Answer:
b. Increase the proportion of unsaturated hydrocarbon tails in its membrane phospholipids.
Explanation:
A bacterial cell in the warm human intestine that has been expelled into the cold world outside has to increase the proportion of unsaturated hydrocarbon tails in its membrane phospholipids to be able to maintain the level of membrane fluidity it experienced in the intestine.
This is because membrane phospholipids are unsaturated fatty acids that have more double bonds and a shorter hydrocarbon tail which prevents them from been tightly packed together thereby increasing the fluidity of the phospholipids in the bacterial cell.
Answer:
b. increase the proportion of unsaturated hydrocarbon tails in its membrane phospholipids.
Explanation:
Fluidity of the plasma membrane depends on the fluctuations of the structure of its cholesterol, length of the fatty acid chains and the extent of saturation.
Thus fluidity is encouraged by;
Increase in unsaturation of the fatty acids tails. The more unsaturated the chain, the' more bent' or 'kink' they are and therefore the less aligned, and therefore more fluidity. Since unsaturation favours fluidity the answer is B.
The short the fatty acid tails; shorter tails are less packaged together they therefore promote fluidity. longer chains gives close packing encouraging rigidity.
Option c, is reducing unsaturation,but promoting saturation which does not favour rigidity, therefore very wrong.
option d, is wrong because increasing cholesterol levels promotes fluidity by preventing close packing. Therefore, decreasing it favours rigidity.
