The given question says that a student has constructed a model of cellular transport using fences and several gates.
This model can be used to demonstrate the cellular transport.
The gates of the fences can be supposed as the protein pumps and the other fence demonstrates the lipid bilayer.
Let’s suppose in the fence, there are many cattles, and outside, there are less cattles, but the student open the gate and bring more cattles inside the fence. In this case, the transport of the cattles is similar to the active transport of the molecules using protein pumps. At cellular level, the energy for the active transport is provided by ATP molecules.
Now, let’s say, the student wants to feed the cattles with some nutrition rich food, which can help in maintaining the health of the cattles. The student fills his car with the cattle food and he enters inside the fence through gates. In this case, the food was not present in the fence, but was abundant in the outside environment, so, the diffusion would occur. But food cannot come self, without help of others, so, the movement is facilitated by the car, as it is done by the carrier proteins. Hence, it is an example of facilitated diffusion.
Domain (Specifically domain Eukarya)
Answer:
Spotted Eagle Rays have two sets of five gills on each side of their body on the ventral side. In order to breathe, A. narinari must be in continuous swimming motion so that oxygen can be absorbed from the water passing through the gills
Explanation:
Answer: The estimated population is 1250 mice
Explanation: The method use was marked and recapture, in which individuals are marked in the first capture and after some time biologist trap a new group of individuals that can be or not marked
Whit this data is possible to estimate the size of a population applying the Peterson method but is important to make some considerations such as:
1. All indivuals have the same probability to be capture
2, The population remain constant in terms of birth and death rate.
Taking this into account, the formula that allows to determine the size of the population is:
N=CM / R
where N is the size of the population, C is number of indivuals trapped in recapture, M is number of individuals marked in the first capture and R is the number of marked animals trapped in recapture
In this case:
N = 250 * 200 / 40 = 1250
