BCHE 5180/6180: Spring Final
(Graded A+)
1 Based on what you have learned about the functioning of enzymes and proteins,
3 points explain why the pH in the cell has to be buffered.
A large group of enzymes show a pH optimum for their activity. This value is
normally close to the pH in the cell. When the pH in the cell changes this can lead to a
protonation or deprotonation of an amino acid that is important for the enzymatic
activity. As a result the activity will decrease. Upon large changes in pH additional
amino acids can be protonated or deprotonated, which can cause the loss of important
ionic interactions and loss of secondary and tertiary structure. Most enzymes and
proteins precipitate at the more extreme pH values.
2 In addition to the pH, also the redox potential needs to stay constant in the cells.
4 points
(a) What species are formed that can higher the redox potential of the cytosol?
(b) What species are involved in buffering the potential and remove these species?
(c) What role does the pentose phosphate pathway play in the removal of these
species?
a) The main species that can shift the potential to a higher more positive value are
oxygen, superoxide, hydrogen peroxide and hydroxyl radicals. In particular the
last three, the so-called reactive oxygen species (ROS) are very dangerous to the
cell if their respective concentration increase too much. Iron-sulfur-cluster-
containing proteins get damaged. Regulatory disulfide sites get oxidized. DNA
gets damaged and eventually the cell will die.
b) Several compounds are present in the cell that can buffer the potential and/or
directly remove ROS. All cells contain glutathione and thioredoxins. Plant
cells also contain ascorbate. In addition, specific proteins will remove the ROS,
like catalase and hydrogen peroxide.
c) Reduced glutathione (GSH) protects the cell by destroying hydroxyl free radicals.
It is also the cosubstrate for glutathione peroxidase that removes hydrogen
peroxide. In both cases the gluthathione gets oxidized (GSSG). Regeneration of
GSH from its oxidized form (GSSG) requires the NADPH produced in the
glucose 6-phosphate dehydrogenase reaction. This reaction is catalyzed by
gluthathione reductase
, 3 For a peptide with the sequence shown below determine the pI.
4 points
Glu–His–Trp–Ser–Gly–Leu–Arg–Pro–Gly
(Hint: If you cannot do that directly, first calculate the charge at different pH values and
use those to find the pI region.)
pH N-term Glu His Arg C-term Net Charge
2 +
H3N– –COOH >NH+ –NH3+ –COOH
+1 0 +1 +1 0 +3
5 +
H3N– –COOˉ >NH+ –NH3+ –COOˉ
+1 -1 +1 +1 -1 +1
7 +
H3N– –COOˉ >N –NH3+ –COOˉ
+1 -1 0 +1 -1 0
10 H2N– –COOˉ >N –NH3+ –COOˉ
0 -1 0 +1 -1 -1
Find the two ionizable groups with pKa values that “straddle” the point at which net
peptide charge = 0 (here, two groups that ionize near pH 8): the amino-terminal α-
amino group of Glu and the His imidazole group.
Thus, we can estimate pI = (8.0 + 6.00)/2 = 7.0
2
(Graded A+)
1 Based on what you have learned about the functioning of enzymes and proteins,
3 points explain why the pH in the cell has to be buffered.
A large group of enzymes show a pH optimum for their activity. This value is
normally close to the pH in the cell. When the pH in the cell changes this can lead to a
protonation or deprotonation of an amino acid that is important for the enzymatic
activity. As a result the activity will decrease. Upon large changes in pH additional
amino acids can be protonated or deprotonated, which can cause the loss of important
ionic interactions and loss of secondary and tertiary structure. Most enzymes and
proteins precipitate at the more extreme pH values.
2 In addition to the pH, also the redox potential needs to stay constant in the cells.
4 points
(a) What species are formed that can higher the redox potential of the cytosol?
(b) What species are involved in buffering the potential and remove these species?
(c) What role does the pentose phosphate pathway play in the removal of these
species?
a) The main species that can shift the potential to a higher more positive value are
oxygen, superoxide, hydrogen peroxide and hydroxyl radicals. In particular the
last three, the so-called reactive oxygen species (ROS) are very dangerous to the
cell if their respective concentration increase too much. Iron-sulfur-cluster-
containing proteins get damaged. Regulatory disulfide sites get oxidized. DNA
gets damaged and eventually the cell will die.
b) Several compounds are present in the cell that can buffer the potential and/or
directly remove ROS. All cells contain glutathione and thioredoxins. Plant
cells also contain ascorbate. In addition, specific proteins will remove the ROS,
like catalase and hydrogen peroxide.
c) Reduced glutathione (GSH) protects the cell by destroying hydroxyl free radicals.
It is also the cosubstrate for glutathione peroxidase that removes hydrogen
peroxide. In both cases the gluthathione gets oxidized (GSSG). Regeneration of
GSH from its oxidized form (GSSG) requires the NADPH produced in the
glucose 6-phosphate dehydrogenase reaction. This reaction is catalyzed by
gluthathione reductase
, 3 For a peptide with the sequence shown below determine the pI.
4 points
Glu–His–Trp–Ser–Gly–Leu–Arg–Pro–Gly
(Hint: If you cannot do that directly, first calculate the charge at different pH values and
use those to find the pI region.)
pH N-term Glu His Arg C-term Net Charge
2 +
H3N– –COOH >NH+ –NH3+ –COOH
+1 0 +1 +1 0 +3
5 +
H3N– –COOˉ >NH+ –NH3+ –COOˉ
+1 -1 +1 +1 -1 +1
7 +
H3N– –COOˉ >N –NH3+ –COOˉ
+1 -1 0 +1 -1 0
10 H2N– –COOˉ >N –NH3+ –COOˉ
0 -1 0 +1 -1 -1
Find the two ionizable groups with pKa values that “straddle” the point at which net
peptide charge = 0 (here, two groups that ionize near pH 8): the amino-terminal α-
amino group of Glu and the His imidazole group.
Thus, we can estimate pI = (8.0 + 6.00)/2 = 7.0
2
