Final Exam

Question 1

Which is the correct order of the central dogma of biochemistry?

Answer 1

DNA -> RNA -> Protein

Question 2

Which of these is not a benefit of enzymatic catalyst?

A. Acceleration under mild conditions
B. High specificity
C. Possibility for regulation
D. Lower activation energy
E. A better route of reaction

Answer 2

D. Lower activation energy

Question 3

Which of these speeds up a reaction?

A. Higher temperatures
B. Higher concentrations of reactants
C. Coupling of reactions
D. Lowering activation barrier via catalysis
E. All the above

Answer 3

E. All the above

Question 4

What kind of isomers contain identical physical properties?

Answer 4

Enantiomers

Question 5

Which are the four main classes of macromolecules?

Answer 5

Lipids, carbohydrates, nucleic acids and proteins

Question 6

Water is a ______ for charged and polar substances and a ______ for nonpolar substances.

Answer 6

Good solvent, poor solvent

Question 7

What kind of solution has a high solute concentration and causes water to be drawn out of the cell?

Answer 7

Hypertonic solution

Question 8

The interaction of polar molecules, such as the attraction between water molecules, causes the non-polar molecules to concentrate is called what?

Answer 8

Hydrophobic effect

Question 9

What is not true of water?

A. Water can serve as a hydrogen donor
B. Water has an unusually high surface tension
C. Water has an usually high boiling point
D. Hydrogen bonding in water is not cooperative
E. Water can serve as a hydrogen acceptor

Answer 9

D. Hydrogen bonding in water is not cooperative

Question 10

When hydronium ions are in water, what two bond types are interchangeable and what function results from this?

Answer 10

Covalent and hydrogen bonds/proton hopping

Question 11

In what ways does column chromatography separate proteins? (select all that
apply)

a. Charge
b. Size
c. Binding affinity
d. A and B
e. B and C

Answer 11

a. Charge
b. Size
c. Binding affinity

Question 12

What is NOT included in the structure of an amino acid

a. An amino group
b. Carboxylic acid group
c. Alpha carbon
d. R-group
e. Phosphate group

Answer 12

Phosphate group

Question 13

What type of amino acid side chain can make a distinct titration curve?

a. Ionizable
b. Polar
c. Non-polar
d. Aromatic
e. All of the above

Answer 13

Ionizable

Question 14

What is the strongest chromophore?

a. Tryptophan
b. Tyrosine
c. Phenylalanine
d. Both a and b
e. Both b and c

Answer 14

Tryptophan

Question 15

chromophore

Answer 15

a group of atoms or electrons in dye molecules that is responsible for the color of the dye

Question 16

Which is not a property amino acids have that makes them well-suited to carry out a variety of biological functions?

a. Capacity to polymerize
b. Acid-base properties
c. Varied physical properties
d. Varied chemical functionality
e. Water insoluble

Answer 16

Water insoluble

Question 17

Which amino acids are commonly found in beta turns?

a. Proline and glycine
b. Tryptophan and glutamic acid
c. Arginine and aspartic acid
d. Serine and Valine
e. Leucine and Alanine

Answer 17

Proline and glycine

Question 18

Which of these are strong helix formers?

a. Alanine
b. Proline
c. Glycine
d. Tryptophan
e. Histidine

Answer 18

Alanine

Question 19

What is the secondary structure of a protein primarily determined by?

a. The sequence of amino acids
b. The hydrogen bonding between the amino hydrogen and carboxyl atoms in the peptide backbone.
c. The covalent binding of side chains
d. The solubility of the protein in water
e. The interaction of disulfide bridges.

Answer 19

The hydrogen bonding between the amino hydrogen and carboxyl atoms in the peptide backbone

Question 20

How many amino acids are in a beta turn?

Answer 20

Four

Question 21

Which of the following best describes the quaternary structure of a protein?

a. The sequence of amino acids in polypeptide chain
b. The folding of a polypeptide into its three-dimensional shape
c. The assembly of individual polypeptides into a larger functional cluster
d. The formation of alpha-helices and beta-sheets within a single polypeptide
e. The covalent bonding of amino acids through peptide bonds

Answer 21

The assembly of individual polypeptides into a larger functional cluster

Question 22

When comparing the T state to the R state, hemoglobin has…

a. higher affinity for O2
b. Lower affinity for O2
c. The same affinity for O2
d. No affinity for O2
e. It does not affect the affinity for O2

Answer 22

Lower affinity for O2

Question 23

Which of the following is a function of a globular protein?

a. Storage of ion
b. Transportation of ions
c. Defense against pathogens
d. Muscle contractions
e. All the above

Answer 23

All the above

Question 24

Which of the following molecules have the strongest binding?

(Picture of Binding strength on slide 7 without the affinity shown)

Answer 24

Biotin-avidin

Question 25

Carbon monoxide is a deadly gas that binds better than oxygen within hemo and myoglobin. Which is the following is NOT a reason for this?

a. It has a similar size as O2
b. Carbon monoxide has a lone pair that can be donated while O2 does not
c. When bound carbon monoxide has more linear bonds
d. Hemoglobin has a protein pocket that decreases affinity for O2, so carbon monoxide binds about 250x better.
e. None of the above

Answer 25

None of the above

Question 26

How do Ligands bind?

a. Non-covalently
b. Covalently
c. Ionic bonds
d. Disulfide bridges
e. LDFs

Answer 26

Non-covalently

Question 27

Describe the difference between endergonic and exergonic reactions.

Answer 27

An exergonic reaction releases energy to the surroundings, meaning the products have less energy than the reactants, while an endergonic reaction absorbs energy from the surroundings, requiring an input of energy to proceed and resulting in products with higher energy than the reactants; essentially, exergonic reactions are spontaneous and release free energy, whereas endergonic reactions are non-spontaneous and require energy input to occur.

Question 28

Explain the difference between eukaryotic and prokaryotic cells.

Answer 28

The primary difference is that eukaryotic cells have a membrane-bound nucleus containing their DNA, while prokaryotic cells lack a nucleus and other membrane-bound organelles

Question 29

Key points about eukaryotic cells:

Answer 29

Nucleus: Contains the cell’s DNA, enclosed by a membrane.
Organelles: Possess various membrane-bound organelles like mitochondria, endoplasmic reticulum, and Golgi apparatus, which perform specialized functions within the cell.
Complexity: Generally more complex than prokaryotic cells.

Question 30

Key points about prokaryotic cells:

Answer 30

No nucleus: Genetic material is free-floating in the cytoplasm.
Limited organelles: Only have ribosomes, which are non-membrane bound structures involved in protein synthesis.
Simpler structure: Lack the complex internal organization of eukaryotes.

Question 31

Describe the difference between cis and trans molecules.

Answer 31

In chemistry, “cis” and “trans” describe the spatial arrangement of similar groups within a molecule, where “cis” indicates that these groups are on the same side of a double bond, while “trans” means they are on opposite sides; essentially, making them geometric isomers with different properties despite having the same molecular formula.

Question 32

Explain the steric favorability between cis and trans molecules.

Answer 32

Trans molecules are considered sterically more favorable than cis molecules because in a trans configuration, the substituents on a double bond are positioned on opposite sides, minimizing steric hindrance (repulsion between electron clouds) compared to a cis configuration where the substituents are on the same side, causing crowding and instability

Question 33

List and explain two ways biological systems overcome unfavorable reactions.

Answer 33

Biological systems overcome unfavorable reactions primarily through coupling the reaction with a favorable one, most commonly using the hydrolysis of ATP (adenosine triphosphate) as the energy source, and by utilizing enzymes to lower the activation energy barrier, effectively making the reaction more likely to occur.

Question 34

Coupling reactions:

Answer 34

This mechanism involves linking an energetically unfavorable reaction with a highly favorable one, such as the breakdown of ATP, where the energy released from the favorable reaction drives the unfavorable one forward. Essentially, the energy “captured” from ATP hydrolysis is used to push the otherwise non-spontaneous reaction to completion.

Question 35

Enzyme catalysis:

Answer 35

Enzymes act as biological catalysts, lowering the activation energy needed for a reaction to proceed. This means that even a thermodynamically unfavorable reaction can occur at a significant rate in the presence of an appropriate enzyme.

Question 36

Describe how water surrounding nonpolar molecules has lower entropy.

Answer 36

Water molecules readily hydrogen bond with each other, creating a dynamic network. When nonpolar molecules are in water, the surrounding water molecules form a more ordered, “cage-like” structure around them, which significantly decreases the entropy of the water because the molecules are forced into a specific arrangement instead of having the freedom to move randomly, a phenomenon known as the “hydrophobic effect.”

Question 37

Hydrogen bonding gives water unique properties such as a high boiling and melting point. What about this type of bonding that makes these properties possible?

Answer 37

Hydrogen bonding in water creates a strong network of intermolecular forces between water molecules as it can act as both a hydrogen donor and acceptor. Due to strong dipole-dipole moments caused by polarity between the positive hydrogen and negative oxygen interacting with nearby water molecules, these molecules require a significant amount of energy to break, which results in their high boiling and melting points. Hydrogen bonds are strongest when the bonded molecules allow for linear bonding patterns. Hydrogen bonding in water is cooperative.

Or when multiple hydrogen bonds form between molecules, the presence of one hydrogen bond strengthens the others nearby, leading to a significantly enhanced overall stability of the system

Question 38

Explain what a buffer is and why they are resistant to changes in pH?

Answer 38

A buffer is a solution that can resist significant changes in pH when small amounts of acid or base are added to it, because it contains a mixture of a weak acid and its conjugate base, which allows it to neutralize added hydrogen or hydroxide ions through a chemical equilibrium reaction, effectively maintaining a stable pH level

Question 39

Describe why hexagonal ice is most common.

Answer 39

Hexagonal ice forms an organized lattice and thus has a low entropy.
It contains maximal hydrogen bonds/ water molecules, forcing the water molecules into equidistant arrangement. Thus:
–ice has lower density than liquid water
–ice floats

Question 40

How does the pH of a solution affect the ionization state of an amino acid?

Answer 40

The pH of a solution affects the ionization state of an amino acid by altering
the protonation of its amino and carboxyl groups. At low pH, the amino group and carboxyl group will be protonated, while at high pH, the amino group is unprotonated and the carboxyl group is deprotonated. At neutral pH levels, it will have a net charge of zero (zwitterion) due to the balance of charges on the two functional groups.

Question 41

How do differences in protein sequences with identical functions from a wide range
of species indicate evolutionary divergences?

Answer 41

As organisms evolve over time, their genes (including those coding for proteins with conserved functions) accumulate mutations, leading to changes in the amino acid sequence of the protein, even if the overall function remains the same; the more differences in the sequence between two species, the further apart they are evolutionarily.

Question 42

What does “essential amino acid” mean?

Answer 42

An “essential amino acid” refers to an amino acid that cannot be synthesized by the body in sufficient quantities and must be obtained through the diet. These amino acids are crucial for various biological processes, such as protein synthesis, enzyme function, and metabolic pathways. Since the body can’t produce them, it is necessary to consume foods that provide them.

The body cannot produce essential amino acids because it lacks the necessary enzymes to synthesize them from other compounds. Each amino acid has a specific metabolic pathway for its production, and for essential amino acids, the required biochemical reactions either don’t occur in the human body or occur at insufficient rates.

The inability to synthesize certain amino acids is a result of evolutionary biology. Over time, humans have evolved metabolic pathways that allow for the synthesis of most amino acids, but some amino acids require more complex or specialized pathways.

Question 43

Hydrophobic effect

Answer 43

Water molecules are polar and form hydrogen bonds with each other. When a nonpolar molecule is introduced into water, the water molecules around it are forced to reorganize to minimize contact with the nonpolar molecule. This leads to an increase in the system’s overall free energy, which is unfavorable.
To minimize this energy increase, the nonpolar molecules tend to aggregate or cluster together, reducing the surface area exposed to water. This clustering helps to minimize the disruption to the water’s hydrogen bonding network.
In biological systems, the hydrophobic effect is crucial for:

Protein folding: Hydrophobic amino acid side chains tend to cluster in the interior of the protein structure, away from the water, while hydrophilic (water-attracting) side chains are exposed to the aqueous environment. This folding is essential for the protein’s three-dimensional shape and function.
Membrane formation: In cell membranes, hydrophobic tails of phospholipids face inward, shielded from water, while the hydrophilic heads are exposed to the aqueous surroundings. This arrangement forms the lipid bilayer structure, a key feature of cell membranes.
The hydrophobic effect is not a force in itself (like hydrogen bonding or van der Waals forces), but rather a result of water molecules’ tendency to minimize the disruption to their hydrogen bonding network by avoiding nonpolar substances.

Question 44

Hydrogen bonds in proteins

Answer 44

Interaction of N−H and C=O of the peptide bond leads to local regular structures such as a-helices and b-sheets.

Question 45

London dispersion

Answer 45

Medium-range weak attraction between all atoms contributes significantly to the stability in the interior of the protein.

Question 46

Electrostatic interactions

Answer 46

–long-range strong interactions between permanently charged groups
–Salt bridges, especially those buried in the hydrophobic environment, strongly stabilize the protein.

Question 47

Explain the differences between parallel vs. antiparallel in alpha helices and beta sheets.

Answer 47

In protein secondary structures, “parallel” refers to when multiple polypeptide chains run in the same direction (N-terminus to C-terminus) within a beta sheet, while “antiparallel” means the chains run in opposite directions (one chain’s N-terminus aligns with the other chain’s C-terminus); importantly, alpha helices cannot be parallel or antiparallel as they are single, coiled strands

Question 48

Explain what it means by primary, secondary, tertiary and quaternary structure.

Answer 48

A. Primary: the structure of the protein is partially dictated by the properties of the peptide bond as a resonance hybrid with unique amino acids.
B. Secondary: refers to a local spatial arrangement of the polypeptide backbone of a-helices, B-sheets and random coils stabilized by hydrogen bonds.
C. Tertiary: refers to the overall spatial arrangement of atoms in a protein. Stabilized by numerous weak interactions between amino acid side chains; hydrophobic interactions, polar interactions and disulfide bonds.
D. Quaternary structure: formed by the assembly of individual polypeptides into a larger functional cluster

Question 49

Explain what affects protein stability.

Answer 49

Protein stability is primarily affected by the protein’s amino acid sequence, which determines the types of interactions (like hydrogen bonds, hydrophobic interactions, and ionic bonds) that can form within the protein structure, but is also significantly impacted by environmental factors such as temperature, pH, ionic strength, presence of denaturants, and solvent composition, all of which can disrupt these interactions and cause the protein to unfold or misfold. This may include chaotropic agents: urea and guanidinium hydrochloride. Other enzymes may assist in refolding or renaturing denatured proteins, such as ribonuclease.

Question 50

What are the five conditions that affect alpha helix stability? Briefly explain them.

Answer 50

1. Hydrogen bonds between the backbone amides of n and n+4 amino acids. Not all polypeptide sequences adopt alpha-helical structures.
2. Small hydrophobic residues such as Ala and Leu are strong helix formers.
3. Pro acts as a helix breaker because the rotation around the N-Ca (φ-angle) bond is impossible. Gly acts as a helix breaker because the tiny R group supports other conformations.
4. Attractive or repulsive interactions between side chains 3 to 4 amino acids apart will affect formation.
5. Charge distribution along the helix (a large dipole moment can destabilize). Positive at the N-terminus and negative at the C-terminus

Question 51

Compare and contrast lock and key fit and induced fit.

Answer 51

The “lock and key” model describes enzyme-substrate interaction as a rigid fit, where the enzyme’s active site (lock) perfectly matches the shape of the substrate (key) before binding.

The “induced fit” model suggests that the enzyme’s active site is flexible and changes shape slightly upon substrate binding to achieve an optimal fit, making it more adaptable to different substrates

Question 52

Explain cooperativity.

Answer 52

Cooperativity in molecules refers to a phenomenon where the binding of a molecule to one site on a protein influences the affinity of other binding sites on the same protein for similar molecules, meaning that the binding of one molecule can either increase or decrease the likelihood of other molecules binding to nearby sites, effectively creating a “cascade” effect in binding behavior; this is often observed in proteins with multiple binding pockets, and is most commonly seen in the context of enzymes and receptors.

Question 53

Explain the difference between homotropic and heterotopic allosteric regulators.

Answer 53

A homotropic allosteric regulator is a molecule that binds to an allosteric site on an enzyme and is the same molecule as the enzyme’s substrate, meaning it directly influences its own binding affinity, while a heterotropic allosteric regulator is a different molecule from the substrate that binds to an allosteric site and affects the enzyme’s affinity for its substrate; essentially, in homotropic regulation, the substrate acts as its own regulator, whereas in heterotropic regulation, a separate molecule acts as the regulator.

Question 54

Explain how a mutation in a residue of the active site can affect the functionality of the enzyme.

Answer 54

A mutation in a residue within an enzyme’s active site can drastically affect its functionality by altering the shape and chemical properties of the active site, which is critical for binding to the substrate and facilitating the chemical reaction; this can lead to decreased enzyme activity, altered substrate specificity, or even complete loss of function depending on the nature of the amino acid change. A mutation in an active site residue can disrupt this induced fit mechanism, hindering the enzyme’s ability to catalyze the reaction. Replacing a charged amino acid in the active site with a neutral one could disrupt electrostatic interactions with the substrate, preventing binding.

Question 55

Explain the difference of binding strength and how its affinity and functionality.

Answer 55

“Binding strength” refers to the overall strength of the interaction between two molecules, while “affinity” specifically describes the strength of a single binding site on a molecule to another molecule, and “functionality” refers to the ability of a bound molecule to produce a specific biological effect once attached.

Question 56

Binding strength:

Answer 56

This is the overall strength of the interaction between two molecules, considering all the non-covalent bonds involved, and is often measured by the dissociation constant (Kd) - a lower Kd indicates a stronger binding strength.

Question 57

Affinity:

Answer 57

This is the strength of the interaction between a single binding site on one molecule and another molecule.

Question 58

Functionality of affinity and binding strength

Answer 58

This describes whether the binding event leads to a specific biological outcome.

Question 59

1. Select the following that all apply for catalysts:
   a. Increase reaction rates without being used up
   b. Increase reaction rates being used up
   c. Greater reaction specificity: avoids side products
   d. Reduces the required activation energy
   e. Delta G (ΔG) is affected by bonding to the substrate

Answer 59

A, C, D

Question 60

2. What part of the enzymatic catalytic reaction is stabilized by enzymes?
   a. Substrate
   b. Primary Structure
   c. Intermediate
   d. Secondary Structure
   e. Product

Answer 60

c. Intermediate

Question 61

3. Select the following change that does not affect enzyme kinetics?
   a. Enzyme concentration
   b. Substrate concentration
   c. Effectors
   d. Temperature
   e. None of the above

Answer 61

a. Enzyme concentration

Question 62

4. Which of the following choices describes the photo shown (attached):
   a. Burke Weave Plot
   b. Competitive inhibition
   c. Mixed inhibition
   d. Uncompetitive inhibition
   e. Sequential kinetic inhibition

Answer 62

d. Uncompetitive inhibition

Question 63

5. Which is not a good use for a biocatalyst:
   a. Create a buffer region for pH
   b. Greater reaction specificity
   c. Milder reaction conditions
   d. Higher reaction rates
   e. Capacity for regulation of biological pathways

Answer 63

a. Create a buffer region for pH

Question 64

1. What is the product when an alcohol attacks an aldehyde
   a. Hemi-acetal
   b. Acetal
   c. Hemi-ketals
   d. No reaction
   e. Ether

Answer 64

a. Hemi-acetal

Question 65

2. Carbohydrates fulfill a variety of functions including
   a. Energy source/storage
   b. structural component of cell walls and exoskeletons
   c. informational molecules in cell to cell signaling
   d. All of the above
   e. none of the above

Answer 65

D

Question 66

Which of the following statements are true for epimers
a) stereoisomers that differ at 1 chiral center
b) stereoisomers that are mirror images
c) epimers are diasteromers
d) Both a and c
e) none of the above

Answer 66

d) Both a and c

Question 67

Two sugar molecules can be joined via a(n) ____ bond
a) glycosidic
b) ionic
c) phosphodiester
d) hydrogen
e) peptide

Answer 67

a) glycosidic

Question 68

5. Which is a diastereomer of glucose?
   a. galactose
   b. sucrose
   c. lactose
   d. beta-galactosidase
   e. cysteine

Answer 68

a. galactose

Question 69

Which of these is found in RNA?
a. deoxyadenosine
b. deoxythymidylate
c. guanosine
d. deoxycytidylate
e. deoxyguanylate

Answer 69

guanosine

Question 70

What type of bond attaches the pentose sugar to the nitrogenous base in nucleotides?
a. hydrogen bond
b. phosphodiester bond
c. n-glycosidic bond
d. covalent bond
e. ionic bond

Answer 70

c. n-glycosidic bond

Question 71

Which is true regarding the phosphate group of a nucleic acid?
a. positively charged at neutral pH
b. negatively charged at neutral pH
c. typically attached to the 5’ position
d. both A & B
e. both B & C

Answer 71

E

Question 72

Which component is absent in a nucleoside compared to a nucleotide?
a. pentose
b. phosphate group
c. nitrogenous base
d. deoxyribose sugar
e. hydrogen bond

Answer 72

b. phosphate group

Question 73

Which scientists discovered the hydrogen bonding within DNA
a. Watson and crick
b. Franklin and wilkins
c. Friedrich Miescher
d. Levene and London
e. Tipson

Answer 73

a. Watson and crick

Question 74

1. Which is not a biological function of lipids
   a. Storage of energy
   b. Insulation from environment
   c. Membrane structure
   d. Signaling
   e. Catalyzing reactions

Answer 74

e. Catalyzing reactions

Question 75

2. A molecule with 4 fused rings and includes a hydroxyl group in the A-ring with various non-polar side chains called:
   a. Lipid
   b. Sterol
   c. Fatty Acid
   d. Triacylglycerol
   e. Lipopolysaccharide

Answer 75

b. Sterol

Question 76

3. Trans fatty acids derive from which of the following?
   a. Saturated fatty acids
   b. Hydrochloric acid
   c. Nitrogenous base pairs
   d. Unsaturated fatty acids
   e. Protein homeostasis regulation.

Answer 76

Unsaturated fatty acids

Question 77

4. Membranes can have a variety of surface properties which comes from diversification of head and tail groups. Diversification comes from which of the following answers. Select all that apply.
   a. Modifying a different backbone
   b. Adjusting the pH of the cytoplasm
   c. Changing the fatty acids
   d. Modifying the head group
   e. All of the above.

Answer 77

A, C, D

Question 78

1. Cholesterol plays a crucial role in the plasma membrane by:
   A. Facilitating ion transport across the membrane.
   B. Providing energy for active transport.
   C. Increasing membrane fluidity at low temperatures and decreasing it at high temperatures.
   D. Acting as a receptor for extracellular signals.
2. What is the main difference between primary and secondary active transport?

Answer 78

C. Increasing membrane fluidity at low temperatures and decreasing it at high
temperatures.

Question 79

2. What is the main difference between primary and secondary active transport?
   A. Primary active transport uses energy directly, while secondary active transport uses energy stored from another transport process.
   B. Primary active transport moves solutes downhill, while secondary active transport moves solutes uphill.
   C. Both types of transport rely only on energy from ATP.
   D. Secondary active transport does not require energy.

Answer 79

A

Question 80

4. What determines the direction in which a charged solute moves across a membrane.
   a. Membrane potential
   B. electrochemical gradient
   c. ion channels
   D. Simple Diffusion

Answer 80

B. electrochemical gradient

Question 81

5. What does the hydropathy index represent?
   A. The degree to which an amino acid side chain binds to other amino acids in a sequence.
   B. The free-energy change associated with the movement of an amino acid side chain from water to a hydrophobic environment.
   C. The free-energy change associated with the movement of an amino acid side chain from a hydrophobic environment to water.
   D. The total molecular weight of the amino acid sequence in a protein.
   E. The number of hydrophobic residues in a polypeptide chain.

Answer 81

C

Question 82

5. All are examples of proteins and enzymes in and on membranes EXCEPT
   a. Transporters
   b. Receptors
   c. ion channels
   d. Adhesion molecules
   e. nucleic acids

Answer 82

E

Question 83

Give a brief description of each of the following catalytic mechanisms:
1. Acid-base catalysis
2. Covalent catalysis
3. Metal ion catalysis

Answer 83

Acid-base catalysis involves the transfer of protons (H⁺) to or from a substrate to stabilize charged intermediates, facilitating the reaction.

General acid catalysis: A proton donor donates a proton to the substrate.
General base catalysis: A proton acceptor abstracts a proton from the substrate.

Covalent Catalysis
Covalent catalysis involves the transient formation of a covalent bond between the enzyme and the substrate. This intermediate is typically more reactive and facilitates the reaction by providing an alternative reaction pathway.
A nucleophilic group on the enzyme reacts with an electrophilic center on the substrate.

Metal Ion Catalysis
Metal ion catalysis involves the use of metal ions to stabilize negative charges, assist in electron transfer, or orient substrates for optimal interaction. Metal ions can act as electrophilic catalysts or participate in redox reactions.

Question 84

How do cofactors and coenzymes assist in enzymatic activation?

Answer 84

Cofactors are non-protein chemical compounds or metallic ions that are required for the biological activity of some enzymes. They can be divided into two main types:
1. Metal ions
2. Inorganic molecules (phosphate)
—-Serve as structural or catalytic components of enzymes.
—-Stabilize enzyme-substrate interactions by neutralizing negative charges.
—-Participate in redox reactions or electron transfer.

Coenzymes are organic molecules that temporarily or permanently associate with an enzyme to facilitate its activity. They often serve as carriers for chemical groups or electrons.
—-Act as group transfer agents
—-Transfer specific chemical groups between molecules by acting as intermediaries
—-Participate in redox reactions
—-Stabilize intermediates
—-Some coenzymes stabilize high-energy intermediates, reducing the activation energy of the reaction.

Question 85

Describe the difference between Uncompetitive Inhibition and Competitive Inhibition.

Answer 85

1. Uncompetitive Inhibition
   Binding site: The inhibitor binds exclusively to the enzyme-substrate complex, not to the free enzyme.
   Mechanism: The inhibitor locks the substrate in the enzyme, preventing the enzyme from releasing the product. This only occurs after the substrate has bound to the enzyme.
   Effect on enzyme kinetics:
   Vmax: Decreases because the enzyme-substrate-inhibitor complex is non-productive, reducing the total number of functional enzymes.
   Km: Decreases because the inhibitor stabilizes the enzyme-substrate complex, making it appear that the enzyme has a higher affinity for the substrate.
2. 2. Competitive Inhibition
      Binding site: The inhibitor competes directly with the substrate for binding to the enzyme’s active site.
      Mechanism: The inhibitor mimics the structure of the substrate and occupies the active site, preventing the substrate from binding.
      Effect on enzyme kinetics:
      Vmax: Remains unchanged because the inhibition can be overcome by increasing the substrate concentration.
      Km: Increases, as more substrate is required to reach half the maximum velocity

Question 86

Explain the functions and differences between irreversible and reversible inhibitors.

Answer 86

Enzyme inhibitors can affect enzyme activity by binding to the enzyme, and they are classified as irreversible or reversible inhibitors based on their mechanism and permanence of binding. Here’s an explanation of their functions and key differences:

Irreversible Inhibitors
Functions:
A substance that permanently blocks the action of an enzyme by forming a stable complex with it. This is done by covalently modifying the enzyme, usually by reacting with an amino acid residue at the active site
Often target critical active site residues, rendering the enzyme incapable of catalysis.
Used in drug design to permanently block harmful enzymes.

Reversible Inhibitors
Functions:
Bind to the enzyme via weak, non-covalent interactions (e.g., hydrogen bonds, ionic bonds, hydrophobic interactions).
Can dissociate from the enzyme, allowing normal enzymatic activity to resume when the inhibitor is removed or its concentration decreases.
Regulate enzyme activity dynamically, often acting as feedback regulators in metabolic pathways.
Types of reversible inhibitors:
Competitive: Compete with the substrate for the active site.
Uncompetitive: Bind only to the enzyme-substrate complex.
Noncompetitive/Mixed: Bind to an allosteric site and can inhibit the enzyme regardless of substrate binding.

Question 87

Describe two ways in which enzymatic activity can be regulated in biological systems.

1. Mechanism
2. Impact
3. Duration

Answer 87

Enzymatic activity can be regulated in various ways in biological systems to ensure proper control of metabolic pathways and maintain homeostasis. The primary mechanisms of enzyme regulation include:

1. Allosteric Regulation
   Mechanism: Allosteric regulation involves the binding of small molecules, called allosteric effectors (which can be activators or inhibitors), to sites on the enzyme that are distinct from the active site. These binding events induce conformational changes that affect the enzyme’s activity.
   Impact on Enzyme: Allosteric modulators bind to an enzyme’s allosteric site (not the active site), which can either enhance or inhibit the enzyme’s catalytic activity by changing the enzyme’s shape, often altering its affinity for the substrate.
   Duration of Effect: The effect of allosteric regulation is usually rapid and reversible. The enzyme returns to its original state once the effector molecule dissociates.
2. Covalent Modification
   Mechanism: Covalent regulation involves the reversible attachment or removal of a chemical group (such as a phosphate, methyl, acetyl, or adenylyl group) to specific amino acids in the enzyme, typically through the action of enzymes like kinases (which add phosphate groups) or phosphatases (which remove them).
   Impact on Enzyme: The covalent modification alters the enzyme’s activity, often by inducing a conformational change in the enzyme’s structure, which can either activate or inhibit its function.
   Duration of Effect: Covalent modifications are usually reversible, but the changes tend to last longer than allosteric regulation, depending on the specific modification. Reversal of the effect generally requires the activity of another enzyme (e.g., phosphatases to remove phosphate groups).

Question 88

Draw Fischer projections of an L-glucose molecule versus a D-glucose molecule.

Answer 88

In D-glucose, the hydroxyl group (-OH) on the chiral center C5 (the carbon furthest from the aldehyde group) is on the right side of the Fischer projection. The specific configuration at each chiral center in D-glucose is as follows:
C1: Aldehyde group (-CHO)
C2: Hydroxyl group on the right (R configuration)
C3: Hydroxyl group on the left (S configuration)
C4: Hydroxyl group on the right (R configuration)
C5: Hydroxyl group on the right (R configuration)
L-Glucose (L-form) Fischer Projection:
In L-glucose, the hydroxyl group (-OH) on the chiral center C5 (the carbon furthest from the aldehyde group) is on the left side of the Fischer projection. The specific configuration at each chiral center in L-glucose is as follows:
C1: Aldehyde group (-CHO)
C2: Hydroxyl group on the right (S configuration)
C3: Hydroxyl group on the left (R configuration)
C4: Hydroxyl group on the right (S configuration)
C5: Hydroxyl group on the left (S configuration)
Key Differences:
D-glucose has the -OH group on the right at C5, while L-glucose has the -OH group on the left at C5.
The configuration at the other chiral centers (C2, C3, C4) is also different, but the key difference is the configuration at C5.

Question 89

Explain what makes a sugar reducing or nonreducing.

Answer 89

A sugar is a reducing sugar if it has a free aldehyde group or a free ketone group that can be oxidized, or if it can form such a group in solution. This allows it to reduce other compounds, including other sugars.
A sugar is non-reducing if it does not have a free aldehyde or ketone group in either its open-chain or cyclic form. This typically occurs when the anomeric hydroxyl group of a sugar is involved in a glycosidic bond, as in many disaccharides and polysaccharides.
In non-reducing sugars, the anomeric carbon (the carbon involved in the aldehyde or ketone group) is part of a glycosidic bond with another sugar molecule. This blocks the ability of the sugar to open and form an aldehyde or ketone group.

Question 90

Beta-L-glucofuranose; explain the components of this molecules name.

Answer 90

The beta (β) designation refers to the anomeric configuration of the sugar when it is in its cyclic form. In the cyclic form of glucose, the anomeric carbon (the carbonyl carbon in the open-chain form) can form two possible configurations:
Beta (β): The hydroxyl group on the anomeric carbon is on the same side of the ring as the CH2OH group.
Since this is beta-glucofuranose, the hydroxyl group on the anomeric carbon is on the same side of the ring as the CH2OH group.

The L refers to the stereochemistry of the molecule, specifically the chirality of the glucose molecule. In this case, L-glucose is the enantiomer of D-glucose. The L- and D-forms of sugars refer to the configuration of the chiral centers in the molecule relative to the standard D- and L-glyceraldehyde. L-glucose has the opposite configuration at each of its chiral centers compared to

The gluco part refers to glucose, which is a six-carbon sugar (hexose) with an aldehyde functional group in its open-chain form. Glucose is the most common monosaccharide in biological systems and can cyclize to form a pyranose or furanose ring structure.

The furanose part of the name indicates the ring structure of the sugar. A furanose ring is a five-membered ring structure formed when the hydroxyl group on carbon 5 reacts with the aldehyde group on carbon 1 in glucose. This forms a cyclic ether ring, specifically a five-membered ring with four carbon atoms and one oxygen atom.

Question 91

Glycosidic bonds join two sugar molecules together. Describe the key features of these bonds and the process in which they are formed.

Answer 91

A glycosidic bond is a covalent bond that links the anomeric carbon of one sugar to a hydroxyl group on another sugar. This bond is formed when a water molecule is eliminated during the reaction, making the bond condensation or dehydration in nature.
The anomeric carbon is the carbon derived from the carbonyl group (aldehyde or ketone) in the open-chain form of the sugar. It is the carbon that becomes chiral when the sugar cyclizes into a ring form. The glycosidic bond forms between the anomeric carbon of one sugar and a hydroxyl group of another sugar.
Alpha (α) glycosidic bond: The bond is formed when the hydroxyl group on the anomeric carbon of the first sugar is on the opposite side of the ring relative to the CH2OH group (on carbon 6). The two sugars are oriented in opposite directions in the bond.
Beta (β) glycosidic bond: The bond is formed when the hydroxyl group on the anomeric carbon of the first sugar is on the same side of the ring relative to the CH2OH group. The two sugars are oriented in the same direction in the bond.
Glycosidic bonds are often described by the linkage number, which refers to the positions of the carbon atoms involved in the bond. For example:
A 1→4 glycosidic bond means that the bond forms between the 1st carbon of the first sugar and the 4th carbon of the second sugar.
A 1→6 glycosidic bond means that the bond forms between the 1st carbon of the first sugar and the 6th carbon of the second sugar.

Process of Glycosidic Bond Formation:
The formation of a glycosidic bond involves a condensation reaction (also called a dehydration reaction). During this process, the hydroxyl group (-OH) on the anomeric carbon of one sugar and the hydroxyl group on another sugar react, releasing a molecule of water (H₂O).
The hydroxyl group (-OH) on the anomeric carbon of the first sugar reacts with the hydroxyl group on the second sugar.
This reaction results in the formation of a covalent glycosidic bond, and the water molecule is eliminated.
The new bond between the sugars is stable and can hold the two sugars together, forming a disaccharide or a larger polysaccharide if additional sugars are added.
The two sugar molecules are now linked via the glycosidic bond, and the anomeric carbon of the first sugar no longer has a free hydroxyl group, making it non-reducing. This is why disaccharides like sucrose (which has a glycosidic bond between glucose and fructose) are non-reducing sugars.
The remaining sugar’s anomeric carbon still has a hydroxyl group, which can be involved in further glycosidic bond formation, making the structure extendable in polysaccharides.

Question 92

Why are 5 and 6 membered carbon chains more stable in a cyclic state.

Answer 92

Carbon atoms in organic molecules are tetrahedral in nature, meaning they have an ideal bond angle of 109.5°.
In a 5-membered ring (like furanose form), the internal angles of the ring are close to this ideal bond angle, resulting in minimal angle strain. Similarly, in a 6-membered ring (like pyranose form), the ring angles also approach 109.5°, making the structure even more stable.

In cyclic forms, the atoms in the ring are spatially arranged in such a way that steric hindrance (the repulsion between atoms or groups that are too close to each other) is minimized.
The 5-membered ring (such as in furanose) and 6-membered ring (such as in pyranose) allow for favorable interactions between atoms, especially the positioning of hydroxyl groups in relation to each other. These interactions help stabilize the ring.

Pyranose rings (6-membered rings) are particularly stable due to their ability to adopt a chair conformation. In this conformation, the atoms are arranged in a way that minimizes steric hindrance and torsional strain. In this chair form, the substituents (like hydroxyl groups) can be in equatorial positions, which are more spatially favorable than the axial positions.

In the cyclic forms of sugars, intramolecular hydrogen bonding can occur, where hydrogen atoms from hydroxyl groups may form weak bonds with oxygen atoms in the ring. These interactions can help stabilize the structure, particularly in the 5- and 6-membered rings.

Question 93

Explain the differences between RNA and DNA, structurally and functionally.

Answer 93

Sugar:
DNA: Contains the sugar deoxyribose, which lacks an oxygen atom on the 2’ carbon (hence the “deoxy” part of its name). The structure is 2-deoxyribose.
RNA: Contains the sugar ribose, which has a hydroxyl group (-OH) attached to the 2’ carbon, making it more reactive than deoxyribose. This extra oxygen is crucial for RNA’s different functions.
Strand Structure:
DNA: Is typically double-stranded, forming a double helix. The two strands are complementary and held together by hydrogen bonds between the nitrogenous bases.
RNA: Is usually single-stranded, although it can fold into complex secondary and tertiary structures. RNA can form double-stranded regions temporarily during processes like hybridization.
Nitrogenous Bases:
DNA: Contains the bases adenine (A), thymine (T), cytosine (C), and guanine (G). Thymine pairs with adenine via two hydrogen bonds.
RNA: Contains the bases adenine (A), uracil (U), cytosine (C), and guanine (G). Uracil replaces thymine and pairs with adenine through two hydrogen bonds. The absence of thymine in RNA makes it more prone to mutations and less stable than DNA.
Stability:
DNA: Is more stable because the 2’ deoxyribose sugar lacks a hydroxyl group, making it less susceptible to hydrolysis. The double-stranded structure also provides additional stability.
RNA: Is less stable due to the presence of the hydroxyl group at the 2’ position of ribose, which makes it more prone to hydrolysis and easier to degrade.
Genetic Information Storage:
DNA: Serves as the primary repository of genetic information in cells. It stores the blueprint for the development, function, and reproduction of organisms. DNA is found in the nucleus (in eukaryotes) or in the cytoplasm (in prokaryotes) and is highly stable for long-term information storage.
RNA: Primarily functions in protein synthesis and various regulatory roles. It acts as an intermediary between DNA and proteins in the form of messenger RNA (mRNA), which carries genetic information from DNA to the ribosome.
Role in Protein Synthesis:
DNA: Does not directly participate in the synthesis of proteins. Instead, DNA encodes the instructions for protein synthesis, which are later transcribed into RNA.
RNA: Plays a direct role in the process of protein synthesis:
mRNA: Carries the genetic code from DNA to ribosomes for translation into proteins.
rRNA: Forms the core structure of ribosomes and aids in protein synthesis.
tRNA: Helps decode the mRNA into a polypeptide chain by carrying amino acids to the ribosome.
Replication and Transcription:
DNA: Undergoes replication to ensure that genetic information is passed on during cell division. The double-stranded nature allows for semi-conservative replication, ensuring accurate copying of genetic material.
RNA: Undergoes transcription, a process where an RNA copy (mRNA) is synthesized from a DNA template. RNA is not typically replicated like DNA; instead, it is transcribed and then degraded or used for protein synthesis.
Types and Variability:
DNA: Exists in several forms (e.g., genomic DNA, plasmids) but is generally a long, continuous molecule.
RNA: Exists in many different forms (mRNA, tRNA, rRNA, small RNA, etc.), each with specific roles in protein synthesis and gene regulation.
Function in Gene Expression Regulation:
DNA: Contains the regulatory regions (e.g., promoters, enhancers) that control gene expression. It serves as the template for the transcription of RNA.
RNA: Involved in regulating gene expression through processes like RNA interference (RNAi), alternative splicing, and microRNAs

Question 94

Explain the difference between pyrimidines and purines, and list the pyrimidines and Purines.

Answer 94

Pyrimidines and purines are two types of nitrogenous bases that make up the nucleotides of nucleic acids (DNA and RNA). They differ in their chemical structure and properties, and they pair in specific ways in the formation of DNA and RNA.
Chemical Structure:
Purines: Purines have a double-ring structure made up of a six-membered and a five-membered ring fused together. This gives them a larger, more complex structure.
Pyrimidines: Pyrimidines have a single six-membered ring. They are smaller compared to purines.
Bases:
Purines include adenine (A) and guanine (G).
Pyrimidines include cytosine (C), thymine (T), and uracil (U).
Base Pairing:
In DNA:
Adenine (A) (purine) pairs with thymine (T) (pyrimidine) via two hydrogen bonds.
Guanine (G) (purine) pairs with cytosine (C) (pyrimidine) via three hydrogen bonds.
In RNA:
Adenine (A) (purine) pairs with uracil (U) (pyrimidine) instead of thymine.
Guanine (G) (purine) pairs with cytosine (C) (pyrimidine), as in DNA.
List of Pyrimidines:
Cytosine (C): Found in both DNA and RNA.
Thymine (T): Found only in DNA.
Uracil (U): Found only in RNA (replaces thymine in RNA).
List of Purines:
Adenine (A): Found in both DNA and RNA.
Guanine (G): Found in both DNA and RNA.

Question 95

Explain how DNA can be denatured, and the conditions with which it can reanneal.

Answer 95

Denaturation refers to the process in which the double-stranded DNA (dsDNA) unwinds and separates into two individual strands, breaking the hydrogen bonds between complementary nitrogenous bases.
How DNA Denatures:
Heat: The most common method of denaturation is to heat the DNA. As the temperature increases, the hydrogen bonds between the complementary bases (A-T and G-C) weaken and eventually break, causing the strands to separate. The temperature required for denaturation depends on the GC content of the DNA (DNA with more G-C pairs requires higher temperatures to denature because G-C pairs have three hydrogen bonds compared to A-T pairs, which have only two).
The melting temperature (Tm) is the temperature at which half of the DNA is denatured. Tm increases with higher G-C content.
Chemical Agents: Certain chemicals, such as urea or formamide, can denature DNA by disrupting the hydrogen bonding between the bases. These chemicals are often used in experiments requiring controlled denaturation at lower temperatures.
pH Changes: Extreme changes in pH (either highly acidic or basic) can also disrupt the hydrogen bonds between bases and denature DNA. However, this is a less common method in typical laboratory protocols.
Effects of Denaturation:
Denatured DNA becomes single-stranded and is more reactive. Single strands are more accessible for processes like replication, transcription, or hybridization.
Denatured DNA absorbs more UV light than double-stranded DNA due to the disruption of base stacking interactions, which can be used to measure the extent of denaturation.

DNA Reannealing:
Reannealing is the process by which the denatured single-stranded DNA strands come back together to form a stable double-stranded DNA molecule. This can occur when the denaturing conditions (such as high heat) are reversed or when appropriate conditions are applied for strand pairing.
Conditions for Reannealing:
Cooling: To reanneal, the temperature must be lowered gradually. As the temperature decreases, the individual DNA strands are able to find their complementary partners and form new hydrogen bonds. This process is highly sequence-dependent, meaning that the strands must be complementary to each other to reanneal successfully.
Salt Concentration: The presence of salt in the solution can help facilitate reannealing by stabilizing the negatively charged phosphate backbone of the DNA. High salt concentrations shield the negative charges on the backbone, reducing repulsion between the strands and promoting hybridization.
A typical buffer for reannealing contains salts such as NaCl or KCl to stabilize the strands.
DNA Concentration: A higher concentration of single-stranded DNA increases the likelihood that complementary strands will encounter each other and reanneal. However, very high concentrations can sometimes cause non-specific binding or aggregation of the DNA strands.
Time: Reannealing is typically a slow process. The rate at which DNA reanneals depends on the temperature, salt concentration, and the complexity of the DNA sequence. Shorter DNA fragments reanneal more rapidly than longer ones due to less complex matching of base pairs.
Factors Affecting Reannealing:
GC Content: DNA with a higher GC content tends to reanneal more slowly, as GC pairs have three hydrogen bonds, making them more stable than the two hydrogen bonds formed between AT pairs.
Sequence Specificity: DNA sequences with a high degree of similarity or perfect complementarity are more likely to reanneal efficiently. Sequences with mismatches or repetitive elements may cause inefficient reannealing or mispairing.

Question 96

Deoxythymidine structure

Answer 96

also known as thymidine, is a chemical compound that is a nucleoside in DNA. It is made up of a deoxyribose sugar and a thymine base.

Question 97

Guanylate structure

Answer 97

A guanylate molecule, also known as guanosine monophosphate (GMP), consists of a phosphate group, a ribose sugar, and the nucleobase guanine

Question 98

Explain the functions of nucleic acids.

Answer 98

Nucleic acids are essential biomolecules that serve a variety of critical functions in living organisms. They are primarily involved in the storage, transmission, and expression of genetic information. The two main types of nucleic acids are DNA (deoxyribonucleic acid) and RNA (ribonucleic acid), each playing distinct roles. Here’s an overview of their key functions:
Functions of DNA (Deoxyribonucleic Acid):
Genetic Information Storage:
DNA is the primary molecule that stores genetic information in living organisms. It contains the instructions for building and maintaining an organism, encoded in the form of sequences of nitrogenous bases (A, T, C, and G).
This genetic code is passed down from one generation to the next during replication and is responsible for the inheritance of traits.
Replication:
DNA replication is the process by which a cell makes a copy of its DNA before cell division. The two strands of the DNA molecule separate, and each strand serves as a template for the formation of a new complementary strand, ensuring that genetic information is faithfully passed on to daughter cells.
Blueprint for Protein Synthesis:
DNA serves as the template for the synthesis of RNA, which in turn is involved in the synthesis of proteins. This process occurs in two stages:
Transcription: The information encoded in DNA is transcribed into messenger RNA (mRNA).
Translation: The mRNA is translated into a specific sequence of amino acids to form proteins, which carry out most of the functions in cells.

Functions of RNA (Ribonucleic Acid):
RNA plays various roles in gene expression, protein synthesis, and regulation. There are several types of RNA, each with specialized functions:
Messenger RNA (mRNA):
mRNA is the transcript of the genetic information encoded in DNA. It carries the genetic instructions from the nucleus (in eukaryotes) to the ribosomes in the cytoplasm, where proteins are synthesized.
mRNA acts as a temporary copy of the genetic code that is used during translation to synthesize proteins.
Transfer RNA (tRNA):
tRNA helps decode the mRNA sequence into a polypeptide chain (protein) by transporting specific amino acids to the ribosome. It “reads” the codons in mRNA and delivers the appropriate amino acid to form the growing protein chain.
Ribosomal RNA (rRNA):
rRNA is a structural and functional component of ribosomes, the molecular machines that synthesize proteins. rRNA helps catalyze the formation of peptide bonds between amino acids during protein synthesis and ensures the correct alignment of mRNA and tRNA on the ribosome.
RNA as Catalysts (Ribozymes):
Some RNA molecules, known as ribozymes, have catalytic properties and can perform chemical reactions, such as cleaving other RNA molecules. Ribozymes are involved in processes like RNA splicing, which removes non-coding regions (introns) from pre-mRNA during processing.

Question 99

Describe how saturated and unsaturated fatty acids have different melting points.

Answer 99

Saturated Fatty Acids:
Structure: Saturated fatty acids have no double bonds between the carbon atoms in their hydrocarbon chain. All the carbon atoms are fully “saturated” with hydrogen atoms, resulting in straight, linear chains.
Packing: Because the chains are straight, saturated fatty acids can pack closely together, allowing for strong van der Waals forces (a type of intermolecular attraction) between the molecules.
Melting Point: The close packing and strong intermolecular interactions require more energy (heat) to break the bonds, so saturated fatty acids typically have higher melting points. At room temperature, many saturated fats are solid, such as butter or lard.
Unsaturated Fatty Acids:
Structure: Unsaturated fatty acids contain one or more double bonds in their hydrocarbon chain. These double bonds introduce kinks or bends in the chain, preventing the molecules from packing tightly together.
Monounsaturated fatty acids have one double bond (e.g., oleic acid).
Polyunsaturated fatty acids have multiple double bonds (e.g., linoleic acid, alpha-linolenic acid).
Packing: The kinks caused by the double bonds make the molecules less able to pack closely together, resulting in weaker van der Waals forces between the molecules.
Melting Point: Due to the less efficient packing and weaker intermolecular forces, unsaturated fatty acids have lower melting points than saturated fatty acids. At room temperature, many unsaturated fats are liquid, such as olive oil or fish oil.

Question 100

Explain what the main difference between saturated and unsaturated fatty acids and how it affects membrane stability.

Answer 100

1. Structure:
   Saturated Fatty Acids: These fatty acids contain no double bonds between the carbon atoms in their hydrocarbon chain. All the carbon atoms are saturated with hydrogen, resulting in straight chains.
   Unsaturated Fatty Acids: These fatty acids contain one or more double bonds between carbon atoms. The presence of double bonds introduces kinks or bends in the hydrocarbon chain, preventing the molecules from packing tightly together.
2. Effects on Membrane Stability:
   The structure of these fatty acids directly affects the fluidity and stability of biological membranes, which are composed primarily of phospholipids that have fatty acid tails.
   A. Saturated Fatty Acids and Membrane Stability:
   Straight chains of saturated fatty acids allow the molecules to pack closely together, forming a tightly packed membrane. This leads to a rigid or less fluid membrane structure.
   Membranes rich in saturated fatty acids tend to be less flexible and more stable at high temperatures. However, they may become too rigid at lower temperatures, affecting the function of membrane proteins and processes like transport.
   B. Unsaturated Fatty Acids and Membrane Stability:
   The kinks in the hydrocarbon chains of unsaturated fatty acids prevent the fatty acid tails from packing closely together. This results in a more fluid membrane.
   The increased fluidity makes membranes with unsaturated fatty acids more flexible and dynamic, allowing better accommodation of membrane proteins and enhanced transport functions. These membranes also perform better at low temperatures, as the fluidity prevents the membrane from becoming too rigid.
   However, too much unsaturation (many double bonds) can make the membrane too fluid and reduce its overall stability.
   Summary:
   Saturated fatty acids lead to more rigid and stable membranes at high temperatures, but can make membranes too rigid at low temperatures.
   Unsaturated fatty acids contribute to more fluid membranes that are more flexible and stable at low temperatures, but excessive unsaturation can compromise membrane stability.

Question 101

Fatty acids can be saturated, monounsaturated and polyunsaturated. Explain the difference between them.

Answer 101

1. Saturated Fatty Acids:
   Structure: Saturated fatty acids have no double bonds between the carbon atoms in the hydrocarbon chain. Every carbon atom is “saturated” with hydrogen atoms.
   Example: Stearic acid (found in animal fats) and palmitic acid.
   Properties:
   These fatty acids have straight chains, which allow them to pack closely together.
   They are typically solid at room temperature (e.g., butter, lard).
   Common in animal fats and tropical oils (such as coconut and palm oil).
2. Monounsaturated Fatty Acids (MUFA):
   Structure: Monounsaturated fatty acids contain one double bond in the hydrocarbon chain. This double bond creates a kink in the chain, preventing tight packing.
   Example: Oleic acid (found in olive oil, avocado, and many nuts).
   Properties:
   The presence of one double bond makes the chain less straight, causing the molecules to pack less tightly compared to saturated fatty acids.
   These fatty acids are typically liquid at room temperature (e.g., olive oil).
   Monounsaturated fats are considered healthier than saturated fats and are commonly found in plant-based oils.
3. Polyunsaturated Fatty Acids (PUFA):
   Structure: Polyunsaturated fatty acids contain more than one double bond in their hydrocarbon chain. Multiple double bonds create multiple kinks in the chain, further preventing tight packing.
   Example: Linoleic acid (an omega-6 fatty acid found in vegetable oils like sunflower oil) and alpha-linolenic acid (an omega-3 fatty acid found in flaxseed oil and walnuts).
   Properties:
   With more than one double bond, these fatty acids have greater fluidity than monounsaturated or saturated fats.
   They are typically liquid at room temperature (e.g., fish oil, corn oil).
   Polyunsaturated fats are essential for human health, as the body cannot synthesize certain polyunsaturated fatty acids (like omega-3 and omega-6 fatty acids), which must be obtained from the diet.

Question 102

Describe how chemical properties of wax make it insoluble in water and have a high melting point.

Answer 102

Waxes are a class of lipids that are made up of long-chain fatty acids linked to long-chain alcohols. The chemical properties of waxes contribute to their insolubility in water and their high melting point. Here’s how these properties are explained:
1. Insolubility in Water:
Hydrophobic Nature: Waxes are primarily composed of long hydrocarbon chains, which are nonpolar. Water, on the other hand, is a highly polar molecule. In order for a substance to dissolve in water, it typically needs to be polar or capable of forming hydrogen bonds with water molecules.
Waxes, being nonpolar, do not form hydrogen bonds with water. Instead, they interact with each other through van der Waals forces, which are much weaker than the hydrogen bonds between water molecules.
As a result, water molecules cannot surround and dissolve the wax molecules effectively, making waxes insoluble in water.
2. High Melting Point:
Long Hydrocarbon Chains: The long-chain fatty acids and alcohols in waxes create a dense, tightly packed structure. The van der Waals forces between the long hydrocarbon chains are substantial, requiring a significant amount of energy (heat) to break these interactions.
High Molecular Weight: Waxes tend to have high molecular weights due to their long hydrocarbon chains. Larger molecules require more energy to overcome the forces holding them together, which results in a higher melting point.
Solid at Room Temperature: Due to the strength of these intermolecular forces, waxes are usually solid at room temperature, as they have a high melting point compared to other lipids like oils or fats

Question 103

At room temperature, olive oil is liquid, butter is soft solid and beef fat is hard solid. Explain why that is.

Answer 103

Saturated fatty acids (beef): Higher melting points, typically solid at room temperature, due to straight chains and tight packing.
Unsaturated fatty acids (olive oil): Lower melting points, typically liquid at room temperature, due to bent chains and weaker intermolecular forces.

Question 104

Describe the process of facilitated diffusion and explain how it differs from simple diffusion.

Answer 104

Facilitated diffusion and simple diffusion are both passive transport mechanisms that allow molecules to move across the cell membrane down their concentration gradient (from high to low concentration) without the use of energy (ATP). However, they differ in how the molecules cross the membrane.

Facilitated Diffusion:
Process: Facilitated diffusion involves the movement of molecules through the cell membrane with the help of transport proteins. These proteins act as channels or carriers that allow specific molecules to cross the membrane more efficiently.
Mechanism:
Channel Proteins: These form open channels through the membrane that allow ions or small molecules to pass through, such as sodium ions (Na⁺) or potassium ions (K⁺).
Carrier Proteins: These proteins undergo conformational changes to transport molecules across the membrane. A carrier protein binds to the molecule on one side of the membrane, changes shape, and releases the molecule on the other side. Examples include glucose transporter proteins.
Specificity: Facilitated diffusion is highly specific to certain molecules due to the selective nature of the transport proteins. For example, glucose can only be transported by specific glucose transporters.
Speed: Facilitated diffusion is faster than simple diffusion for certain molecules because the transport proteins can accelerate the movement across the membrane.

Simple Diffusion:
Process: Simple diffusion is the movement of molecules directly through the lipid bilayer of the membrane without the need for transport proteins.
Mechanism: Small, nonpolar molecules (such as oxygen (O₂), carbon dioxide (CO₂), and lipid-soluble substances) move through the hydrophobic interior of the lipid bilayer, down their concentration gradient. Larger or polar molecules cannot pass through the lipid bilayer easily without assistance.
Specificity: Simple diffusion is non-specific, meaning that any molecule that is small and nonpolar enough to pass through the lipid bilayer can diffuse freely.
Speed: Simple diffusion is generally slower than facilitated diffusion, especially for larger or polar molecules.

Question 105

Explain the steps of membrane fusion during neurotransmitter release.

Answer 105

The process of membrane fusion during neurotransmitter release is a crucial part of synaptic transmission. It involves the fusion of a synaptic vesicle containing neurotransmitters with the presynaptic membrane, leading to the release of neurotransmitters into the synaptic cleft. Here are the key steps of this process:

1. Action Potential Arrival:
   An action potential (electrical signal) travels down the axon to the axon terminal of the presynaptic neuron.
   The depolarization of the presynaptic membrane caused by the action potential leads to the opening of voltage-gated calcium (Ca²⁺) channels.
2. Calcium Influx:
   When the voltage-gated calcium channels open, calcium ions (Ca²⁺) flow into the presynaptic terminal from the extracellular space.
   The increase in intracellular calcium concentration is a key trigger for neurotransmitter release.
3. Vesicle Docking:
   The synaptic vesicle, which contains neurotransmitters (such as dopamine, serotonin, acetylcholine, etc.), is tethered to the presynaptic membrane near the release site by proteins in the active zone.
   SNARE proteins (such as synaptobrevin on the vesicle and syntaxin and SNAP-25 on the presynaptic membrane) play a crucial role in bringing the vesicle and membrane close together.
4. Vesicle Priming:
   Priming refers to the preparation of the vesicle for fusion. This step involves the assembly of the SNARE complex, which is necessary for the fusion of the vesicle with the membrane.
   Synaptotagmin, a calcium-sensitive protein on the vesicle, plays a key role in sensing the increase in calcium and initiating the fusion process.
5. Membrane Fusion:
   Calcium binding to synaptotagmin causes a conformational change in the protein, which triggers the vesicle to fuse with the presynaptic membrane.
   The SNARE complex helps to mediate this fusion by bringing the lipid bilayers of the vesicle and the presynaptic membrane close together.
   Fusion pores form as the vesicle and membrane merge, allowing neurotransmitters to flow from the vesicle into the synaptic cleft.
6. Neurotransmitter Release:
   As the vesicle fuses with the presynaptic membrane, the neurotransmitters inside the vesicle are released into the synaptic cleft through the fusion pore.
   These neurotransmitters will then bind to receptors on the postsynaptic membrane, transmitting the signal to the next neuron or muscle cell.
7. Vesicle Recycling:
   After the neurotransmitter release, the vesicle membrane is retrieved through a process known as endocytosis. This involves the vesicle being pulled back into the presynaptic neuron and either being recycled to be refilled with neurotransmitters or degraded.

Question 106

What is the “fluid mosaic” model of biological membranes, and how does it relate to membrane permeability?

Answer 106

The “fluid mosaic model” is a widely accepted concept that describes the structure and behavior of biological membranes, particularly the phospholipid bilayer that forms the basis of the cell membrane.

Key Features of the Fluid Mosaic Model:
Phospholipid Bilayer:
The membrane is primarily composed of phospholipids, which have hydrophilic (water-loving) heads and hydrophobic (water-fearing) tails.
The hydrophilic heads face outward toward the aqueous environment (inside and outside the cell), while the hydrophobic tails are oriented inward, away from water. This arrangement forms the lipid bilayer.
Fluidity:
The term “fluid” refers to the dynamic nature of the membrane. The phospholipids and embedded proteins can move laterally within the bilayer, allowing the membrane to be flexible and self-healing.
The membrane is not a rigid structure, but instead behaves more like a fluid, where the lipids and proteins can shift positions, change shape, or even rotate.
The degree of fluidity depends on factors like the temperature and the composition of fatty acids in the phospholipids (e.g., saturated versus unsaturated fatty acids).
Mosaic of Proteins:
The “mosaic” part of the model refers to the patchwork arrangement of proteins within the lipid bilayer.
Integral (or transmembrane) proteins span across the membrane, while peripheral proteins are associated with the surface of the membrane.
These proteins serve various functions, including transport, signal transduction, and cell recognition.
Carbohydrates may also be attached to proteins (glycoproteins) or lipids (glycolipids), forming structures such as glycocalyx, which play a role in cell-cell interaction.

Membrane Permeability and the Fluid Mosaic Model:
Selective Permeability: The fluid mosaic model explains the selective permeability of biological membranes, meaning that only certain substances can pass through while others are restricted. This is crucial for maintaining homeostasis in the cell.
The phospholipid bilayer itself acts as a barrier to the movement of large, polar molecules (e.g., glucose) and ions.
Hydrophobic molecules (e.g., oxygen, carbon dioxide) and small, nonpolar molecules can diffuse freely across the membrane because they can interact with the hydrophobic interior of the lipid bilayer.
The fluidity of the membrane allows membrane proteins to facilitate the transport of specific molecules that cannot pass through the lipid bilayer by themselves (e.g., using channels or carrier proteins for ions or glucose).
Influence of Membrane Fluidity on Permeability:
The fluidity of the membrane plays a key role in its permeability. At higher temperatures, the membrane becomes more fluid, which can increase the permeability of certain molecules. Conversely, at lower temperatures, the membrane becomes more rigid, decreasing permeability.
The unsaturated fatty acids in phospholipids create kinks in the tail chains, preventing close packing and increasing membrane fluidity, which can affect how easily molecules pass through.

Question 107

What is the primary function flippases in the plasma membrane.

Answer 107

Flippases are enzymes that play a crucial role in maintaining the asymmetry of the phospholipid bilayer in the plasma membrane. The primary function of flippases is to catalyze the movement (flip) of phospholipids between the two leaflets of the bilayer, specifically from the outer leaflet to the inner leaflet or vice versa.

Functions of Flippases:
Maintaining Membrane Asymmetry:
Phospholipids are not distributed symmetrically between the two leaflets of the bilayer. For example, phosphatidylserine and phosphatidylethanolamine are more commonly found on the inner leaflet, while phosphatidylcholine and sphingomyelin are more abundant on the outer leaflet.
Flippases help maintain this asymmetry by flipping specific phospholipids from one side of the bilayer to the other.
Membrane Functionality:
The asymmetry of the plasma membrane is essential for its functionality, including membrane fluidity, signaling, and cell recognition. For example, phosphatidylserine on the inner leaflet is exposed on the outer leaflet during apoptosis (programmed cell death), signaling to phagocytic cells that the cell is undergoing apoptosis.
Vesicle Formation:
Flippases also play a role in the formation of vesicles during processes like endocytosis and exocytosis, as proper lipid distribution is necessary for membrane curvature and vesicle formation.
Regulation of Lipid Composition:
By flipping lipids between the leaflets, flippases help regulate the lipid composition of the two sides of the membrane, ensuring proper membrane integrity and function.
How Flippases Work:
Flippases are membrane-bound transport proteins that use ATP hydrolysis to move specific phospholipids against their concentration gradient. This energy-dependent process allows them to flip lipids from one leaflet to the other.

Question 108

Describe the three general classes of transport systems.
1. Cotransport
2. Antiport
3. Symport

Answer 108

These systems typically involve membrane proteins that facilitate the transport of ions, nutrients, or other substances. While they all involve the movement of molecules either in or out of the cell, the main difference between them lies in how the molecules are transported relative to each other.

1. Cotransport:
   Definition: Cotransport is a broad term that refers to any transport mechanism where two or more molecules are transported across the membrane simultaneously. Cotransport includes both symport and antiport systems, as both involve the movement of multiple molecules at once.
   Mechanism: In cotransport, one of the molecules typically moves down its concentration gradient (from high to low concentration), which provides the energy to transport another molecule against its concentration gradient (from low to high concentration). This is often coupled to the movement of ions, such as Na⁺ (sodium) or H⁺ (protons).
2. Antiport:
   Definition: An antiport is a type of cotransport system where two molecules are transported in opposite directions across the membrane.
   Mechanism: In antiport, one molecule is moved into the cell while another is moved out of the cell. This transport is often driven by the electrochemical gradient of ions like Na⁺ or H⁺, which provide the energy for the active transport of the other molecule.
3. Symport:
   Definition: A symport is another type of cotransport system where two molecules are transported in the same direction across the membrane.
   Mechanism: In symport, both molecules move either into or out of the cell, often driven by the gradient of one molecule (such as Na⁺ or H⁺) that provides the energy to transport the second molecule against its concentration gradient.