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SCIENCE BTH2741collection-of-past-year-paper-questions-and-answers
1. Hydrogen Bonding
Hydrogen bonds are formed due to an attraction between dipoles. These bonds are largely
dependent on the electronegativities of the atoms carrying the dipole charges; the difference
in electronegativities between two atoms must be greater than 1 or hydrogen bonds cannot
form. The most common hydrogen bonds formed to stabilize the tertiary structure of
biological macromolecules are between hydrogen (H) and oxygen (O), hydrogen (H) and
nitrogen (N), and hydrogen (H) and sulfur (S). This is because hydrogen is very weakly
electronegative (electronegativity of 2.1), but oxygen, nitrogen and sulfur are highly
electronegative (electronegativities of 3.5, 3.0 and 2.5 respectively).
Hydrogen bonds are most commonly seen between polar R groups of amino acids and the H
in H2O.
Hydrogen bonds are highly directional – the atomic centers of the atoms must be aligned,
otherwise a hydrogen bond will not form. The distance between the two atoms engaged in a
hydrogen bond is also very important – it must be roughly 0.2nm. If this distance gets shorter
(roughly 0.1nm), electrons will be shared between the two atoms and they will be covalently
bonded.
Two amino acids that can participate in hydrogen bonding are serine and threonine as
both these amino acids have hydroxyl groups, where the oxygen can form hydrogen bonds
with water.
2. Electrostatic interactions
Electrostatic interactions are observed between two charged atoms – these atoms can be
oppositely charged, ie. one carrying a positive and one carrying a negative charge in which
case they would experience electrostatic attraction, or they can be similarly charged, ie. both
carrying a positive charge or both carrying a negative charge, in which case they would
experience electrostatic repulsion. These interactions are inversely proportional to the
distance between charged centers, meaning that the further away the two atoms are, the
stronger the interaction.
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However, these interactions are easily disrupted by changes in the pH as the atom(s) may
become ionized and carry a different charge. This is especially seen in amino acids, which are
easily ionized.
Two amino acids that can participate in electrostatic interactions are aspartic acid and lysine.
3. Van Der Waal’s interactions
Van Der Waal’s interactions are extremely weak (10 times weaker than hydrogen bonds),
however, they are significant when collective in large macromolecules. They are observed
due to attractions between mini-dipoles of opposing charges when individual atoms of
molecules are closely packed together. The distribution of electrons is not intense and the
charge distribution within atoms is asymmetrical. These interactions are highly distance-
dependent, if the distances between the atomic centers is too long, there will be no Van Der
Waal’s interactions. If the distance is too short, repulsion between the mini-dipoles will
occur.
Two amino acids that can participate in Van Der Waal’s interactions are tyrosine and
phenylalanine (all amino acids can participate in this).
4. Hydrophobic interactions
These interactions allow for the aggregation of non-polar groups in aqueous solution – this is
particularly useful in the isolation of a specific compound (for example, a specific protein)
from solution. Hydrophobic interactions occur between the non-polar R groups of amino
acids. The polar groups of a particular compound will be exposed to water and may form
hydrogen bonds, while the non-polar group will face the inner area of the molecule and
participate in hydrophobic interactions. This structure is thermodynamically favourable as it
minimizes the disruption of hydrogen bonds between water molecules. It also stabilizes the
aggregation of non-polar R groups in aqueous solution due to the net gain of entropy of
surrounding water molecules.
Two amino acids that can participate in hydrophobic interactions are isoleucine and valine as
they are non-polar and neutral.
Lecture 3: Body Water
a) Sketch the orientations that fatty acids and phospholipids might adopt in a
hydrophobic solvent.
Fatty acids are hydrophobic; therefore, they will orient themselves towards the hydrophobic
solvent and form hydrophobic interactions with it. Phospholipids are hydrophilic, so, they
will orient themselves away from the solvent and instead, form hydrogen bonds with one
another between the phosphorus atom (electronegativity of 2.1) and the oxygen atom
(electronegativity of 3.5 – difference in electronegativity is greater than 1). They will
precipitate out of the solvent.
b) Discuss any stability problems that may be encountered in the formation
of aggregates and suggest how these could be overcome.
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A large amount of energy will have to be used in order for non-polar groups to disrupt
hydrogen bonds between water molecules and form aggregates. This is thermodynamically
unfavourable and the aggregates will therefore be unstable. To combat this stability problem,
non-polar groups will engage in hydrophobic interactions with each other and reduce their
contact with water, thereby maximizing hydrogen bonding between water molecules. Van
Der Waal’s forces further stabilize this aggregation. This is then thermodynamically
favorable and the aggregates will be stable.
Lecture 4: pH and Buffers
1. List five features you should consider when choosing a buffer system for use in a
particular laboratory experiment. (5 marks)
1. The buffer must be able to act as a buffer in the appropriate pH range – this is typically a
pH of 6-8, but it can vary according to the reagents/compounds used in the experiment. This
is especially important in experiments involving amino acids as they are very sensitive to pH
changes and will ionize extremely quickly if the pH is not appropriate.
2. The buffer must be minimally affected by temperature ranges. During an experiment, it
is likely that reactions will involve temperature changes. The buffer should not denature
when exposed to temperature changes; it should be able to buffer the solution regardless of
temperature changes.
3. The buffer should not absorb light in the visible range. Otherwise, it can interfere in
spectrophotometric analyses and the results of the experiment will not be accurate.
4. The buffer must not form complexes or precipitates with other ions in the reaction mixture.
The buffer is only there to keep the pH of the reaction mixture relatively constant so the
reaction can proceed; it must not react with the reagents/compounds themselves.
5. The buffer should be water soluble, so that it can dissociate into the acid and conjugate
base/base and conjugate acid and be able to buffer the reaction mixture efficiently.
2. Why does breathing rate affect plasma pH? (5 marks)
When you breathe, you inhale oxygen and exhale carbon dioxide. Basically, plasma pH is
maintained by the bicarbonate buffer system and the action of lungs. The bicarbonate buffer
system remains in equilibrium with atmospheric air as this is an open system. Carbon dioxide
gas is dissolved in blood to be present in aqueous form. It is then quickly converted to
carbonic acid (H2CO3) in the presence of water in the body. It then dissociates to the
bicarbonate ion, HCO 3- and the hydrogen ion H+. All these reactions are reversible; therefore,
these compounds exist in equilibrium in the body:
CO2 (g) ⇌ CO2 (aq) + H2O (l) ⇌ H2CO3 (aq) ⇌ HCO 3- (aq) + H+ (aq)
According to Le Chatelier’s principle, if the concentration of reactants (on the left-hand side
of the reaction) increases, the equilibrium will shift to the right so as to be stabilized and the
concentration of the products (on the right-hand side of the reaction) will also increase, and
vice versa.
Plasma pH is proportional to the ratio of [HCO3-] / pCO2 in the blood.
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SCIENCE BTH2741collection-of-past-year-paper-questions-and-answers
1. Hydrogen Bonding
Hydrogen bonds are formed due to an attraction between dipoles. These bonds are largely
dependent on the electronegativities of the atoms carrying the dipole charges; the difference
in electronegativities between two atoms must be greater than 1 or hydrogen bonds cannot
form. The most common hydrogen bonds formed to stabilize the tertiary structure of
biological macromolecules are between hydrogen (H) and oxygen (O), hydrogen (H) and
nitrogen (N), and hydrogen (H) and sulfur (S). This is because hydrogen is very weakly
electronegative (electronegativity of 2.1), but oxygen, nitrogen and sulfur are highly
electronegative (electronegativities of 3.5, 3.0 and 2.5 respectively).
Hydrogen bonds are most commonly seen between polar R groups of amino acids and the H
in H2O.
Hydrogen bonds are highly directional – the atomic centers of the atoms must be aligned,
otherwise a hydrogen bond will not form. The distance between the two atoms engaged in a
hydrogen bond is also very important – it must be roughly 0.2nm. If this distance gets shorter
(roughly 0.1nm), electrons will be shared between the two atoms and they will be covalently
bonded.
Two amino acids that can participate in hydrogen bonding are serine and threonine as
both these amino acids have hydroxyl groups, where the oxygen can form hydrogen bonds
with water.
2. Electrostatic interactions
Electrostatic interactions are observed between two charged atoms – these atoms can be
oppositely charged, ie. one carrying a positive and one carrying a negative charge in which
case they would experience electrostatic attraction, or they can be similarly charged, ie. both
carrying a positive charge or both carrying a negative charge, in which case they would
experience electrostatic repulsion. These interactions are inversely proportional to the
distance between charged centers, meaning that the further away the two atoms are, the
stronger the interaction.
Verspreiden niet toegestaan | Gedownload door Ivy Ong
, lOMoARcPSD|1825843
However, these interactions are easily disrupted by changes in the pH as the atom(s) may
become ionized and carry a different charge. This is especially seen in amino acids, which are
easily ionized.
Two amino acids that can participate in electrostatic interactions are aspartic acid and lysine.
3. Van Der Waal’s interactions
Van Der Waal’s interactions are extremely weak (10 times weaker than hydrogen bonds),
however, they are significant when collective in large macromolecules. They are observed
due to attractions between mini-dipoles of opposing charges when individual atoms of
molecules are closely packed together. The distribution of electrons is not intense and the
charge distribution within atoms is asymmetrical. These interactions are highly distance-
dependent, if the distances between the atomic centers is too long, there will be no Van Der
Waal’s interactions. If the distance is too short, repulsion between the mini-dipoles will
occur.
Two amino acids that can participate in Van Der Waal’s interactions are tyrosine and
phenylalanine (all amino acids can participate in this).
4. Hydrophobic interactions
These interactions allow for the aggregation of non-polar groups in aqueous solution – this is
particularly useful in the isolation of a specific compound (for example, a specific protein)
from solution. Hydrophobic interactions occur between the non-polar R groups of amino
acids. The polar groups of a particular compound will be exposed to water and may form
hydrogen bonds, while the non-polar group will face the inner area of the molecule and
participate in hydrophobic interactions. This structure is thermodynamically favourable as it
minimizes the disruption of hydrogen bonds between water molecules. It also stabilizes the
aggregation of non-polar R groups in aqueous solution due to the net gain of entropy of
surrounding water molecules.
Two amino acids that can participate in hydrophobic interactions are isoleucine and valine as
they are non-polar and neutral.
Lecture 3: Body Water
a) Sketch the orientations that fatty acids and phospholipids might adopt in a
hydrophobic solvent.
Fatty acids are hydrophobic; therefore, they will orient themselves towards the hydrophobic
solvent and form hydrophobic interactions with it. Phospholipids are hydrophilic, so, they
will orient themselves away from the solvent and instead, form hydrogen bonds with one
another between the phosphorus atom (electronegativity of 2.1) and the oxygen atom
(electronegativity of 3.5 – difference in electronegativity is greater than 1). They will
precipitate out of the solvent.
b) Discuss any stability problems that may be encountered in the formation
of aggregates and suggest how these could be overcome.
Verspreiden niet toegestaan | Gedownload door Ivy Ong
, lOMoARcPSD|1825843
A large amount of energy will have to be used in order for non-polar groups to disrupt
hydrogen bonds between water molecules and form aggregates. This is thermodynamically
unfavourable and the aggregates will therefore be unstable. To combat this stability problem,
non-polar groups will engage in hydrophobic interactions with each other and reduce their
contact with water, thereby maximizing hydrogen bonding between water molecules. Van
Der Waal’s forces further stabilize this aggregation. This is then thermodynamically
favorable and the aggregates will be stable.
Lecture 4: pH and Buffers
1. List five features you should consider when choosing a buffer system for use in a
particular laboratory experiment. (5 marks)
1. The buffer must be able to act as a buffer in the appropriate pH range – this is typically a
pH of 6-8, but it can vary according to the reagents/compounds used in the experiment. This
is especially important in experiments involving amino acids as they are very sensitive to pH
changes and will ionize extremely quickly if the pH is not appropriate.
2. The buffer must be minimally affected by temperature ranges. During an experiment, it
is likely that reactions will involve temperature changes. The buffer should not denature
when exposed to temperature changes; it should be able to buffer the solution regardless of
temperature changes.
3. The buffer should not absorb light in the visible range. Otherwise, it can interfere in
spectrophotometric analyses and the results of the experiment will not be accurate.
4. The buffer must not form complexes or precipitates with other ions in the reaction mixture.
The buffer is only there to keep the pH of the reaction mixture relatively constant so the
reaction can proceed; it must not react with the reagents/compounds themselves.
5. The buffer should be water soluble, so that it can dissociate into the acid and conjugate
base/base and conjugate acid and be able to buffer the reaction mixture efficiently.
2. Why does breathing rate affect plasma pH? (5 marks)
When you breathe, you inhale oxygen and exhale carbon dioxide. Basically, plasma pH is
maintained by the bicarbonate buffer system and the action of lungs. The bicarbonate buffer
system remains in equilibrium with atmospheric air as this is an open system. Carbon dioxide
gas is dissolved in blood to be present in aqueous form. It is then quickly converted to
carbonic acid (H2CO3) in the presence of water in the body. It then dissociates to the
bicarbonate ion, HCO 3- and the hydrogen ion H+. All these reactions are reversible; therefore,
these compounds exist in equilibrium in the body:
CO2 (g) ⇌ CO2 (aq) + H2O (l) ⇌ H2CO3 (aq) ⇌ HCO 3- (aq) + H+ (aq)
According to Le Chatelier’s principle, if the concentration of reactants (on the left-hand side
of the reaction) increases, the equilibrium will shift to the right so as to be stabilized and the
concentration of the products (on the right-hand side of the reaction) will also increase, and
vice versa.
Plasma pH is proportional to the ratio of [HCO3-] / pCO2 in the blood.
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