is there a mathematical model of hepatitis viral b?which is it?
85020
There was, and continues to be, a lot of work on mathematical modelling of various aspects of viral infection and spread. \n\nThis started apparently with the work of Daniell Bernoulli on smallpox in 1760, and continues to this day. A special 5 year effort on mathematical modelling of viruses by DIMACS is described in one of the references below. There is probably a lot of material associated with mathematical modeling of viruses that one could find in connection with this DIMACS effort. \n\nI also include a reference to a book that describes some mathematical modelling of the hepatitis B virus, and some other references.
704347
Hepatitis B surface antigen (HBsAg) is a protein antigen produced by HBV. This antigen is the earliest indicator of acute hepatitis B and frequently identifies infected people before symptoms appear. HBsAg disappears from the blood during the recovery period. In some people (particularly those infected as children or those with a weak immune system, such as those with AIDS), chronic infection with HBV may occur and HBsAg remains positive.\nFor more info, check out the link below.
1022070
it doesnt....it just doesnt reproduce. Now, the body can get rid of certain infections on its own which is where the virus load actually decreases.Hope that was helpful. I'm sure you were expecting some big long drawn out answer which I could give you about how antibodies in the body work....but thats not what you asked.
1044811
Application 1 : Exponential Growth - Population\nLet P(t) be a quantity that increases with time t and the rate of increase is proportional to the same quatity P as follows\n\nd P / d t = k P\n\n\nwhere d p / d t is the first derivative of P, k > 0 and t is the time.\n\nThe solution to the above first order differential equation is given by\n\nP(t) = A ek t\n\n\nwhere A is a constant not equal to 0.\n\nIf P = P0 at t = 0, then\n\nP0 = A e0\n\nwhich gives A = P0\n\nThe final form of the solution is given by\n\nP(t) = P0 ek t\n\n\nAssuming P0 is positive and since k is positive, P(t) is an increasing exponential. d P / d t = k P is also called an exponential growth model.\n\nApplication 2 : Exponential Decay - Radioactive Material\nLet M(t) be the amount of a product that decreases with time t and the rate of decrease is proportional to the amount M as follows\n\nd M / d t = - k M\n\n\nwhere d M / d t is the first derivative of M, k > 0 and t is the time.\n\nSolve the above first order differential equation to obtain\n\nM(t) = A e- k t\n\n\nwhere A is non zero constant.\n\nIt we assume that M = M0 at t = 0, then\n\nM0 = A e0\n\nwhich gives A = M0\n\nThe solution may be written as follows\n\nM(t) = M0 e- k t\n\n\nAssuming M0 is positive and since k is positive, M(t) is an decreasing exponential. d M / d t = - k M is also called an exponential decay model.\n\nApplication 3 : Falling Object\nAn object is dropped from a height at time t = 0. If h(t) is the height of the object at time t, a(t) the acceleration and v(t) the velocity. The relationships between a, v and h are as follows:\n\na(t) = dv / dt , v(t) = dh / dt.\n\nFor a falling object, a(t) is constant and is equal to g = -9.8 m/s.\n\nCombining the above differential equations, we can easily deduce the follwoing equation\n\nd 2h / dt 2 = g\n\nIntegrate both sides of the above equation to obtain\n\ndh / dt = g t + v0\n\nIntegrate one more time to obtain\n\nh(t) = (1/2) g t + v0 t + h0\n\nThe above equation describes the height of a falling object, from an initial height h0 at an initial velocity v0, as a function of time.\n\nApplication 4 : Newton's Law of Cooling\nIt is a model that describes, mathematically, the change in temperature of an object in a given environment. The law states that the rate of change (in time) of the temperature is proportional to the difference between the temperature T of the object and the temperature Te of the environment surrounding the object.\n\nd T / d t = - k (T - Te)\n\n\nLet x = T - Te so that dx / dt = dT / dt\n\nUsing the above change of variable, the above differential equation becomes\n\nd x / d t = - k x\n\n\nThe solution to the above differential equation is given by\n\nx = A e - k t\n\nsubstitute x by T - Te\n\nT - Te = A e - k t\n\nAssume that at t = 0 the tempearture T = To\n\nTo - Te = A e 0\n\nwhich gives A = To - Te\n\nThe final expression for T(t) i given by\n\nT(t) = Te + (To - Te)e - k t\n\nThis last expression shows how the tempearture T of the object changes with time.\n\nApplication 5 : RL circuit\nrl circuit for application 5\nLet us consider the RL (resistor R and inductor L) circuit shown above. At t = 0 the switch is closed and current passes through the circuit. Electricty laws state that the voltage across a resistor of resistance R is equal to R i and the voltage across an inductor L is given by L di/dt (i is the current). Another law gives an equation relating all voltages in the above circuit as follows:\n\nL di/dt + Ri = E , where E is a constant voltage.\n\nLet us solve the above differential equation which may be written as follows\n\nL [ di / dt ] / [E - R i] = 1\n\nwhich may be written as\n\n- (L / R) [ - R d i ] / [E - Ri] = dt\n\nIntegrate both sides\n\n- (L / R) ln(E - R i) = t + c , c constant of integration.\n\nFind constant c by setting i = 0 at t = 0 (when switch is closed) which gives\n\nc = (-L / R) ln(E)\n\nSubstitute c in the solution\n\n- (L / R) ln(E - R i) = t + (-L/R) ln (E)\n\nwhic
731490
Assuming the hepatitis does not kill the person, then their life span won't necessary be connect to the hepatitis.\nHealth and life span may be considered as independent factors.
45317
The simplest model of population growth is based on the straightforward quantities of four properties of population change:\n\nB - Births or Natality \nD - Deaths or Mortality \nI - for Immigration and \nE - for Emigration\n\n\n\nNt+1 = Nt + B - D + I - E\n\n[Equation 3.1]\n\n\n\n\nThe net rate at which the population is increasing or decreasing is called the net rate of population growth - Ro. The net rate of population growth is simply the ratio of the size of the population at some time Nt+1 compared to the size of the population during some the previous moment in time Nt: \n\n\n\n\n[Equation 3.2]\n\n \n\nIf birth rates (b) and death rates (d) in a population are determined for a specific time interval we can calculate another measure of population growth call the intrinsic rate of increase - r from these two population parameters: \n\n\nr = b - d\n\n[Equation 3.3]\n\n\n\n\nThe rate of change of population size dN/dt is then given by the differential equation: \n\n\n\n\n[Equation 3.4]\n\n\n\n\nwhere N is the size of the population at a particular instant in time. To obtain the size of the population Nt at some time t, is integrated to give: \n\n\n\n\n[Equation 3.5]\n\n\n\n\nwhere No is the initial size of the population at some time designated zero, e is a constant = 2.718 (the base of natural logarithms), r is the intrinsic rate of increase and t is the time elapsed since zero.\n\nThe population growth described by equation [3.5] is geometric, which means the population size goes on doubling (or halving if d is greater than b) at a constant rate. This situation is clearly unrealistic, since the growth of all species populations must be limited by the availability of some finite resource.
1143698
No,dear..It's not possible..Unfortunately,Hepatitus B and C are very resistant and there is only one option,in the first stage of the infection,if the immune system of the patient can cope by itself with it. It's not probable to be reversed the both forms and there are specific treatment,most of them based on Interferone..
935430
Birth -- an infected mother can transmit the Hepatitis B virus and Hepatitis C virus to the baby during delivery or shortly thereafter. \n Infants born of mothers who either currently have acute hepatitis B or who have had the infection receive a special immunization series to prevent viral transmission. This includes administering hepatitis B immune globulin and a hepatitis B immunization within 12 hours of birth. Please consult your doctor.\nPlease see the webpages for more details on Hepatitis.
52075
Some people can clear the virus naturally (immune system) and have evidence of past infection, but do not have ongoing viral replication (detectable hepatitis viral load). In fact, most people infected as adults are able to clear the virus (about 85% clear the infection when infected as an adult). Newborns and children are often unable to clear the infection, leading to risk of disease progression.\n\nThere are vaccines available for Hepatitis B so the numbers of new infections are declining.\n\nThere are treatments for Hepatitis B - these include Pegasys, a pegylated interferon medicine that is injected under the skin once weekly or oral agents. The oral agents include Entecavir (Baraclude) or Adefovir (Hepsera). A new agent is being studied - Telbivudine.\n\nThere are different types of Hepatitis B: eAg-positive and eAg-negative. This impacts how long patients receive treatment. Typically eAg-positive patients are treated until the virus is no longer detectable plus about 6-12 months and then the medicine can be stopped. eAg-negative patients usually are treated for life.\n\nTalk with your doctor to find more information.
575342
There are currently several treatments for chronic hepatitis B that can increase a person's chance of clearing the infection. Treatments are available in the form of antivirals such as lamivudine and adefovir and immune system modulators such as interferon alpha. There are several other antivirals under investigation. Roughly, all of the currently available treatments, when used alone, are about equally efficacious. However, some individuals are much more likely to respond than others. It does not appear that combination therapy offers any advantages.[4] In general, each works by reducing the viral load by several orders of magnitude thus helping a body's immune system clear the infection. Treatment strategies should be individualized by a doctor and patient. Considerations include the risks associated with each treatment, a person's likelihood of clearing the virus with treatment, a person's risk for developing complications of persistent infection, and development of viral resistance with some of the treatments.\n\nOn March 29, 2005, the US FDA approved Entecavir for the treatment of hepatitis B.\n\nOn February 25, 2005, the EU Commission approved PEGASYS for the treatment of hepatitis B making it the first pegylated interferon to be approved for hepatitis B.\n\nChronic carriers should be strongly encouraged to avoid consuming alcohol as it increases their risk for cirrhosis and hepatocellular carcinoma (liver cancer).\n\nInfants born to mothers known to carry hepatitis B can be treated with antibodies to the hepatitis B virus (hepatitis B immune globulin or HBIg). When given with the vaccine within twelve hours of birth, the risk of acquiring hepatitis B is reduced 95%. This treatment also allows a mother to safely breastfeed her child.\n\nAn individual exposed to the virus who has never been vaccinated may be treated with HBIg immediately following the exposure. For instance, a health care worker accidentally stuck by a needle used in a hepatitis B carrier would qualify. Treatment must be soon after exposure, however.
883446
Hi, it depends on particular organisms. Check the link below for basic mathematical formulas you may be able to use.\n\nA study is done in which the number of bacteria in a liquid culture is monitored over a period of several hours. Ten observations were made:\n\n\nTwo questions need to be answered:\n\nIs this data consistent with an exponential growth model: N = N0ekt, where \nN0 = the number of bacteria present initially (at t = 0)\nk = a growth rate constant\nN = the number of bacteria present at a given time t,\nand e is the fundamental mathematical constant forming the base for natural logarithms. \nIf the answer to the first question is "yes", then what are the values of N0, k, and t2 (the so-called doubling time) for this bacterial system?\nThe experimental measurements listed above indicate the number of bacteria are growing with time. If we plot the observed values of N (vertically) against the elapsed time, t, in hours, we get a graph (available on web page below).\n\nYou can draw a smooth curve through the experimental points as shown in this figure, and the points all seem to lie relatively close to this curve. Further, this curve looks quite a bit like the graph of an exponential function, but from what we see, it might just as well be part of a parabola or of one of a number of other curves that result when various algebraic forms are graphed. This graph suggests that the answer to the first question may be "yes", but we would usually want stronger evidence than this to draw a definite conclusion. The trouble here is that it is difficult to distinguish precisely and reliably between the curves are the graphs of quite different mathematical functions.\n\nThere is, of course, one curve that is easily recognized by everyone - the straight line! In this situation, the logarithmic function can be used to rewrite the exponential growth formula into a form that will allow us to create a plot that does give a straight line when data is the result of an exponential growth (or decay) process. Start with the exponential growth formula:\n\nN = N0 ekt\n\nSince this involves an exponential with base e, it is "natural" to use natural logarithms - those based on the number e. Using the more common notation ‘ln’ for ‘loge’, we simply take the logarithm of both sides of the previous equation:\n\nln N = ln(N0 ekt)\n\n= ln N0 + ln(ekt)\n\n= ln N0 + kt ln e\n\n= ln N0 + kt\n\nHere, we have used two of the three basic properties of logarithms to manipulate the right-hand side, namely that\n\nlog (AB) = log A + log B \n\nlog (Am) = m log A \n\nFurther, in the last step we took advantage of the fact that by choosing to work with natural logarithms, ln e = logee = 1. Thus, the exponential growth formula can be written in the logarithmic form\n\nln N = kt + ln N0\n\nIf the exponential growth formula is correct and we were to plot ln N along the vertical axis of our graph (rather than N itself), and t along the horizontal axis, then the equation of the plot we are preparing will have the generic form\n\nvertical quantity = constant x horizontal quantity + another constant \n\nBut this is just the equation of a straight line. The plot that results using the data given above is available on web site below.\n\nAlthough the values written on the vertical axis are values of N, this graph is ruled vertically so that the vertical distances are actually proportional to ln N. Since one scale is logarithmic (the vertical one) whereas the other graph scale, the horizontal one, is the ordinary uniform spacing type, this is called a ‘semilogarithmic plot’.\n\nAt any rate, the figure indicates that the points are scattered about a straight line, the scatter due to small random errors in measurement. Since the plot of ln N vs t gives a straight line, we have confirmed that\n\nln N = kt + ln N0 (equation *) \n\nis a valid formula for this data, and therefore, so is its equivalent, N = N0 ekt, valid for this data.
769374
This sounds like it came out of a textbook. What is the name of the various propagation models?
736908
Hepatitis B is highly sexually transmitted. It is very possible to pass it onto the mother, who could in turn pass it onto the child. There is post exposure medication that could be given, but it would have to be given within a certain time frame.  The mother would have to be tested to see if she has HBV, and then to see if it is chronic, or if it resolved on it's own.  80% of people who become exposed to it clear it on their own. The rest go on to have chronic liver disease and will need treatment.  Treatment can help people clear it, but I'm not sure what the success rate is (I'm more familiar with Hep C statistics). Best wishes.
1013268
If you are looking for a treatment for Hep B\n\ncheck this link\n\nhttp://www.hivandhepatitis.com/hep_b/treatment.html
375178
Hepatitis B surface antigen (HBsAg) is a protein antigen produced by HBV. This antigen is the earliest indicator of acute hepatitis B and frequently identifies infected people before symptoms appear. HBsAg disappears from the blood during the recovery period. In some people (particularly those infected as children or those with a weak immune system, such as those with AIDS), chronic infection with HBV may occur and HBsAg remains positive. \nSometimes, HBV goes into “hiding” in the liver and other cells and does not produce new viruses that can infect others, or produces them in such low amounts that they cannot be found in the blood. People who have this form are said to be carriers. In other cases, the body continues to make viruses that can further infect the liver and can be spread to other people. In both these cases, HBsAg will be positive. The next test is helpful for distinguishing these two states. \n\nHepatitis B e-antigen (HBeAg) is a viral protein associated with HBV infections. Unlike the surface antigen, the e-antigen is found in the blood only when there are viruses also present. When the virus goes into “hiding,” the e-antigen will no longer be present in the blood. HBeAg is often used as a marker of ability to spread the virus to other people (infectivity). Measurement of e-antigen may also be used to monitor the effectiveness of HBV treatment; successful treatment will usually eliminate HBeAg from the blood and lead to development of antibodies against e-antigen (anti-HBe). There are some types (strains) of HBV that do not make e-antigen; these are especially common in the Middle East and Asia. In areas where these strains of HBV are common, testing for HBeAg is not very useful. \n\nAnti-HBe is an antibody produced in response to the Hepatitis B e antigen. In those who have recovered from acute hepatitis B infection, anti-HBe will be present along with anti-HBc and anti-HBs. In those with chronic hepatitis B, usually anti-HBe becomes positive when the virus goes into hiding or is eliminated from the body. In strains that do not make HBe antigen, anti-HBe is also positive.\nwww.labtestsonline.org/understanding/analytes/hepatitis_b/test.html \nSo from this and my slight background knowledge, it depends on your Hepatitis B e-antigen (HBeAg) reveals.\n\nDiscuss your concerns with your MD.
356040
elementary mode analysis Is also known as extreem path way analysis. It defines the limiting bondaries of the operation of a metabolic network. The normal approach is to use singular value decomposition to determine the constraints of a metabolic network.\n\nTo be honest its a bit over rated since all it does is define the feasable space of operation of a cell. You have to use other principles such as cyberbnetics (maximise growth rate and then use linear programming) or kinetic modeling to determine the expected operation of a cell. \n\nTry a search for Palson - but dont cite or talk to him because he is an arrogant self promoting w****r who doesnt cite competing groups work.
336891
It's complex.  Basically the virus infects cells, and the viral protein is present on the cell surface by MHC molecules.  The circulating T cells then can latch on to and identify the virus-infected cell as foreign.  That specific T-cell then proliferates, and starts hunting down infected cells.  There is alot more going on, including antibody production and release by B cells.\n\nWell described in this book chapter.