Answer:
thermodynamically unstable but kinetically stable.
Explanation:
The complete question is as follows:
Under physiological conditions, peptide bond formation and degradation both require enzymes, but only formation requires coupling to GTP hydrolysis. Based on this information, peptide bonds under physiological conditions are:
A. both thermodynamically and kinetically stable.
B.thermodynamically unstable but kinetically stable.
C.thermodynamically stable but kinetically unstable.
D. both thermodynamically and kinetically unstable.
- The term thermodynamically unstable refers to the fact that the peptide bonds are prone to breakage under physiological conditions.
- The reason why one can conclude the thermodynamic instability of the peptide bonds under physiological condition is that there is a need for a source of energy i.e. GTP hydrolysis for the formation of the peptide bond.
- The fact that the breakage of peptide does not require any input of energy but the only formation does confirms the fact that under physiological conditions they are thermodynamically unstable.
- Even though they are thermodynamically unstable, they are kinetically stable because both the formation and degradation require enzymes.
- The function of enzymes is to decrease the activation energy and hence, increase the rate of reaction. This means that if the enzymes are absent the rate of breakage of peptide bonds would be really slow this points out to the fact that they are kinetically stable under physiological conditions.
Answer:
48g
Explanation:
Volume of rock=16cm^3 (1 ml=1 cm^3)
Density=3 g/cm^3
Mass=?
d=m/v
3g/cm^3=m/16cm^3
3g/cm^3*16cm^3 =m
Therefore, m= 48g
If you are asking if that is true or false, it is false because all fungus like protists are heterotrophs, not autotrophs
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