Answer:
C. parvum takes energy from glucose which is present in the digestive tract after the process of glycolysis.
Lactate dehydrogenase which is responsible for the conversion of lactate into pyruvate molecule.
Explanation:
C. parvum is a protozoa that lives as a parasite in the digestive tract of animals. They take nutrients from the cell which are present in the form of glucose. C. parvum uses a specific type of enzyme i. e. lactate dehydrogenase which is responsible for the conversion of lactate into pyruvate and also helps in the production of ATP through glycolysis process. In this process, the glucose molecule is broken down into two molecules of pyruvate, two molecules of ATP, two molecules of NADH, and two molecules of water. So C. parvum takes ATP from that way from the host cells.
Vacuoles store water, food, and waste. The endoplasmic reticulum (ER) is a series of tunnels throughout the cytoplasm. They transport proteins from one part of the cell to another. Ribosomes are the protein factories of the cell.
<h3>Hope it helps uh...</h3>
The right answer is the activation energy.
The catalyst increases the reaction rate by introducing new reaction paths (mechanism), and lowering its activation energy, or activation Gibbs free energy. By doing this it can increase the speed, or lower the temperature of the reaction. It is important to note that the catalyst does not alter the total Gibbs free energy of the reaction which is a system state function and therefore has no effect on the equilibrium constant.
Hey there mate!
Based on my information to this, I believe sense (it would be a closed system), this would most likely be considered to be (equal to) the total energy after.
But, if it was not a (closed system), it would not be unrelated because the energetic system would not be that greatly increased.
So, it really depends whether the system closes up, or whether it would be free of energy.
So, in this case, your correct option to this answer would be (option A).
<span>In a closed system, the total energy prior to an energy transformation is unrelated to the total energy after.
I hope this helps you!</span>
The most appropriate microscope for studying the internal parts of the cell is ELECTRON MICROSCOPE. The correct option is B.
Electron microscope, which is a compound microscope is usually used to study the internal environment of the cell because of the easiness with which one can easily distinguish the cell organelles from one another. Electron microscope gives great clarification of the cellular components and allow scientists to observe minute details about cellular structures. Electron microscope usually magnify specimens to one million times of their real sizes.
