Explain the difference in size between Plasmodium and Amoeba by comparing their lifestyles?
They are asking about how their lifestyle (what they eat, where they live) effects their size. If you look up Amoeba and Plasmodium in Wikipedia you will fast see that Ameoba engulf their prey and Plasmodium such as malaria act as parasites living within a cell of the host. It should be simple to explain why these differences would necessitate a difference in size. Hope that helps.
what are the signs of amoeba? does the local park lakes test for it everyday?
Amoeba can be found in many places,I doubt whether they would test waterways etc unless an actual outbreak of illness occured nearby.Read this link article,it will give you details,but get your son off to the doctor to be sure.
How does the Amoeba use its pshyical features to live in its particular habitat and life style?
Amoebas do indeed live in different aquatic habitats. Its main adaptation is called a contractile vacuole. This organelle allows the amoeba to "pump" in or out water, depending on its needs. If the amoeba is placed in an isotonic solution, or the concentration of solute outside is the same as the amoeba, then nothing will happen. When placed into a hypertonic solution, say, saltwater, then water will rush into the amoeba and it will pump water out using the contractile vacuole.
If p is the probability that a single amoeba's descendants will die out eventually, the probability that N amoebas' descendants will all die out eventually must be p^N, since each amoeba is independent of every other amoeba. Also, the probability that a single amoeba's descendants will die out must be independent of time when averaged over all the possibilities. At t=0, the probability is p, at t=1 the probability is 0.25(p^0+p^1+p^2+p^3), and these probabilities must be equal. Extinction probability p is a root of f(p)=p. In this case, p = sqrt(2)-1.
The generating function for the sequence P(n,i), which gives the probability of i amoebas after n minutes, is f^n(x), where f^n(x) == f^(n-1) ( f(x) ), f^0(x) == x . That is, f^n is the nth composition of f with itself.
Then f^n(0) gives the probability of 0 amoebas after n minutes, since f^n(0) = P(n,0). We then note that:
f^(n+1)(x) = ( 1 + f^n(x) + (f^n(x))^2 + (f^n(x))^3 )/4
so that if f^(n+1)(0) -> f^n(0) we can solve the equation.
The generating function also gives an expression for the expectation value of the number of amoebas after n minutes. This is d/dx(f^n(x)) evaluated at x=1. Using the chain rule we get f'(f^(n-1)(x))*d/dx(f^(n-1)(x)) and since f'(1) = 1.5 and f(1) = 1, we see that the result is just 1.5^n, as might be expected.