You are an oncologist specializing in colon cancer. A patient presen.pdf

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You are an oncologist specializing in colon cancer. A patient presents to you with disease that has spread to his bones and liver and there is no treatment available. What stage cancer is this?2345 Solution Answer is Stage IV (4) Stages of colon cancer are: Stage 0 It is very early cancer and it will be affecting the innermost layer of the intestine Stage I In stage I, cancer is seen in the inner layers of the colon Stage II In stage II, cancer is seen spreading through the muscle wall of the colon Stage III In stage III, cancer is seen spreading to the lymph nodes Stage IV In stage IV, cancer is seen spreading to other organs outside the colon In case, of colon cancer, at stage IV, it will be spreading to the distant location in the body (liver, lungs, bones, distant lymph nodes or other sites). At this stage there is no cure for this cancer or patient will be left with very few options. Stage 0 It is very early cancer and it will be affecting the innermost layer of the intestine Stage I In stage I, cancer is seen in the inner layers of the colon Stage II In stage II, cancer is seen spreading through the muscle wall of the colon Stage III In stage III, cancer is seen spreading to the lymph nodes Stage IV In stage IV, cancer is seen spreading to other organs outside the colon.

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JAVA LAB #8 - ARRAY BASED LISTS The next exercise is based on this implemetation for an UnorderedArrayList of integers: //Interface: ArrayListADT //works for int public interface ArrayListADT { public boolean isEmpty(); //Method to determine whether the list is empty. public boolean isFull(); //Method to determine whether the list is full. public int listSize(); //Method to return the number of elements in the list. public int maxListSize(); //Method to return the maximum size of the list. public void print(); //Method to output the elements of the list. public boolean isItemAtEqual(int location, int item); //Method to determine whether item is the same as the item in the list at location. public void insertAt(int location, int insertItem); //Method to insert insertItem in the list at the position public void insertEnd(int insertItem); //Method to insert insertItem at the end of the list. public void removeAt(int location); //Method to remove the item from the list at location. public int retrieveAt(int location); //Method to retrieve the element from the list at location. public void replaceAt(int location, int repItem); //Method to replace the element in the list at location with repItem. public void clearList(); //Method to remove all the elements from the list. public int search(int searchItem); //Method to determine whether searchItem is in the list. public void remove(int removeItem); //Method to remove an item from the list. } //Class: ArrayListClass implements //Interface: ArrayListADT public abstract class ArrayListClass implements ArrayListADT { protected int length; //to store the length of the list protected int maxSize; //to store the maximum size of the list protected int[] list; //array to hold the list elements //Default constructor public ArrayListClass() { maxSize = 100; length = 0; list = new int[maxSize]; } //Alternate Constructor public ArrayListClass(int size) { if(size <= 0) { System.err.println(\"The array size must be positive. Creating an array of size 100.\"); maxSize = 100; } else maxSize = size; length = 0; list = new int[maxSize]; } public boolean isEmpty() { return (length == 0); } public boolean isFull() { return (length == maxSize); } public int listSize() { return length; } public int maxListSize() { return maxSize; } public void print() { for (int i = 0; i < length; i++) System.out.print(list[i] + \" \"); System.out.println(); } public boolean isItemAtEqual(int location, int item) { if (location < 0 || location >= length) { System.err.println(\"The location of the item to be compared is out of range.\"); return false; } return list[location]== item; } public void clearList() { for (int i = 0; i < length; i++) list[i] = 0; length = 0; System.gc(); //invoke the Java garbage collector } public void removeAt(int location) { if (location < 0 || location >= length) System.err.println(\"The location of the item to be removed is out of range.\"); else { for(int i = location; i < length - 1; i++) list[i] = list[i + 1.

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In the year 2000, a halophile was isolated from a 250 million year old salt crystal and grown up in a university laboratory in Westchester PA. What did the scientists isolate and why do you think it survived 250 million years? Solution Halophile is salt loving, They are extremophiles which live in high saline environment. They adapt different strategies to survive in the extreme salt conditions. These adaptations are- Hence the halophile survived 250 million years.

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In in the image seen here, in which direction would you expect water to move if the membrane is not permeable to the solute? It would move equally between both sides. Right to left Left to right Solution The process shown here is called as Osmosis.. According to which the movement of water molecules take place from the region higher concentration to a region of lower water concentration when separated by the semi - permeable Membrane.. In this the left side contains more water molecules concentration then right side as the right side contains more solute molecules... Soo, the movement will be from Left Side to Right side...!!!!.

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How does the movement of DNA fragments on the gel and the movement of the ethilisim troemide in the eel affect the appearance of the DNA fragments? Solution Answer: the movement of DNA happens from a negative charged electrode to positively charged electrode as the DNA is negatively charged. By applying the electric field through the agarose gel matrix DNA is seperated on the basis of size. Ethidium bromide is a flourescent dye which intercalates into circular DNA and can sunsequently change the charge, length, as well as the superhelicity of the DNA molecule, EtBr presence in gel during electrophoresis can affect movement of DNA. Agarose gel electrophoresis can be used to resolve circular DNA with different supercoiling topology. as more EtBr binds between the DNA bases, the molecule stiffens and hence mobility is decreased as compared to same frangment being run without EtBr. Ethedium bromide is positively charged and runs from the positive electrode towards the negative electrode. i.e opposite direction of movement of DNA. hence when the two encounter, the intercalaton happens, leading to flourescence of orange red colour under UV light..

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Genetics Homework Problem 1: You are mapping genes in a species of fruit fly. You cross a wild type fly to a fly from a true breeding mutant strain with green eyes (g), short wings (s w), and hairy legs (hl). The F1 progeny all appear wild type (red eyes, long wings, smooth legs). You then testcross a female F1 fly to a male from the mutant strain, and obtain 200 progeny a. What are the genotypes of the two tiles in the test cross?_____ times ______ b. How many progeny of each phenotype would be expected if the genes were assorting independently? c. Can you reject independent assortment at P=0.05? Use statistical evidence to back up your claim. d. Assuming these genes are all linked on the same chromosome, draw a genetic map showing the order and map distances between the three genes. Solution Answer: a). The genotypes of the two flies in the test cross: Gg Swsw Hlhl X gg swsw hlhl b). If the genes were assorting independently, 8 types of phenotypes are expected in equl proportions 1:1:1:1:1:1:1:1 Green, short, hair: 1 Red, long, smooth: 1 Green, long, smooth: 1 Green, short, smooth: 1 Red, short, hairy: 1 Red, long, hairy: 1 Green, long, hairy: 1 Red, short, hairy: 1 c). Phenotype Observed(O) Expected (E) O-E (O-E)2 (O-E)2/E Green, short, hair: 44 25 19 361.00 14.44 Red, long, smooth 42 25 17 289.00 11.56 Green, long, smooth 20 25 -5 25.00 1.00 Green, short, smooth 5 25 -20 400.00 16.00 Red, short, hairy 20 25 -5 25.00 1.00 Red, long, hairy 7 25 -18 324.00 12.96 Green, long, hairy 30 25 5 25.00 1.00 Red, short, hairy 32 25 7 49.00 1.96 200 200 59.92 Chi-square value = 59.92 The independent assortment at P=0.05 is rejected. d). Explanation Shorthand Number of progeny g sw hl 44 G Sw Hl 42 g Sw Hl 20 g sw Hl 5 G sw hl 20 G Sw hl 7 g Sw hl 30 G sw Hl 32 Imagine the order is G SW Hl / g sw hl 1. If single cross over (SCO) occurs between G & Sw Normal order= G---------Sw & g-----sw After cross over= G-----sw & g------Sw G-----sw recombinants are 20+ 32 = 52 g----- Sw recombinants are 20+ 30 = 50 Total recombinants = 102 RF = (102/200.)*100 =51% 2. If single cross over (SCO) occurs between Sw & Hl Normal order= Sw---------Hl & sw----hl After cross over= Sw-----hl & sw------Hl Sw-----hl recombinants are 30 + 7 = 37 sw------Hl recombinants are 32+5 = 37 Total recombinants = 74 RF = (74/200)*100 = 37% 3. If single cross over (SCO) occurs between G & Hl Normal order= G---------Hl & g------hl After cross over= G-----hl & g------Hl G-----hl recombinants are 20 + 7=27 g------Hl recombinants are 20+5=25 Total recombinants = 52 RF = (52/200)*100 = 26% % RF = Map unit distance The order of genes is ----- G------26 m.u.--------Hl------37 m.u.---------Sw Gene for color of eyes and length of wings are not linked. The gene for legs is linked with gene for wings & gene for eye. Phenotype Observed(O) Expected (E) O-E (O-E)2 (O-E)2/E Green, short, hair: 44 25 19 361.00 14.44 Red, long, smooth 42 25 17 289.00 11.56 Green, long, smooth 20 25 -5 25.00 1.00 Green, sho.

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For f(x) =2x, find a formula for the Riemann sum obtained by dividing the interval [2.5] subintervals and using the right hand endpoint for each ck. Simplify the sum and take the limit as n--> infinity to calculate the area under the curve over [2,5] please show all of your work as be as descriptive as you can I appreciate your help thank you! Solution a=2, b =5, x =(b-a)/n =(5-2)/n =3/n f(x)=2x xi=a+ ix =2+i(3/n) f(xi)=2(2+3(i/n)) the area under the curve over [2,5] is =limn->[i =1 to n]f(xi)x =limn->[i =1 to n]2(2+3(i/n))(3/n) =limn->[i =1 to n]6((2/n)+3(i/n2)) =limn-> 6((2n/n)+3(n(n+1)/2n2)) =limn-> 6(2 +3((1+(1/n))/2)) = 6(2 +3((1+0)/2)) =6(2 +3(1/2)) =6(2 +(3/2)) =6(7/2) =3*7 =21 the area under the curve over [2,5] is 21.

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Discuss the recruitment and selection procedure done by human resource management? Solution The procedure of recruitment and selection done by the human resource management involves five distinct and basic aspects which are as follows: 1. CRITERIA DEVELOPMENT: Those individuals which are involved in the process of hiring should be trained enough to develop criteria , to conduct interview , to review the resumes , to develop the questions for interview , to weigh the candidates. In this step the basic thing is to plan the interview and develop the criteria as which of the sources are to be used during interview and how to use these sources to score better in the interview.The criteria should be job focussed and also the focus should be on the personality. In this process the involvement of discussion of the skills, abilities and some personal characteristics takes place that are required to have to the job.Before reviewing the resumes by developing this criteria can be sure that he is fair in selecting the applicants to interview.The application should include the details of the candidate , previous job experience certificates and educational qualification certificates. 2. APPLICATION AND REVIEWING OF RESUME : After the completion of step 1 the applications can now be reviewed. There are different ways to go through this process but there are various computer based programs that can search for the keywords in the resumes and shortlist the number of resumes that can b looked into. 3. INTERVIEW PROCESS : After shortlisting those applicants who have met the minimum criteria.The HR manager selects these applicants for the interview.Due to lack of time sometimes interview is being done on phone. 4. TEST ADMINISTRATION : Before the hiring decision is made any number of test are administered. The tests may be physical tests ,drug tests, response test, credit report checks, personality tests, etc. 5. OFFER MAKING PROCESS : This is the final step in which the choosen candidate is offered the job position. The offer can b developed through email or through the job letter.All the benefits and compensation are discussed in the offer letter..

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Do The statement of stockholder\'s equity explains changes in equity from net income? Solution Statement of equity is a summary of changes in share holders equity during the reporting period. The statemnt includes all changes including issue of new stocks, payment of dividend, issue of bonus shares, buy back of shares. It also includes changes to retained earnings, reserves and stock reserve. Distribution of dividend from net income and transfer of remaining balance to reserves are shown at the bottom of P&L account. The remaining balance of profits transferred to reserves are reflected in Statement of changes in equity as addition to retained earnings. Hope I have answered the question, for any further clarification please comment..

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Describe what is meant by “radical cure” and identify the drug that is produces a radical cure for malaria. Solution Radical cure refers to the elimination of malaria parasites completely from the body.It specifically refers to the elimination of hypnozoites of P.vivax and P. ovale which are the parasitic stages in the liver and could result in relapse of the infection. A combination of two drugs i.e chloroquine and primaquine produces a radical cure for malaria..

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Decasper and Spence (1986) measured babies sucking rates as they liscenced to a story they had heard in the womb (the cat in the hat) and to a story they had not heard (the king, the mice, and the cheese) the study concluded that: a.) amniotic fluid blocks sound waves from reaching the fetus b) increased heart rate showed that an infant can learn even before it is born c) different sucking styles and rates showed that an infant can learn even before it is born d) the cadence of the reading material not previous experience altered sucking style and rate Solution different sucking styles and rates showed that an infant can learn even before it is born.

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Corhin Eckles and Connor Hayes have recieved walkie talkies for Christmas. If they leave from the same point at the same time, Corhin walking north at 2.5 mph and Connor walking east at 3 mph, how long will they be able to talk to each other if the range of the walkie-talkies is 4 mi? Solution Corhin DATA: distance= x miles; rate=2.5 mph; time=d/r= x/2.5 Connor DATA: distance= y miles; rate=3 mph; time=y/4 EQUATIONS: 1) x^2+y^2=4^2 2) x/2.5 = y/4 Using 2) you get y=4x/2.5 Substitute into 1) to get x^2+(4x/2.5)^2=16 x^2+(16x^2/6.25)=16 6.25x^2+16x^2=16(6.25) 22.25x^2=100 x^2=100/22.25=4.49... x=2.12 hrs. x/2.5= 0.85 hrs. = 50.87 minutes This is Corhin\'s time, but it is the same for Connor..

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Consider three genes in a hypothetical bird. One controls the color of feathers on the top of the head (red or brown), one controls the color of feathers on the chest (black or brown) and one controls the number of syllables in the song (3 or 4). You notice that birds frequently have either red heads and black chests or brown heads and brown chests, but very few have red heads and brown chests or brown heads and black chests. However, there doesn\'t seem to be any association between the syllables in the bird song and color. In other words, red heads are equally likely to sing songs with 3 or 4 syllables, and the same goes for the brown heads. Based on what you know about how the genome is organized and passed down to the next generation, explain this observation. Solution The observation can be attributed to the mechanism of genetic linkage. It is a well known fact that genes get passed from one generation to another. Mendel\'s second law of inheritance states that during this process, the inheritance of one gene is independent of the inheritance of another gene. Howeve, this law holds true only when the different genes are sufficiently spaced apart from each other. Inother words, if the genes are close to each other, it is difficult for them to be inherited in an independent manner. If one of the genes is inherited, then the other gene (by virtue of its proximity to the first gene) also tends to get inherited. This phenomenon is called as genetic linkage, and the genes involved are said to be Linked Genes. The same mechanism is in action in the above case. The genes for Red Head and Black Crest are linked, as are the genes for Brown Head and Brown Crest. Therefore, these two genes are inherited in a linked manner, and not in an independent manner. The two genes therefore behave as One Single Gene. The gene for song syllable is at a sufficient distance from the linked genes, and follows Mendel\'s law of independent assortment and is not dependent on the presence or absence of the former two traits..

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are organic compounds that bind very tightly to iron ions for transport into cells. Solution Organisms like bacteria secrerte substances called siderophores which bind to iron for transport inside the cell. Iron is one of the important micronutrients for the bacterial metabolism. Ferrichrome,Desferrioxamine B and Desferrioxamine E are some of the important siderophores secreted by various bacteria..

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Book Business Its Legal, Ethical, Global Environment 10th Edition.pdf

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According the United States (U.S.) Centers for Disease Control and Prevention, in 2015 influenza and pneumonia combined were the 8th leading causes of death (55,227 deaths in total) and the only communicable diseases ranked among the top ten killers in the U.S. What were the 10 causes of mortality in the U.S. in 1900 and 2015? Since communicable disease caused a relatively few deaths in the U.S. in 2015, why should public and public health professionals be concerned about them anymore and devote resources in their prevention and control? Solution Top 10 leading causes of deaths in US: Hearts diseases Cancer Airway diseases which are non infectious Cardio vascular disorders Accidents Alzheimer\'s disease Diabetes Pneumonia or influenza Nephropathys Suicide. The period of communicability is less for the communicable diseases, hence they are prevention of disease in early stages is possible. Prevention and control of influenza: Good ventilation of public buildings Encouraging sufferers to cover their faces with a handkerchief when coughing and sneezing Stay in home at the first sign of influenza Vaccination against the influenza, may be killed or level attenuated vaccines Prevention of exposure to influenza strains by hygiene practice Antiviral drugs like amantadine or rimantidine used Prevention and control of pneumonia: Pneumococcal pneumonia vaccination, PPV 23 AND PCV7. Education of mother to observe the sings of pneumonia. History taking and clinical examination are very important for management Physical examination for, breath count, chest indrawing, listen for stridor, wheezing and for severe malnutrition, cyanosis Any signs detected, they are treated early by antibiotics and other measures Improving the primary medical care services and development..

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9. For some non-Mendelian inheritance patterns, the genotype of an individual is not sufficient to predict the phenotype. For the following traits, what else do you need to know to predict the phenotype? a) Snail coiling direction b) Angelman syndrome due to deleted UBE3A gene Solution The inheritance mechanisms of both snail coiling and Angelman syndrome involve strong maternal influences. Snail coiling direction is governed by the genotype of the oocyte, not of the zygote, so much so that the snail coiling direction can be predicted from the maternal genotype alone. Similarly, Angelman syndrome (due to deletions in ube3A) is governed by the maternal copy of the gene alone. If the copy of the gene derived from the mother is functional, the individuals are healthy. If the maternal copy is defective or deficient, the individual suffers from the Angelman syndrome..

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