ATP or the Adenosine Triphosphate is referred to as the basic energy unit of the body. In the case of doing an exercise, what happens is that, the more someone increases his or her activity, the more energy is required and the body should compensate for this need. Therefore, the answer for this would be option D.
The double fortification process involves adding iodine and iron to salt. It is a method used to fight micronutrient deficiencies in developing countries. Iron and iodine are two of the most important micronutrients involved in cognitive function, maternal and infant survival and human productivity. This is a cost-effective method that ensures that the population receives these nutrients without having to change their eating habits.
Answer:
The correct options are MACROPHAGES AND NEUTROPHILSE.
Explanation:
Majority of the white blood cells in humans are specialized phagocytic cells, examples of these are macrophages, neutrophilse, monoctyes, mast cells and dendritic cells. The major functions of phagocytic cells is to protect the human body from disease pathogens. They do this by ingesting foreign bodies that are found in the body. Macrophages and neutrophilse are the major phagocytic cells in the body, they are the principal effector of non-specific host defense and inflammation.
Answer: D. Digestion increases as the volume of the enzyme increases.
Explanation:
Enzymes speed up reactions by reducing the activation time of a reaction. If there are more enzymes therefore, the reaction will move faster.
Pepsin is an enzyme that prefers acidic conditions so it can work with dilute HCL. It can also work in temperatures of 40 °C without getting denatured. As more Pepsin is added therefore, the reaction will move faster and digestion will increase.
Answer:
Null Hypothesis -
The observed frequency is approximately equal to the expected frequency of phenotype.
Explanation:
Pure Breeding Cross - TTww x ttWW
Genotype of offspring in F2 generation - TtWw
Null Hypothesis -
The observed frequency is approximately equal to the expected frequency of phenotype.
The chi square analysis is attached
The degree of freedom for this question is 3
The p value for X^2 estimated through chi square test is 0.5
Hence the null hypothesis is accepted.
