Answer:
One way in which complement activation destroys pathogens is by C3b binding to the surface of microbes, which then causes inflammation through histamine and heparin release.
Explanation:
C3b binds to the surface of microbes (opsonin), and functions as a component of C3 and C5 convertases while C3a stimulates inflammation.
The alternative pathway of complement activation is triggered by the deposition of C3b on the surface of a microbe. The microbe- bound C3b binds another protein called Factor B, which is then broken down by a plasma protease called Factor D to generate the Bb fragment.
This fragment remains attached to C3b, and the C3bBb complex functions as an enzyme, called C3 convertase, to break down more C3. The C3 convertase is stabilized by properdin, a positive regulator of the complement system.
As a result of this enzymatic activity, many more C3b and C3bBb molecules are produced and become attached to the microbe. Some of the C3bBb molecules bind an additional C3b molecule, and the resulting C3bBb3b complexes function as C5 convertases, to break down the complement protein C5 and initiate the late steps of complement activation.
The main effectors of the complement system are opsonization, cell lysis and inflammation. It also stimulates B cell responses and antibody production.
Answer:
<em>The correct option is D) Ethyl alcohol is produced.</em>
Explanation:
Fermentation can be described as a process by which energy is extracted by organisms such as bacteria and yeast.
Sugars are converted into alcohol by this metabolic process occurring in certain microorganisms. There are two main types of fermentation:
Fermentation which takes place in the presence of oxygen is termed as aerobic fermentation.
The process of fermentation which occurs in the absence of oxygen is termed as anaerobic respiration.
Hence, as alcohol will be produced from both types of fermentation, optic C is correct.
Formula: Molarity = <span>Amount
of substance (in moles)/Volume of solution (in litres)</span>
40.0⋅g<span>/40</span>⋅<span>g⋅</span><span><span>mol</span><span>−1</span></span> ÷ (1.50⋅L) = 0.667 <span>−1</span>
sodium hydroxide.
The Molarity of
40 grams of Sodium Hydroxide in 1.50L of solution is 0.667 mol*L-1
<span> </span>
Answer:
The results show that the increase in the substrate concentration is increasing the rate of the reaction. We can see that the reaction rate has increased to 2.88mM/min. After this even if we increase the substrate concentration
<span>An animal with these such features would be well suited to live in environment like the rain forest. Its long limbs and tail would allow it to swing from tree to tree, and also reach up into the trees to gather food. The hand-like paws would make climbing easier, and allow the animals to grab food from higher places.</span>
