In Exercises 19-24, find the volume of the solid generated by revolvi.pdf

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In Exercises 19-24, find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Solution Area = Integral(ydx) from x= 0 to 3 = int[1/sqrt(1+x) dx] from x= 0 to 3 = 2 sqrt(1+x) from x = 0 to 3 = 2 [sqrt(4) - sqrt(1)] = 2 Volume when rotated along x axxis = Area * 2 *pi = 2*2*pi = 4pi.

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John Rawls argued that society\'s goal should be to lessen the welfare of the most well off property rights should be equally distributed among all citizens. Solution Rawls refers to “the two principles of Justice as Fairness,” but the second has two parts. These principles address two different aspects of the basic structure of society: the “First Principle” addresses the essentials of the constitutional structure. It holds that society must assure each citizen “an equal claim to a fully adequate scheme of equal basic rights and liberties, which scheme is compatible with the same scheme for all.” So answer here will be B..

John Rawls argued that societys goal should be to lessen the welfar.pdf

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In which step of meiosis, the members of tetra separate Solutio.pdf

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Identify the distinguishing characteristics of member of Phylum Cnidaria, including symmetry, number of embryonic tissue layers, unique structures, skeleton and polymorphism (polyp and medusa), place Cnidaria on a cladogram Compare and contrast how a general polyp and a medusa feed Describe why a nerve net was such a major evolutionary adaptation Describe the characteristics of the four classes of Cnidarians Label the following on a drawing: tentacles, mouth, anus, gastrovascular cavity, cnidocytes, nerve net Solution Answer1) Symmetry- Radially symmetrical body Number of embryonic tissue layers- two (endoderm and ectoderm) Unique structures- Stinging cells (Cnidocytes) Skeleton- Hydrostatic skeleton Polymorphism (polyp and medusa)- polyp is a flower like solitary form and medusa is a free- floating miniature of cnidarians. Answer 2) a tubular shaped body form is called as polyp. They are sessile and are found as attached bottom (basal plate) to a solid surface with the opened mouth at the top. while medusa is an umbrella or bell shape form, having their mouth facing downwards. And they are free floating. Being a sessile, polyp filter their food from the water by their tiny-beating cilia. While medusas are free floating and motile so they can trap their feedings, and need not to filter the water all the time though these structures were more related to the sexual reproduction. Answer 3) Nerve net is a diffused framework of neurons in the body which can transmit signals/impulse in different directions from a single point of stimulus. This became a major evolutionary adaptation because it allowed the organism to sense the environmental stimulus from any region of its body and respond accordingly. This allowed the readily development and motion of different body parts such as limb like structures. Answer 4) These are the four classes of cnidarians. Hydrozoa are diploblastic and multicellular animals. Which has only mouth opening. They have a fixed polyp stage and the reproductive medusa stage. They have the gonads as well. Scyphozoan are present in medusoid forms. They have tentaculocysts as Sense organs. They have Velarium in endodermal canal (Acraspedote). They endodermal gonads. Anthozoan are present only in polyp forms. Their medusa stage is absent. They have Mesenteries which bear nematocyst. They also possess endodermal Gonads. Cubozoan looks like the jelly fish and called as jelly boxes. They possess box shaped medusa and they share typical characteristics of scyphozoans..

Identify the distinguishing characteristics of member of Phylum Cnid.pdf

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If most evolutionary change occurs during and immediately after speciation, could speciation rates and extinction rates explain major patterns in macroevolution? In turn, what factors may influence speciation and extinction rates (are they all biological, or can geological / meteorological / astronomical factors also play a role)? Solution In the course of evolution, some mutations keep on occurring with every replicative step. These mutations lead to differences in alleles, which further result in different genotypes. Some genotypes have higher fitness while others have lower fitness. Natural selection selects the phenotypes which have higher fitness, and slowly discards the others during the course of evolution. So, extinction rate depends on fitness and natural selection. As new phenotypes are selected, they result in formation of new species; the process is known as speciation. Speciation further results in evolution. Higher the speciation rate, higher will be the rate of evolution. So, you can say that most evolutionary changes occur during and after speciation. Overall the biological factors affecting speciation and extinction rates are: It is important to note that that apart from the above two factors, geological/astronomical factors also affect speciation, extinction and evolution. Changes in the earth since times immemorial ultimately resulted in changes like:.

If most evolutionary change occurs during and immediately after spec.pdf

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Imagine that someone picking up roses in a garden gets a scratch from a thorn. Some bacteria (Staphyloccocus aureus and Streptococcus pyogenes) take advantage of the wound to gain access to the deeper layers of the skin and attempt to cause an infection. Describe the different stages of the host response (in particular the types of cells and receptors involved) and how the bacteria will be recognized and eliminated. Solution The superficial staphylococcal and streptococcal infections are very common in skin. The physiological condition of the skin is much suitable to these bacteria. Both these bacteria are potent toxin and enzyme producing strains. S.auresus infection on the skin immediately stimulates the inflammatory response and macrophages and neutrophils are proliferated at the site of infection. The organism can enter the body through openings in the skin barrier like hair follicles and can cause infections like folliculitis, pimples, furuncle which is type of abscess. The furuncles also damage the neighboring tissue by invading deep in to the tissue. This is known as carbuncle. Staphylococci cause a highly contagious skin infection in children known as impetigo. The impetigo is two types-nonbullous impetigo is caused by both S.aureus and S.pyogenes; bullous impetigo is caused by staphylococcal toxin. It is also known as staphylococcal scalded skin syndrome. At later stages of toxic shock syndrome the scalded skin syndrome is the characteristic feature. Streptococci also secrete toxin and enzymes. The toxins are called hemolysins that lyse the red blood cells. Various hemolysins like alpha, beta and gamma are produced by various streptococcus strains. S.pyogenes is an important beta hemolytic streptococcus. Streptococcus skin infections are generally superficial but some times may reach the deeper tissues also. S.pyogenes infects the dermis of the skin and also causes a disease known as erysipelas in which the skin erupts in to reddish patches. The skin scraping is cultured and stained to observe the bacteria under microscope. Staphylococci and streptococci both are gram positive and round shaped bacteria. The staphylococci occur as grape-like clusters (Staphylococci name indicates bunch of grapes in Greek). The streptococci are arranged in chains or pairs. Topical antibiotics like Mupirocin and oral antibiotics like first generation cephalosporins and erythromycin are used to eliminate the bacteria..

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I need help working out this question . woman has normal vision although her maternal grandfather (her mother’s father) had red- green colorblindness, a sex-linked recessive trait. Her maternal grandmother and the woman’s own father are assumed to not possess a copy of the mutant allele. The woman marries a man with normal vision although his father was colorblind. What is the probability that the first child of this couple will be colorblind? Solution As it is x linked inheritance, the genotype of women is either xx = Normal or xxc =Carrier for disease. If she is carrier (xxc) and married normal man, then the probability of 1st chaild getting colour blindness is 1/4 = 0.25 Color blind ness is x linked and it inherit criss cross manner. That mean, male transfer color blindness to her female children and. female transfer her male children. XXc x XY ----------------> XX, XY, XXc & XcY XcY progeny is only colour blind. So probability is 1\\4.

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I have a Programming is Binary Tree Search (BTS) for strings with pointers, but I do not know how to change BTS for strings using arrays. Can you help me to change the method, please? This is programming: #include #include #include using namespace std; struct myNode { string data; struct myNode *lNode; struct myNode *rNode; } *roNode; class myBTS { public: void seek(string, myNode **, myNode **); void add(myNode *, myNode*); void rem(string); void caseOne(myNode *, myNode *); void caseTwo(myNode *, myNode *); void caseThree(myNode *, myNode *); void print(myNode *, char); myBTS() { roNode = NULL; } }; // Main method int main () { int myOptions; string rData; myBTS mybst; myNode *tmp; while (1) { cout << \"-----------------------------------------\" << endl; cout << \"Welcome to Binary Search Tree Programming\" << endl; cout << \"-----------------------------------------\" << endl; cout << \"1. Add Data \" << endl; cout << \"2. Remove Data \" << endl; cout << \"3. Print Data \" << endl; cout << \"4. Exit \" << endl; cout << \"Enter Option : \"; cin >> myOptions; switch(myOptions) { case 1: tmp = new myNode; cout << \"Enter data to add : \"; cin >> tmp->data; mybst.add(roNode, tmp); break; case 2: if (roNode == NULL) { cout << \"Tree empty!!!\" << endl; continue; } else { cout << \"Enter data to remove: \"; cin >> rData; mybst.rem(rData); } break; case 3: cout << \"Print Tree:\" << endl; mybst.print(roNode,1); cout << endl; break; case 4: exit (1); default: cout << \"Wrong Option!!!\" << endl; } } } // Data in BTS void myBTS::seek(string sData, myNode **myPar, myNode **myLoc) { myNode *ptr1, *ptr2; if (roNode == NULL) { *myLoc = NULL; *myPar = NULL; return; } if (sData == roNode-> data) { *myLoc = roNode; *myPar = NULL; return; } if (sData < roNode-> data) ptr1 = roNode->lNode; else ptr1 = roNode-> rNode; ptr2 = roNode; while (ptr1 != NULL) { if (sData == ptr1->data) { *myLoc = ptr1; *myPar = ptr2; return; } ptr2 = ptr1; if (sData < ptr1 -> data) ptr1 = ptr1 ->lNode; else ptr1 = ptr1 -> rNode; } *myLoc = NULL; *myPar = ptr2; } // Add Data into BTS void myBTS::add(myNode *myTree, myNode *myNewNode) { if (roNode == NULL) { roNode = new myNode; roNode -> data = myNewNode ->data; roNode -> lNode = NULL; roNode -> rNode = NULL; cout << \"Root data is added successfully\" << endl; return; } if (myTree -> data == myNewNode -> data) { cout << \"Data already in BTS\" << endl; return; } if (myTree -> data > myNewNode -> data) { if (myTree ->lNode != NULL) { add (myTree -> lNode, myNewNode); } else { myTree -> lNode = myNewNode; (myTree -> lNode) -> lNode = NULL; (myTree -> lNode) -> rNode = NULL; cout << \"Data added to the Left successfully\" << endl; return; } } if (myTree -> rNode != NULL) { add(myTree -> rNode, myNewNode); } else { myTree -> rNode = myNewNode; (myTree -> rNode) -> lNode = NULL; (myTree -> rNode) -> rNode = NULL; cout << \"Data added to the Right successfully\" << endl; return; } } //Remove data from BTS void myBTS:: rem(string sData).

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How is ATP produced in mammalian cells under aerobic conditions? 1. photophosphorylation 2. oxidative phosphorylation 3. substrate-level phosphorylation 4. phosphorolysis a. 1 b. 2,3 and 4 c. 2 d. 1 and 2 e. 2 and 3 Solution Hi, 4.Phosphorolysis: in this process cleavage of a compound where attacking group is phosphate. Hence by analysing above explanation it is obvious that only option c. 2 (oxidative phosporylation) is correct. I hope my answer is satisfactory. Thanks!!.

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How con osmosis work for and against homeostasis? How do osmosis and diffusion relate to dialysis? (Short paragraph o fine) Solution Multi cellular organisms should maintain a balance between internal and external environments to carry out proper metabolic activities. The maintenance of relatively stable environment internally is known as homeostasis. All the body functions carried out by various systems of the body contribute to the homeostasis to maintain a stable environment in the body for its survival and function. Homeostasis is much required and the vital force for the each cell and through performing specialized activity, each cells maintain a stable internal environment. The net diffusion of water down the concentration gradient through the selectively permeable membrane is called osmosis..

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Enzymes and KE. What\'s the relationship in terms of substrate molecules Enzymes and KE. What\'s the relationship in terms of substrate molecules Solution ANS: Enzyme KE(kinetic energy)and substrate molecules relationship: When the temperature raises the kinetic energy of reactants is also raised, increasing in temperature results in higher activation energy molecules and more molecular (enzyme and substrate) collision and interaction for the reaction to proceed faster. The enzymes present in the human body usually need 37.5 o C (optimal temperature). The optimum temperature for most of the enzymes between 40-45o C. In this condition rate is increased. Increasing substrate concentration means an increasing rate at which the enzyme and substrate molecules encounter one another In general, the enzymes are exposed to a temperature above 50 o C, denaturing leading to derangement in native structure. Majority of enzymes became inactive at higher temperature above 70 o C In enzymes at higher temperatures intra and intermolecular bonds are broken down, so the enzyme molecules gain even more kinetic energy. After reaching the maximum point, the rate of the reaction suddenly fall down due to the enzyme degradation..

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Explain membranes. Include the definition and name the four types and which tissues each type contains. Solution Answer : Membranes are the flat sheets of pliable tissue that cover or line a part of the body. The combination of an epithelial layer and an underlying connective tissue layer constitutes an epithelial membranes. These includes, Mucous membranes : These membranes lines the body cavity that opens directlyto the exterior. Mucous membranes lines the entire digestive, respiratory and reproductive tracts and much of the urinary tract. Serous Membranes : A serous membrane or Serosa lines a body cavity that does not open directly to the exterior and it covers the organs that lie within the cavity.serous membranes consist of areolar connective tissue covered by mesothelium and also consists of two layers such as Parietal and visceral layer. Cutaneous Membrane : These membarnes covers the surface of the body and consists of a superficial portion called the epidermis and a deeper portion called the dermis. Synovial Membranes : Like serous membranes, synovial membranes line structures that do not open to the exterior. They lack in epithelium and are composed of discontinuous layer of cells called synoviocytes, which secretes synovial fluid. Synovial fluid lubricates and nourishes the cartilage covering the bones at movable joints and contains macrophages that moves microbes and debris from the joint cavity..

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Define 1. Reciprocal cross – 2. True or pure-breeding – Fill in the Blank 3. When trying to calculate the probability that two independent events will occur together you should _____________________ the individual probability of each event. 4. When crossing dihybrid peas, Mendel noticed that there were four phenotypic types amongst the resulting F2 progeny; two had the phenotype of the parental generation and the other two were ___________________. True or False 5. The dominant trait is that which is masked in the F1 generation resulting from a cross between true breeding parents. 6. Mendel’s law of segregation says that during gamete formation different pairs of alleles segregate independently of each other. Short Answer 7. Calculate the probability of A.) drawing a 4 from a deck of playing cards, B.) Drawing a black 4 and a red 8 from a deck of playing cards, C.) drawing a king, queen, or jack and the ace of spades from a deck of playing cards. (Show your work.) 8. A. is this an autosomal, X linked or Y linked trait? B. Is the transmission of this trait dominant or recessive? C. what is the genotype of individual II-2? 2 2 3 4 2- 1 ! 1 1 Solution 1. Reciprocal cross: a pair of crosses between a male of one strain and a female of another, and vice versa. 2. True or pure-breeding:True breeding means that the parents will also pass down a specific phenotypic trait to their offspring. Remember that a phenotype is the outward appearance of something. True bred organisms will have a pure genotype (genetic expression of a trait) and will only produce a certain phenotype. 3. When trying to calculate the probability that two independent events will occur together you should specify size of the individual probability of each event. 4. When crossing dihybrid peas, Mendel noticed that there were four phenotypic types amongst the resulting F2 progeny; two had the phenotype of the parental generation and the other two were genotype. 5. The dominant trait is that which is masked in the F1 generation resulting from a cross between true breeding parents is True. 6. Mendel’s law of segregation says that during gamete formation different pairs of alleles segregate independently of each other is True..

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Clinical immunology question Mr. R.M., a 60-year-old Caucasian man, was stung by a bee while gardening. He had been stung once before earlier in the summer. Within a few seconds, his hand began to itch and he began to experience abdominal cramping. He subsequently began experiencing difficulty breathing. Fortunately, he was able to reach a first aid kit in his garage. Inside of the kit was an EpiPen (injectable epinephrine) that his wife kept for herself because she was allergic to bee venom. He immediately had his wife drive him to the hospital. He was asymptomatic on arrival at the hospital. R.M. had no history of adverse reactions to bee venom or antibiotics. Because of the nature of the incident, a diagnosis of anaphylactic shock due to bee venom sensitivity was made. An IgE level was ordered. The results indicated a level more than twice the (normal) reference range value. In addition, a follow-up skin test was performed. The patient was extremely positive for bee venom. Questions: 1.Which of the 4 types of hypersensitivity reactions does this case represent? 2.What is the immunologic mechanism involved in anaphylaxis? 3.What types of agents can induce anaphylactic shock? 4.Describe 2 different types of testing to diagnosis this patient. 5.What is the risk in performing a skin test? Solution Answer: 1) There are four types of hypersensitivity reactions, they include: 1) Type - I ; Allergy , 2) Type- II ; Cytotoxic, antibody-dependent, 3) Type III; Immune complex disease 4) Type IV; Delayed-type hypersensitivity. 2) Anaphylaxis is an Type-I reaction ; allergic reaction. During allergic reaction, IgE react with antigen, and it activates basophils and mast cells, due this activation, histamines are released. 3) Anaphylactic shock could be due to the food agents such as milk, eggs and shellfish. In few cases, insect stings can cause shock. 4) Diagnosis includes, 1) Allergy testing (skin allergy test), 2)Differential diagnosis. 5) Skin test is available only to test effect of penicillin drug allergy.so, it is risk for allergies..

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As shown in the pedigree below, a pair of first cousins (I and K) marry an unrelated pair of first cousins (J and L). Their children M and N produce child O. 1. Calculate the inbreeding coefficient (f) of O (express answers to 1 and 3 to three decimal places). Assume that f= 0 for all ancestors. (Hint: To help you out, I have shaded individuals who cannot contribute to autozygosity.) To show your work, use the letter names to indicate each path/loop that contributes to autozygosity. For example: path 1 = M-I-E… 2. Interpret your result. What does this value of f indicate? 3. Recalculate f for child O when f=1/2 for great-great grandmother D (f=0 for all other ancestors). Hint: After you have calculated all the loops above, you’ll adjust just the loop that includes ancester D to account for her ancestral inbreeding using the formula presented in class (and text). As shown in the pedigree below, a pair of first cousins (I and K marry a unrelated pair of first cousin s and L). Their children M and N produce child O. 1. Calculate the inbreeding coefficient (f) of O (express answers to 1 and 3 to three decimal places). Assume that f 0 for all ancestors. (Hint: To help you out, I have shaded individuals who cannot contribute to autozyBosity.) o show your work, use the letter names to indicate each path/loop that contributes to autoz osit For example: path 1 M-1-E 2. Interpret your result. What does this value of findicate? 3. Recalculate f for child o when f-1/2 for great-great grandmother D (f 0 for all other ancestors). Hint: After you have calculated all the loops above, you\'ll adjust just the loop that includes ancester D to account for her ancestral breeding using the formula presented in class (and text) Solution 1. Path 1: So, the probability that a copy of gene in A makes the way down to O via E and a copy of gene in B makes the way down to O via F =( ½)n = (½)7=0.00781. Here n= node=7 Path 2: Similarly like the first one the total probability in path 2 will be same that is 0.00781. Thus inbreeding co-efficient = 0.00781+0.00781=0.01562. 2. Generally, the higher the level of inbreeding closer the coefficient of relationship approaches a value of 1 and approaches a value of 0 for individuals with remote common ancestors. As here the value of inbreeding co-efficient is aproximately 0 so we can conclude that 3. When a common ancestor is iself inbreed the inbreeding coefficient is = (1/2)n (1+ inbreeding co-efficient of common ancestor) , The probability that a copy of gene in D makes the way to O = (1/2)7 = 0.00781 Therefore inbreeding co-efficient is = 0.00781 * (1 + 1/2 )=0.01172.

As shown in the pedigree below, a pair of first cousins (I and K) ma.pdf

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An individual with genes A, H, Se and lele has which of the following phenotypes? AH, Le(a-b-) ABH, Le(a+b-) ABH, Le(a-b-) AH, Le(a+b-) The Rh phenotype results for a patient are: anti-C = 4+, anti-c = 3+, anti-D = NEG, anti-E = NEG, anti-e = 4+. The patients most probable genotype using the Wiener nomenclature is? r\'r R1r r\"r rr With respect to the Lewis blood group system, which statement is false. Lewis antibodies occur frequently in the sera of pregnant women. Lewis blood group antigens are made by tissue cells, secreted into body fluids and plasma, and adsorbed onto the red cell membrane. The Le gene codes for L-fucosyltransferase which adds L-fucose to the type 1 chain. Lewis antigens are well developed at birth. Lewis antibodies are generally IgM, and are therefore not a cause of HDN. The major immunoglobulin class of anti-B in a group A individual is: IgM. None of the alternatives IgA. IgG. Antibody screening cells may not detect antibodies against low frequency antigens. Which of the following is most likely to go undetected? K Kpa M k e Rh antigens are inherited as? Codominant alleles. X linked. Sex-linked genes. Autosomal recessive alleles. AH, Le(a-b-) ABH, Le(a+b-) ABH, Le(a-b-) AH, Le(a+b-) Solution An individual with genes A, H, Se and lele has which of the following phenotypes? answer: AH, Le(a-b-) . if B antigen is not found. if B antigen is present then ABH, Le(a-b-) The Rh phenotype results for a patient are: anti-C = 4+, anti-c = 3+, anti-D = NEG, anti-E = NEG, anti-e = 4+. The patients most probable genotype using the Wiener nomenclature is? answer is R1r positive sign shows that antigens are present. so we can see here that D,d, C,c,e are present on RBC. D and C are homo and heterozygous whereas e is only homozygous. combinations are DCe/dce, Dce/dCe, Dce/DCe. when they are converted into fischer race chart. among these phenotypes,R1r is the most probable one. With respect to the Lewis blood group system, which statement is false. answer: Lewis antigens are well developed at birth. they cannot be detected until 2 years of birth. The major immunoglobulin class of anti-B in a group A individual is answer: IgM. Major immunoglobulin class for anti-b in group A individual or anti-a in group B individual is IgM. Antibody screening cells may not detect antibodies against low frequency antigens. Which of the following is most likely to go undetected? answer: Kp-a, low frequency antigens are K,Kp-a, Le-a and Js-a Rh antigens are inherited as? answer: codominant alleles which are present on autosomes.

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8.8 Program: Playlist (Java) You will be building a linked list. Make sure to keep track of both the head and tail nodes. (1) Create two files to submit. SongEntry.java - Class declaration Playlist.java - Contains main() method Build the SongEntry class per the following specifications. Note: Some methods can initially be method stubs (empty methods), to be completed in later steps. Private fields String uniqueID - Initialized to \"none\" in default constructor string songName - Initialized to \"none\" in default constructor string artistName - Initialized to \"none\" in default constructor int songLength - Initialized to 0 in default constructor SongEntry nextNode - Initialized to null in default constructor Default constructor (1 pt) Parameterized constructor (1 pt) Public member methods insertAfter() (1 pt) setNext() - Mutator (1 pt) getID() - Accessor getSongName() - Accessor getArtistName() - Accessor getSongLength() - Accessor getNext() - Accessor printPlaylistSongs() Ex. of printPlaylistSongs output: (2) In main(), prompt the user for the title of the playlist. (1 pt) Ex: (3) Implement the printMenu() method. printMenu() takes the playlist title as a parameter and outputs a menu of options to manipulate the playlist. Each option is represented by a single character. Build and output the menu within the method. If an invalid character is entered, continue to prompt for a valid choice. Hint: Implement Quit before implementing other options. Call printMenu() in the main() method. Continue to execute the menu until the user enters q to Quit. (3 pts) Ex: (4) Implement \"Output full playlist\" menu option. If the list is empty, output: Playlist is empty (3 pts) Ex: (5) Implement the \"Add song\" menu item. New additions are added to the end of the list. (2 pts) Ex: (6) Implement the \"Remove song\" method. Prompt the user for the unique ID of the song to be removed.(4 pts) Ex: (7) Implement the \"Change position of song\" menu option. Prompt the user for the current position of the song and the desired new position. Valid new positions are 1 - n (the number of nodes). If the user enters a new position that is less than 1, move the node to the position 1 (the head). If the user enters a new position greater than n, move the node to position n (the tail). 6 cases will be tested: Moving the head node (1 pt) Moving the tail node (1 pt) Moving a node to the head (1 pt) Moving a node to the tail (1 pt) Moving a node up the list (1 pt) Moving a node down the list (1 pt) Ex: (8) Implement the \"Output songs by specific artist\" menu option. Prompt the user for the artist\'s name, and output the node\'s information, starting with the node\'s current position. (2 pt) Ex: (9) Implement the \"Output total time of playlist\" menu option. Output the sum of the time of the playlist\'s songs (in seconds). (2 pts) Ex: Solution SongEntry.java package playlist; public class SongEntry { private String uniqueID; private String songName; private String artistName.

8.8 Program Playlist (Java)You will be building a linked list. Ma.pdf

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5. Cells with an intracellular osmolarity of 300mosm/L are suspended in the following solutions. Describe the sequence of events that take place. (a)0.3M urea (membrane permeable) (b)0.3M glucose (membrane impermeable) (c)0.6M urea (d)0.6M glucose Solution Osmolarity of a solution refers to the concentration of the solution in terms of how strongly it attracts water molecules across a semipermeable membrane. It is given the intracellular osmolarity of the cell is 300 mosm/L. It can also be expressed as 0.3 osm/L or 0.3 M. (a) 0.3 M Urea (membrane permeable) This solution is isosmotic to the intracellular osmotic pressure. But, the cell membrane being selectively permeable, allows the entry of urea until both the solutions become isotonic. This leads to bursting ofcell as along with urea water enters the cell. (b) 0.3 M Glucose (membrane impermeable) This solution is also isosmotic to the intracellular osmotic pressure. But given that the membrane is impermeable, glucose cannot enter to cell and an isotonic condition cannot be reached. Hence the cell would be unstable and would distort due to the uneven tonicity. (c) 0.6 M urea This solution is hypertonic and hyperosmotic as compared to the cell\'s intracellular fluid. Thus, water will move out through the pores of the cell membrane causing the cell to shrink or become flaccid. (d) 0.6 M glucose This solution is hypertonic and hyperosmotic as compared to the cell\'s intracellular fluid. Thus, water will move out through the pores of the cell membrane causing the cell to shrink or become flaccid..

5. Cells with an intracellular osmolarity of 300mosmL are suspended.pdf

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Which symptom of inflammation is NOT paired with its source?A) swelling: accumulation of fluid in connective tissueB) heat: due to increased blood flowC) redness: blood leaking into surrounding tissuesD) pain: pressure of fluid on nerve endingsE) pain: release of inflammatory chemicals affect nerve endings directlyF) All of the listed responses are correctly paired. Solution Option C is wrong. Because both redness and heat is mainly due to increased blood flow to the inflamed area..

Which symptom of inflammation is NOT paired with its sourceA) swell.pdf

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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. Click the icon to view the histogram. I and III I, II and III I only I and II III only Solution (I) The data shows that the shape of this histogram is bell-shaped. Also,if a line is drawn down the middle of the graph, the two sides will mirror each other. So,the distribution is approximately symmetric. (II) In the given histogram where the values of data occur at regular frequencies. So, the mean, median and mode occur at the same point. (III) The IQR is the difference between the third and the first quartile. IQR makes the values by estimate of outliers where mean and standard deviation are not effected against outliers. So, the median and IQR is not summarize the data well. Answer is (I) and (II).

Which is true of the data shown in the histogram I. The distribution.pdf

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Which of the following is a member of the domain ArchaeaA. The ce.pdf

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