Answer:
50 micro mol/min
Explanation:
So, we have the following parameters which are going to help us in solving this particular Question or problem;
=> For the wild type enzyme, the value for the Maximum velocity = 100 micromol/min.
=> For the mutant type enzyme, the value for the Maximum velocity = 1 micromol/min.
So, we can determine or calculate the value for the initial velocity for each of the enzyme type by using the formula below;
Initial velocity = (substrate concentration × maximum velocity) / substrate concentration + Km.
Therefore, for the wild type enzyme; the Initial velocity = (substrate concentration × maximum velocity) / substrate concentration + Km.
Initial velocity =( 10mM × 100micromol/min) ÷ ( 10mM + 10 mM ) = 50 micro mol/min.
Initial velocity for the wild type enzyme = 50 micro mol/min.
All of the liquid molecules in cell membranes amphipathic (or amphiphilic) -that is, they have a hydrophilic ("water loving") or polar end and a hydrophobic ("water fearing") or nonpolar end. The most abundant membrane lipids are the phospholipids. These have a polar head group and two hydrophobic hydrocarbon tails.
Hope that Helps :)
Answer:
c) There will be a net movement of salt from side B to side A
Explanation:
According to the given information, the solution at side A is hypotonic to the solution at side B. The separating membrane is permeable to salt and would allow the movement of salt from the hypertonic side B to the hypotonic side A. Transport of substances down their concentration gradient is a passive movement and occurs from the region of higher concentration of the substance to that of its lower concentration.
I have only heard a few phrases for spine
Backbone
Vertebrae (vertebral column)
Spinal column
