What atom is the site of covalent attachment of amc to the model tetrapeptide used in the studies?.
1 answer:
The experiment referenced here involves the use of Y32P-ATP. The atom involved in the site of the covalent attachment is the N atom in Aminomethylcyclohexane (AMC).
<h3>The explanation in relation to the attachment of AMC to the Moel Tetrapeptide</h3>
Given that Aminomethylcyclohexane (AMC) had been linked around the carboxyl side, it, therefore, means that that type of linkage called the Amide-Linkage (That is N-acylation) can be utilized in the covalent bonding of the fluorophore into the peptide
See the link below for more about Covalent Attachements:
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Answer:
177.993 MeV
Explanation:
Nuclear fission refers to a process where a nucleus is broken into smaller ones.
m(Am) = 242.0595490 u, m(Sr) = 89.9077387 u, m(La) = 148.934733 u and m(n) = 1.008665 u. To find the mass of X, we sum up the masses on both side and subtract them according to the equation given
242 + X = 90 +149 + 4
X = 243 - 242 = 1
so X is a neutron
next we calculate the Δm ( change in mass)
Δm = mass reactant - mass of product
(242.0595490 + 1.008665 ) - ( 89.9077387 + 148.934733 + 4 (1.008665))
= 243.068214 - 242.8771317 = 0.1910823 u
using the formula
E = (Δm) c² = 0.1910823 u × c² × ×
= 177.993 MeV