bacterial growth curve?
476381
Actually, your teacher is wrong.  It all depends on the growth phase of the bacterial culture.   There are 3 phases in the growth of a bacterial culture; the lag phase, the log phase, and the stationary phase.  The lag phase is exactly what it describes...the bacteria divide veeeeerrrryyy slowly.  In simple terms, this is mainly due to low amounts of certain "growth signals".  It usually takes bacteria several hours to get out of lag phase and into log phase.  \n\nLog phase is the stage where the bacteria are dividing at a constant rate.  This is when you're going to take measurements for the growth rate.  Say, for example, that your cells have a doubling time of 1hr.  So if your beginning OD is 0.8 your end OD after an hour would be 1.6.  In this case B is twice b.  However using the same example, if we grow for 2 hours instead of 1, you've gone thorugh 2 doubling times and the final OD would be 3.2 verses 0.8 (B is 4x b).  So the relation between b and B is strictly dependant on the time (which is why "t" is a component of the equation) and growth rate of the cells.\n\nThe final phase of bacterial growth is the stationary phase.  This is where the number of bacteria increases to such a high density that further division is inhibited (by physical limitations such as nutrients, pH, cell/cell inhibition, etc) and the growth substatially slows.\n\nI hope this clearly answers your question.
943178
A = A_(0)*exp(k*t) \n\nA represents the amount at a given time t.\n\nAo represents the initial amount of the growing entity.  Note that this is the amount when t = 0.\n\n\nk is a constant that represents the growth rate.  It is POSITIVE when talking in terms of exponential GROWTH.\n\n\nt is the amount of time that has past.  If the information for time is given in dates, you need to convert it to how much time has past since the initial time.  For example, if the model is set up at an initial year of 2000 and you need to find out what the value is in the year 2010, t would be 2010 - 2000 = 10 years.\n\nYou can use this formula to find any of its variables, depending on the information given and what is being asked in a problem.  For example, you may be given the values for Ao and t and you need to find the amount A after the given time.  Or, you may be given the final amount A and the initial amount Ao and you need to find the time t.\n\nSome examples of exponential growth are population growth and financial growth.  The information found, can help predict what a population for a city or colony would be in the future or what the value of your house is in ten years.  Or you can use it to find out how long it would take to get to a certain population or value on your house.
258097
Is a simple culture enough?\nthe anerobic won't grow in air the aerobic will grow and give off O2. Yeasts will give off CO2.\nYou may need an autoclave to maintain temperature.\n{sorry not to be more helpful}
806438
The exponential formula for growth would be\n\nn(t) = n(0)* e^(t/tc), where n0 is the initial number of cells, and tc the "time constant" for growth.  This can also be written\n\nn(t)/n(0) = e^(t/tc)\n\nYou are given that when t = 6mo, n(6)/n(0) = 2\n\ntherefore 2 = e^(6/tc);   solve this for tc:\n\nln(2) = 6/tc, and tc  =  6/ln(2) =  8.656\n\nThe equation is now n(t)/n(0) = e^t*/8.656\n\nwhere t is in months.   For t = 3 years, that is 36 months, so n(36)/n(0) = e^(36/8.656) = 64.0   For t = 6 years, or 72 months n(72) = e^(72/8.656) = 4.10*10^3
741317
This link is one of the better microbiology websites that I've used.\n\nhttp://www.textbookofbacteriology.net/
13245
It depends on the bacterium. For E. coli (common lab bacteria) 20-30 minutes is an average doubling time, but different strains vary.\n\nI heard something somewhere about a weird form of bacteria that lives miles underground in granite formations and only divides once every ten thousand years, or something crazy like that. I can't give you a source, it's just a freaky thing off the top of my head that I haven't gone to the trouble to confirm. My contention is that if reincarnation is true, that would be the *worst* thing to come back as. So mind your karma.\n\nPart of the reason it would be the worst is that bacteria reproduce by one cell dividing into two, so as long as there are any of that strain still alive, it hasn't really died.\n\nThey can die though. In a liquid culture you can tell because of a lot of turbidity (cloudiness) some of which is from cells and some from debris from dead cells. Or, you could have agar plates that get really nasty and dried up, and most of those bacteria are probably dead. Or you could tell because they look really crappy under a microscope. If you try to streak it and grow it on a plate and it doesn't grow, it's probably dead.
338000
I only had time to do the first one, btu the answer is \ne-2.  Check your email for the work.
255920
The factor of time dilation depends on the square root of (1-v^2/c^2). Taking the speed of light as 1 and the speed of the bacteria as .9950, the time dilation factor is\n\nsqrt(1 - .9950^2) = just slightly less than .1\n\nThus, time slows for the bacteria and they think they are in space for 100 days, even though 1000 days pass on earth. They will double 5 times, once every one of "their" 20 days, to 4, 8, 16, 32, and finally 64 upon their return to earth.
913686
This is a simple ratio problem. The equation would be Dgalaxy/Dearth=Dearth/x so x=(Dearth)^2/Dgalaxy. Asssuming a parsec is 3.6 light years, and a light year is 5.87 trillion miles and the diameter of the Earth is 8,000 miles, then the Earth would be 0.001 millimeter or one micron in size, which is the size of an average bacterium, maybe? Very cool. Oops! I re-did my calculation and I see that I made a factor of ten error. The correct answer is about 0.1 microns, or the size of some viruses. But tell me how you would make such a model of the Earth?
869195
I will assume because it is not clear that the number of bacteria is given by the formula:\n\nf(t) = 720*(3/2)^t\nWhen t = 0, this formula will give\nf(0) = 720 * (3/2)^0 = 720 * 1 = 720, which is consistent with the data.  I will also assume that t is given in hours. \n\nAt 11 am t = 3 hours \n\nPutting the time in the formula:\n\nf(3) = 720 * 1.5^3 = 2430\n\nIf the formula is different from what I understood, just put t=3 in the right formula.
100001
First you have to determine the bacteria strain.
687306
Check the Michaelis-Menten equation.It's too long to write it all down for you ;)
964800
Bacteria is constantly mutating. Sooner or later it'll mutate into a strain that is resistant to antibiotics
96543
A boy was 1.4 meter.\nLater his length was 1.5 meter.\nHis abolute grow was 0.1 meter\nHis relative grow was 0.1 m / 1.4 m = 0.071 = 7.1%.
86677
Okay, hopefully my quick math is correct....\n\nApril 1st we have 47 rats\nJune 1st we have 900 rats\n\nGrowth model N=No e^k*t  could be re written as:\n\nln(N/No)=k*t\n\nln = natural logarithm\nN = End result of rats\nNo = Beginning amount of rats\nk = growth constant\nt = time\n\nSo, our equation we have \n\nln(900/47) = k*61   (61 is the days from April 1 - June 1)\n\nWhich gives us a growth model of\n\nN=No e^(0.04839*t)\n\nAlrighty...From April 1 - Dec 1 it's approx 240 days\n\nPlugin the numbers  x=amount of rats on Dec 1\n\nln(x/47)=0.04839*240 = 5,197,808.236\n\nWe get approx. 5,197,808 rats on Dec 1st  (alot of rats)\n\nNow you ask when will there be 10000 rats...we plug in our numbers....    t=how many days it'll take\n\nln(10000/47)=0.04839*t = 110.7706 days\n\nSo, approx 111 days from April 1st to reach 10000 rats
924623
If you mean how do you culture, pure cultures of bacteria, then one of the easiest ways is to get a culture plate of generic media (broad appeal media, not specific to certain bacteria types), clean a nichrome wire in a flame, dip it into whatever has bacteria you wanna try and culture (old mayonnaise, rotten meat, lake water), streak (run it back and forth along one side of the media),  clean the wire in the flame, alter the media by 90 degs, streak again, making sure the wire passes into the area previously streaked a few times, repeat one more time, place the lid on the plate, turn upside down and place in a incubator for 24 to 48 hours. Pure cultures (those grown from a single bacterium) will have grown up in the second and third streaking areas.
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