, also called summer, is a digital circuit that performs addition of n-bit binary numbers. 6. A multiplexer (or mux) is a device that selects one of several analog or digital input signals and forwards the selected input into a single line. A multiplexer of 2n inputs has n select lines, which are used to select which input line to send to the output..

What is the definition of the words, and state the diffe.pdf

What is the definition of the words, and state the diffe.pdf

What is the definition of the words, and state the diffe.pdf

Two sources of radio waves (call them 1 and 2) are separated by a distance d along the a-axis in the plane, and emit sinusoidal waves of the same wavelength A and frequency f in the N (+y), S (-y), E (+x), and W (-x) directions. Suppose that we want to arrange the sources to broadcast a signal in the E direction, but not the W. Can this be accomplished? If so, how? In not, why? Solution both the sources are identicle and emit in all direction N,S,E,W. by introducing a path difference equal to half wave length the waves will have distructive interference and signal is destroyed but this happens in both the direction E nd W equally. Hence it is not possible to arrnge the sources such that we have signal only in E nd not in W, the only way to achieve this is to shield the source in W direction or put reflectors in W direction so that all W signal is reflected back to E.

Two sources of radio waves (call them 1 and 2) are separated by a dis.pdf

Two sources of radio waves (call them 1 and 2) are separated by a dis.pdf

Two sources of radio waves (call them 1 and 2) are separated by a dis.pdf

Use the information in the table to select the image that represents the structure of B-form DNA. Solution First one represents the B DNA The structure identified by Watson and Crick, is known as B DNA and is believed to be the idealised form of the structure adopted by virtually all DNA in vivo. It is characterized by a helical repeat of 10 bp/turn by the presence of base pairs lying on the helix axis and almost perpendicular to it, and by having well defined, deep major and minor grooves..

Use the information in the table to select the image that represents .pdf

Use the information in the table to select the image that represents .pdf

Use the information in the table to select the image that represents .pdf

Triggers can only be used to update table values. PL/SQL blocks have a section used to declare variables. True False Solution Please find the answers below: 1) False; triggers are not only used to update the table values, but also they are being used in many places like: generating derived column, enforcing referential integrity, imposing security authorizations, etc... 2) True; PL/SQL has a dedicated section where the variables are being declared. The blocks name is DECLARE block, which comes before the actual code of statements which is defined in a BEGIN - END blocks..

Triggers can only be used to update table values. PLSQL blocks have.pdf

Triggers can only be used to update table values. PLSQL blocks have.pdf

Triggers can only be used to update table values. PLSQL blocks have.pdf

TB is a major worldwide disease with new cases arising at an alarming rate, What factors can account for this and what lays behind the trend of multidrug resistant strains of mycobacterium? please cite reference. Solution Tuberculosis is caused by mycobacteriam tuberculosis. TB is one of the most common cause of death due to infection of the mycobacteriam species.various factor have been found to develop resistance against M. tuberculosis. resistance means simply afterwards that drug potency in terms of pharmacological response will be decresed gradually and at certain level it will not active against the strain. There are two types of resistance 1. natural resistance and second is aquired resistance. natural resistance is found on wild strain. multidrug resistance is the phase in which strain is not responsive against primary line therapy of disease. factor which induced resistance gene factor In a recent year study suggest that the patient who shows the presence of HLA -DRB1 13 and HLA-DRB 1 14 has two fold risk to develop resistance against strain.(Park et al.) Incomplete and inadequate mangment Review of literature suggests that the most powerful factor of MDR-TB is a history of treatment of tuberculosis(diagnosis). TB patients get treated with DOTS regimens not only through the Revised National Tuberculosis Control Programme (RNTCP), but also receive treatment from private medical doctors. Irregular, incomplete, inadequate treatment is the commonest mean of acquiring drug resistant due to improper mangment of drug mostly organism develop the resistnacce against drug eg. use of single drug to treat tb is the fetal cause to develope resistance against organisms.it may be due to ignorance. anothe common error in prescription error it may like additions syndromes. Inadequate treatment compliance : the change over form has complied a significant effece in development of drug resistance it was observed that only 44 per cent of the patients receiving short-course treatment (n=2300) and 36 per cent of those receiving standard chemotherapy (n=1000) completed 81 per cent or more of their treatment. Noncompliance with prescribed treatment is often underestimated by the physician and is difficult to predict. On the other hand, certain factors such as psychiatric illness, alcoholism, drug addiction and homelessness do predict noncompliance .The drug defaulter, just like placebo reactor is not a consistent or readily identified person demographic factors such as ,sex, age, marital status,socio-economic status and education levelhave not been found to correlate with the degree of compliance..

TB is a major worldwide disease with new cases arising at an alarmin.pdf

TB is a major worldwide disease with new cases arising at an alarmin.pdf

TB is a major worldwide disease with new cases arising at an alarmin.pdf

7. How did printing technology impact literacy and education? Solution 7. How did printing technology impact literacy and education? Printing technology impacted literacy and education. Printing transformend the world and made very important changes. Previously people don\'t have much access to information. They used to write. So, the books are expensive. In 15th century Johannes Gutenberg invented a printing machine and made it available to world. Not all are rich enough to buy hand written books till 15th century. As books are expensive so does literacy. To learn to read and write, need to refer to books. When this printing machine came to availability, books became cheap and affordable. Then on wards the barrier of high cost had gone. Source of knowledge are hand written papers or books besides people. Information is not so much available. Again because of expensive things. When printing is available news papers came into picture and from then on people had one more source of knowledge which is cheaper. In this way printing technology impact literacy and education..

7. How did printing technology impact literacy and educationSol.pdf

7. How did printing technology impact literacy and educationSol.pdf

7. How did printing technology impact literacy and educationSol.pdf

Question 1////Theories of Law Subject __________ is a collaborative effort of two or more agencies that provides resources, expertise, and information with the goal of maximizing its capabilities to detect, prevent and respond to terrorist activities. State patrol Federal investigations units Fusion centers DHS 2 points Question 2 By 2008, the TSA had supplied about 43,000 __________ in an effort to secure air transportation systems. police officers U.S. coast guard agents border patrol agents security personnel 2 points Question 3 The average life expectancy of a police officer is __________ years old. 62 49 57 40 2 points Question 4 The Department of Justice oversees all of the following agencies EXCEPT: the Federal Bureau of Investigation. the U.S. Marshals Service. the Drug Enforcement Administration. the Financial Crimes Network. 2 points Question 5 The most controversial method of ending job discrimination is the concept of __________. affirmative action reverse discrimination adverse impact double marginality 2 points Question 6 On average, about __________of full-time sworn personnel in a local police department are female. 88.7% 10.2% 2.7% 11.3% 2 points Question 7 The police attempt to prevent crime by trying to create a sense of __________. omniscience omnipresence patrolling fear 2 points Question 8 Klockars defines the basic functions of policing as: dealing with all those problems that may require the use of coercive force. patrolling on foot throughout local neighborhoods. arresting persons for crimes of vandalism. arresting persons for violent offenses. 2 points Question 9 The position of the Patrol Sergeant falls under the __________ in the chain of command from Chief to Patrol officer. Patrol Lieutenant Operations Captain Chief of Police First Lieutenant 2 points Question 10 The Supreme Court ruled that police may stop automobiles based on reasonable suspicion in all of the following ways EXCEPT: a crime was being committed. a traffic violation occurred. a random routine stop. all citizens are subject to the stop. 2 points Question 11 The __________ is considered one of the most important units in investigating terrorism in the U.S. FBI-Joint Terrorist Task Force ATF Department of Homeland Security National Counterterrorism Center 2 points Question 12 Studies have indicated that laws establishing administrative license revocations have helped reduce alcohol-related crashes by __________ percent. 30 40 20 60 2 points Question 13 Which of the following is NOT one of the FBI\'s post-9/11 priorities? To protect civil rights To combat terrorism against the United States To protect the U.S. against cyber-based attacks and high-technology crimes To monitor the flow of trade and travel 2 points Question 14 __________ is the process of running an organization so that the organization can accomplish its goals. Bureaucracy Management Structure Coordinating 2 points Question 15 In the early days of policing in England, the watchmen in a.

Question 1Theories of Law Subject__________ is a collaborative.pdf

Question 1Theories of Law Subject__________ is a collaborative.pdf

Question 1Theories of Law Subject__________ is a collaborative.pdf

Q: Describe the 2 types of homologs. (1) Orthologs (2) Paralogs Solution 1. Orthologs: These are genes found in different species that resulted from speciation of a common ancestral gene. During this event of evolution, the orthologs retain their original function. 2. Paralogs: These genes are resulted from duplication event with a genome. And during the evolution, paralogs evolve to attain new function unlike orthologs..

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Q Describe the 2 types of homologs. (1) Orthologs (2) ParalogsS.pdf

Q Describe the 2 types of homologs. (1) Orthologs (2) ParalogsS.pdf

Please fix the java code (using eclipse) package hw4p1; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.net.URL; import java.util.ArrayList; import java.util.Collections; import java.util.Scanner; public class MeanMedianHWv2 { public static void main(String[] args) throws IOException { String[] arrDate = { \"\", \"\", \"\",\"\" }; Scanner in = new Scanner(System.in); Boolean day = true; // Year do { System.out.println(\"Enter a year (2016)\"); arrDate[0] = in.next(); if (!arrDate[0].equalsIgnoreCase(\"2016\")) { System.out.println(\"Wrong Input. \'2016\' only! Try again.\"); } else { arrDate[0] = \"2016\"; } } while(!arrDate[0].equalsIgnoreCase(\"2016\")); // Month do { System.out.println(\"Enter a month (01)\"); arrDate[1] = in.next(); if (!arrDate[1].equalsIgnoreCase(\"01\")) { System.out.println(\"Wrong Input. \'01\' only! Try again!\"); } else { arrDate[1] = \"01\"; } } while(!arrDate[1].equalsIgnoreCase(\"01\")); // Day do { System.out.println(\"Enter a Day (04-10)\"); arrDate[2] = in.next(); //if (arrDate[2] > 10 || arrDate[2] < 4) if (!arrDate[2].equalsIgnoreCase(\"04\")) { System.out.println(\"Wrong Inpit. \'04\' - \'10\' only! Try again!\"); } //else if (arrDate[2] == 4) else if (!arrDate[2].equalsIgnoreCase(\"05\")) { System.out.println(\"Error! Invalid day. Try again!\"); } else if (!arrDate[2].equalsIgnoreCase(\"06\")) { System.out.println(\"Error! Invalid day. Try again!\"); } else if (!arrDate[2].equalsIgnoreCase(\"07\")) { System.out.println(\"Error! Invalid day. Try again!\"); } else if (!arrDate[2].equalsIgnoreCase(\"08\")) { System.out.println(\"Error! Invalid day. Try again!\"); } else if (!arrDate[2].equalsIgnoreCase(\"09\")) { System.out.println(\"Error! Invalid day. Try again!\"); } else if (!arrDate[2].equalsIgnoreCase(\"10\")) { System.out.println(\"Error! Invalid day. Try again!\"); } else { arrDate[2] = \"04\"; day = false; arrDate[2] = \"05\"; day = false; arrDate[2] = \"06\"; day = false; arrDate[2] = \"07\"; day = false; arrDate[2] = \"08\"; day = false; arrDate[2] = \"09\"; day = false; arrDate[2] = \"10\"; day = false; } } while (day); do { System.out.println(\"which data? \'w\' for wind, \'b\' for baro pressure, \'t\' for temp\"); arrDate[3] = in.next(); if (!arrDate[3]. equalsIgnoreCase (\"w\") && !arrDate[3].equalsIgnoreCase(\"b\") && !arrDate[3].equalsIgnoreCase(\"t\")) { System.out.println(\"Wrong input. \'w\' / \'b\' / \'t\' only! Try again\"); } //while(!arrDate[2].equalsIgnoreCase(\"04\") && !arrDate[2].equalsIgnoreCase(\"05\") && !arrDate[2].equalsIgnoreCase(\"05\")); if (arrDate[3].equalsIgnoreCase(\"w\")) { System.out.println(\"w\"); } else if (arrDate[3].equalsIgnoreCase(\"b\")) { System.out.println(\"b\"); } else if (arrDate[3].equalsIgnoreCase(\"t\")) { System.out.println(\"t\"); } } while (!arrDate[3].equalsIgnoreCase(\"w\") && !arrDate[3].equalsIgnoreCase(\"b\") && !arrDate[3].equalsIgnoreCase(\"t\")); in.close(); String Date = arrDate[0] +.

Please fix the java code (using eclipse)package hw4p1;import jav.pdf

Please fix the java code (using eclipse)package hw4p1;import jav.pdf

Please fix the java code (using eclipse)package hw4p1;import jav.pdf

Overview You are tasked with writing a program called Social Security that prompts the user to enter the data (name and date-ofbirth)for a collection of people. The program will then process this collection (array) of People (objects we create in another file) to determine who qualifies for Social Security benefits. Specifications For our Social Security program we will need two java files 1. Person java The person class which stores the data for a specific individual 2. SocialSecurity.java (The driver) Details for these two class files appear below. Person ava UML diagram name: String birthdate: LocalDate Person0 Person (newName: String, month: int, day: int, year: int) NetName newName: String) void getName(): String set BirthDate month int, day: int, year int) void get BirthDate 0 LocalDate toString(): String calculate Age(myBirth LocalDate) int Person ava method description Two constructors 1. A Default constructor that creates a Person whose name and birthdate are set to null. 2. A parameterized constructor that accepts a String representation of the name, and three integers corresponding to the month, day, and year in which the person was born. Setters and getters for the two instance variables. A toString0 method that prints the person in the form dob Solution import java.time.LocalDate; import java.time.Period; import java.time.format.DateTimeFormatter; import java.util.Scanner; class Person { String name; LocalDate birthdate; public Person() { super(); } public Person(String name, int month, int day, int year) { super(); this.name = name; this.birthdate = LocalDate.of(year, month, day); } public String getName() { return name; } public void setName(String name) { this.name = name; } public LocalDate getBirthdate() { return birthdate; } public void setBirthdate(int month, int day, int year) { this.birthdate = LocalDate.of(year, month, day); } @Override public String toString() { DateTimeFormatter formatter = DateTimeFormatter.ofPattern(\"MM/dd/yyyy\"); birthdate.format(formatter); return \"Person [name=\" + name + \", birthdate=\" + birthdate + \"]\"; } public int calculateAge() { return Period.between(birthdate, LocalDate.now()).getYears(); } public void qualifySSN() { if(this.calculateAge()>66) System.out.println(\"Qualifies for Social Security Benifits\"); else System.out.println(\"Doesnot qualifies for Social Security Benefits\"); } } public class SocialSecurity { public static void main(String[] args) { if (args.length > 1) System.out.println(\"Invalid number of command line arguments\"); int num = Integer.parseInt(args[0]); Person[] p = new Person[num]; Scanner in = new Scanner(System.in).useDelimiter(\"\\\ \"); ; for (int i = 0; i < num; i++) { System.out.println(\"enter persons name\"); String name = in.next(); System.out.println(\"enter date of birth in the format Month Day Year\"); String dob = in.next(); String[] temp = dob.split(\" \"); p[i] = new Person(name, Integer.parseInt(temp[0]), Integer.parseInt(temp[1]), Integer.pa.

Overview You are tasked with writing a program called Social Security.pdf

Overview You are tasked with writing a program called Social Security.pdf

Overview You are tasked with writing a program called Social Security.pdf

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