Answer:
Number of peptide fragments resulting from cleaving with cyanogen bromide? A: Three peptide fragments
Number of peptide fragments resulting from cleaving with trypsin? A: Four peptide fragments
Which of these reagents gives the smallest single fragment (in number of amino acid residues)? A: CnBr, a dipeptide fragment consisting of AL (Alanine-Leucine)
Explanation:
Cyanogen bromide cleaves the methionine C-terminus, then we have a first fragment of 8 amino acids: DSRLSKTM, a second fragment of 15 aas YSIEAPAKLDWEQNM, and a last fragment of only 2 aas is produced, AL
Trypsin cuts the C-terminus of Arginine and Lysine, then we'll have a first fragment of 3 aas DSR, a second fragment consisting of also 3 aas LSK, a third fragment of 10 aas TMYSIEAPAK, and a last fragment of 9 aas LDWEQNMAL. All produced in three cut sites.
I believe that what happens here is that when
those cells carrying those intercellular signals do no pass through the
connections, actually the signal is passed through or are transferred to adjacent
cells for signal transmission, hence the answer is:
<span>B</span>
Answer:
Explanation:
ΔG° = - 8.03 kJ/mol & ΔG = -45.28 kJ/mol
Thus:
Overall ΔG = (- 8.03 + 45.28) kJ/mol
ΔG = 37.25 kJ/mol
ΔG = 37250 J/mol
Temperature = 27 °C
= (27 + 273)K
= 300 K
Therefore :
ΔG = -RTInK
In K =
In K =
In K =
In K = 14. 93
K =
K =
the standard free energy of the first reaction at 27 °C =