Question

Escherichia coli O157:H7 is a pathogenic strain of E. coli associated with consumption of raw or undercooked foods. When exposed to temperatures above its normal growth temperature (temperatures that do not kill it), it may change its membrane lipid composition to counteract the rise in temperature. Which of the following changes would be most likely?

Select one:

a. A greater proportion of the unsaturated 16-carbon palmitoleate

b. A greater proportion of the saturated 12-carbon laurate

c. Generating a thinner membrane

d. A greater proportion of the saturated 16-carbon palmitate

e. A greater proportion of the saturated 14-carbon myristate

Answer 1

Phospholipid membrane thinning is a process by which antimicrobials enter their cell membrane. So to keep the pathogenic E.coli intact membrane thinning is not the option. Also increase in membrane fluidity is a process by which cells can cope up better with adverse conditions. The increased fluidity helps the cells to insert the membrane proteins properly and allows easy diffusion of the phospholipids into the bilayer after synthesis. It allows proper replication of the cells by the proper distribution of membrane molecules. Thus increased fluidity of the membrane allows the pathogen to live better. This fluidity is brought about by the increase in the incorporation of unsaturated fatty acids in the bilayer. The saturated fatty acids make the membrane more stiff.

So the change made by the bacteria to counter changes in temperature was to incorporate a greater proportion of the unsaturated 16-carbon palmitoleate(option A).

The other options have saturated fatty acids.