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.
1. Action potential reaches the axon terminal and depolarizes it.
2. Depolarization opens voltage-gated calcium channels, enabling influx of Ca into the neuron.
3. Calcium binds to specialized proteins on vesicles (containing pre-made acetylcholine) and triggers them to fuse with the neuron membrane at the synapse.
4. Exocytosis of acetylcholine into the synaptic cleft occurs.
5. Acetylcholine diffuses across the synapse and binds to nicotinic receptors on the end plate of the myocyte.
6. Activated nicotinic receptors, themselves ion channels, cause cation influx into the myocyte and generate an end plate potential. This eventually gives rise to the full depolarization within the myocyte that enables contraction.
Answer:
The correct answer would be -
Membrane A - Hypotonic solution - the movement of water towards inside the cell
Membrane B - Isotonic soltion - there will be no movement of water
Membrane C - Hyertonic solution - the movement of water towards outside of the cell
Explanation:
This experiment deals with tonicity as this solution will affect the tonicity of the egg membrane. In membrane A there are more solutes inside the cell than outside the cell so it is hypotonic solution so the movement of water will be towards inside the cell.
In membrane B the solutes are equal in both sides so there will be no movement as its isotonic condition while in membrane C the solution is in hypertonic situation as the solutes are more outside than inside.
Thus, the behavior of the membranes are-
Membrane A - Hypotonic solution - the movement of water towards inside the cell
Membrane B - Isotonic soltion - there will be no movement of water
Membrane C - Hyertonic solution - the movement of water towards outside of the cell
<span>A joint united by dense fibrocartilaginous tissue that usually permits a slight degree of movement is a symphysis.</span>
Answer:
E. 1/600
Explanation:
Hint:
The probability of fixation of a new neutral mutation is 1/(2N)
Given N as 300
= 1/(2×300)
=1/600
Therefore,
1/600 gives a sure fixation of one allele from the large population
