A) energy is converted from one form to another
<u>Given:</u>
Surface area at the narrow end, A1 = 5.00 cm2
Force applied at the narrow end, F1 = 81.0 N
Surface area at the wide end, A2 = 725 cm2
<u>To determine:</u>
Force F2 applied at the wide end
<u>Explanation:</u>
Use the relation
F1/A1 = F2/A2
F2 = F1*A2/A1 = 81.0 N * 725 cm2/5.00 cm2 = 11,745 N
Ans: (b)
The force applied at the wide end = 11,745 N
Answer:
0.35 milli moles of ethanol can be theoretically be produced under these conditions.
Explanation:
Moles of glucose = 
milli mole
Moles of ADP = 0.35 milli mole
Moles of Pi = 0.35 milli mole
Moles of ATP = 0.70 milli mole
As we can see that ADP and Pi are in limiting amount which means tat they are limiting reagent. So, the moles of ethanol produced will depend upon the moles of ADP and Pi.
According to reaction, 2 moles of ADP gives 2 moles of glucose.
Then 0.35 milli moles of ADp will give :
of ethanol
0.35 milli moles of ethanol can be theoretically be produced under these conditions.
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.
