Biological Molecules — Diagnostic Tests

Unit Tests

UT-1: Alpha-Glucose vs Beta-Glucose and Polysaccharide Formation

Question:

Alpha-glucose and beta-glucose are both hexose monosaccharides with the molecular formula $\text{C}_6\text{H}_{12}\text{O}_6$, yet they form polysaccharides with very different properties.

(a) Describe the precise structural difference between alpha-glucose and beta-glucose.

(b) Starch is formed from alpha-glucose. Describe the formation of a glycosidic bond between two alpha-glucose molecules, naming the specific atoms involved and the molecule eliminated.

(c) Cellulose is formed from beta-glucose. Explain why cellulose molecules are straight and form hydrogen-bonded fibres, whereas starch molecules are coiled.

(d) Humans can digest starch but not cellulose. Explain this difference with reference to enzyme specificity.

Solution:

(a) The difference lies in the position of the hydroxyl ($-$OH) group on carbon 1. In alpha-glucose, the $-$OH group on C1 is below the plane of the ring; in beta-glucose, the $-$OH group on C1 is above the plane of the ring. All other $-$OH positions are identical.

(b) A condensation reaction occurs between the hydroxyl group on C1 of one alpha-glucose molecule and the hydroxyl group on C4 of another alpha-glucose molecule. This forms a glycosidic bond (specifically a 1,4-glycosidic bond) and eliminates a molecule of water ($\text{H}_2\text{O}$). The oxygen atom in the glycosidic bond originates from C1 of the first glucose molecule.

(c) In beta-glucose, the alternating orientation of $-$OH groups above and below the ring means that each successive beta-glucose molecule must rotate 180 degrees relative to its neighbour to form a 1,4-glycosidic bond. This produces a straight, unbranched chain. The straight chains align parallel to each other, allowing extensive hydrogen bonding between the $-$OH groups of adjacent chains, forming strong, insoluble microfibrils. In alpha-glucose, the $-$OH groups all point in the same relative direction, so successive molecules do not need to rotate — the chain naturally coils into a helix, which is compact and less accessible for hydrogen bonding between chains.

(d) The enzyme amylase, produced in human saliva and the pancreas, has an active site complementary in shape to the alpha-glycosidic bond. Cellulase, the enzyme that hydrolyses beta-1,4-glycosidic bonds, is not produced by humans. The active site of amylase cannot bind to the beta-glycosidic bond because the spatial arrangement of atoms in the beta linkage does not match the active site's three-dimensional shape. This is an example of the lock-and-key model of enzyme action (though induced fit also applies more broadly — the key point is that the active site is specific to the alpha linkage).

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UT-2: Levels of Protein Structure and Enzyme Active Site Models

Question:

A newly discovered enzyme, protease X, catalyses the hydrolysis of peptide bonds in proteins at pH 7.4. When the pH is lowered to pH 5.0, the rate of reaction falls to near zero. Raising the temperature from 25 degrees Celsius to 45 degrees Celsius increases the rate, but above 60 degrees Celsius the rate drops to zero and does not recover when the temperature is returned to 25 degrees Celsius.

(a) Describe the four levels of protein structure (primary, secondary, tertiary, quaternary), including the types of bonds and interactions that maintain each level.

(b) Explain the effect of lowering the pH to 5.0 on protease X's activity, making reference to the induced fit model of enzyme action.

(c) Explain why the loss of activity above 60 degrees Celsius is irreversible, whereas the loss of activity at low temperature is reversible.

(d) Explain the difference between the lock-and-key model and the induced fit model of enzyme action, and state which model is currently preferred with a reason.

Solution:

(a) Primary structure: the sequence of amino acids in the polypeptide chain, held together by peptide bonds (covalent). Secondary structure: the folding of the polypeptide chain into alpha-helices and beta-pleated sheets, stabilised by hydrogen bonds between the $-$C=O of one amino acid and the $-$N-H of another amino acid four residues away (in an alpha-helix). Tertiary structure: the overall three-dimensional shape of a single polypeptide chain, stabilised by disulfide bridges (covalent, between cysteine residues), ionic bonds (between $-$NH$_3^+$ and $-$COO$^-$ groups), hydrogen bonds, and hydrophobic interactions (between non-polar R-groups that cluster in the interior of the protein away from water). Quaternary structure: the arrangement of multiple polypeptide subunits into a functional protein (e.g., haemoglobin has four subunits), held together by the same types of bonds as tertiary structure. Not all proteins have quaternary structure.

(b) At pH 5.0, the increased concentration of H$^+$ ions causes amino acid side chains in the active site to gain or lose protons. This alters the charge distribution and the three-dimensional shape of the active site. Under the induced fit model, the active site is flexible and changes shape slightly when the substrate binds. If the active site's shape is altered by pH change, the substrate can no longer induce the correct complementary fit, so the enzyme-substrate complex does not form efficiently and the rate falls to near zero. The bonds broken are ionic bonds and hydrogen bonds in the tertiary structure.

(c) Above 60 degrees Celsius, the increased kinetic energy breaks the hydrogen bonds, ionic bonds, and hydrophobic interactions maintaining the tertiary structure. The enzyme denatures — the active site permanently loses its specific shape. Since this involves breaking bonds in the tertiary structure, the change is irreversible (the protein cannot spontaneously refold correctly). At low temperature, the kinetic energy of enzyme and substrate molecules is reduced, so fewer successful collisions occur per unit time. No bonds in the protein structure are broken, so when the temperature is raised again, the enzyme resumes normal activity — this is reversible.

(d) The lock-and-key model proposes that the active site has a rigid, fixed shape that is exactly complementary to the substrate, like a key fitting a lock. The induced fit model proposes that the active site is flexible and changes shape slightly upon substrate binding to mould around the substrate, improving the fit and placing strain on the substrate to lower the activation energy. The induced fit model is currently preferred because X-ray crystallography and spectroscopic evidence show that enzyme active sites do change conformation upon substrate binding, and because the induced fit model better explains why enzymes catalyse only one reaction (specificity) and can exclude molecules of very similar structure.

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UT-3: DNA vs RNA Structure and Enzyme Inhibition

Question:

A student investigates the effect of a competitive inhibitor on the enzyme catalase, which breaks down hydrogen peroxide into water and oxygen. The student measures the initial rate of reaction at different hydrogen peroxide concentrations, with and without the inhibitor.

The results are shown below:

[H$_2$O$_2$] (mol dm$^{-3}$)	Rate without inhibitor (arbitrary units)	Rate with inhibitor (arbitrary units)
0.05	3.2	1.6
0.10	5.6	3.2
0.20	8.0	5.6
0.40	9.6	8.0
0.80	10.2	9.6

(a) Explain why the rate with the inhibitor approaches the rate without the inhibitor at high substrate concentrations.

(b) Describe three structural differences between DNA and RNA.

(c) ATP is a nucleotide derivative. Describe the structure of ATP and explain why the hydrolysis of ATP to ADP and inorganic phosphate is described as exergonic.

(d) Explain how a non-competitive inhibitor differs in its mechanism of action from a competitive inhibitor, and describe the effect of increasing substrate concentration on the activity of an enzyme affected by a non-competitive inhibitor.

Solution:

(a) A competitive inhibitor has a similar molecular shape to the substrate and binds reversibly to the active site, blocking substrate access. At high substrate concentrations, the probability that a substrate molecule rather than the inhibitor occupies the active site increases, because substrate molecules outcompete the inhibitor molecules for the active site. Therefore, the $V_{\max}$ is the same with and without the inhibitor, but the apparent $K_m$ is increased (the Michaelis constant — the substrate concentration at half $V_{\max}$ — is higher with the inhibitor present, as seen by the rightward shift of the curve).

(b) Three structural differences between DNA and RNA:

1. Sugar: DNA contains deoxyribose (lacks an oxygen on C2); RNA contains ribose (has an $-$OH on C2).
2. Bases: DNA contains adenine, thymine, cytosine, and guanine; RNA contains adenine, uracil, cytosine, and guanine (uracil replaces thymine).
3. Structure: DNA is typically double-stranded (forming a double helix); RNA is typically single-stranded. (Additional acceptable differences: DNA is a very long molecule; RNA is relatively short. DNA has a regular helical structure; RNA can fold into complex 3D shapes.)

(c) ATP (adenosine triphosphate) consists of three components: adenine (a nitrogenous base), ribose (a pentose sugar), and three phosphate groups. The hydrolysis of the terminal phosphoanhydride bond between the second and third phosphate groups releases energy (approximately 30.5 kJ mol$^{-1}$) because the products (ADP + P$_i$) are more stable than ATP — the negative charges on the phosphate groups in ATP repel each other, making the bond strained and high-energy. The reaction is exergonic because it has a negative change in Gibbs free energy ($\Delta G \lt 0$); energy is released to the surroundings.

(d) A non-competitive inhibitor binds to the enzyme at a site other than the active site (the allosteric site). This binding causes a conformational change in the enzyme's tertiary structure that alters the shape of the active site, preventing substrate binding. Unlike competitive inhibition, the inhibitor does not compete with the substrate for the active site. Increasing substrate concentration does not overcome non-competitive inhibition because the inhibitor and substrate bind at different sites — the inhibitor can bind whether or not the substrate is present. Therefore, $V_{\max}$ is reduced but $K_m$ is unchanged (the substrate still binds with normal affinity to any uninhibited enzyme molecules).

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Integration Tests

IT-1: Protein Synthesis and Enzyme Specificity (with Genetics and DNA)

Question:

Cystic fibrosis is caused by a mutation in the CFTR gene, resulting in the production of a faulty CFTR protein that functions as a chloride ion channel in cell membranes.

(a) Describe how the information in the CFTR gene is used to produce the CFTR protein, including the processes of transcription and translation.

(b) The most common CF mutation is the deletion of three nucleotides (coding for phenylalanine at position 508). Explain why this deletion does not cause a frameshift mutation, but still results in a non-functional protein.

(c) Explain how the change in the primary structure of the CFTR protein (loss of phenylalanine at position 508) leads to a change in the tertiary structure and loss of function.

(d) CFTR protein functions as a channel in the cell surface membrane. With reference to the fluid mosaic model, explain how proteins are positioned in the cell membrane.

Solution:

(a) Transcription: The enzyme RNA polymerase binds to the promoter region of the CFTR gene on DNA. The DNA double helix unwinds (hydrogen bonds between complementary bases break). RNA polymerase catalyses the formation of mRNA using one strand of DNA as a template, following complementary base pairing (A-U, T-A, C-G, G-C). The pre-mRNA undergoes post-transcriptional modification: a 5' cap is added, a poly-A tail is added, and introns are removed by splicing to produce mature mRNA, which exits the nucleus through nuclear pores.

Translation: The mature mRNA binds to a ribosome. Transfer RNA (tRNA) molecules, each carrying a specific amino acid and with an anticodon complementary to a codon on the mRNA, bind to the mRNA at the ribosome. The ribosome moves along the mRNA (5' to 3'), reading codons and forming peptide bonds between adjacent amino acids (catalysed by peptidyl transferase in the large ribosomal subunit). Translation continues until a stop codon is reached. The polypeptide chain is released and folds into its functional three-dimensional structure.

(b) The deletion of three nucleotides removes exactly one codon (since each codon is three nucleotides). The reading frame of the mRNA downstream of the deletion is not shifted — every subsequent codon is still read in the correct triplet. This is why it is not a frameshift mutation. However, the amino acid sequence is altered (phenylalanine is missing at position 508), which affects the protein's folding and function.

(c) The primary structure (amino acid sequence) determines the secondary and tertiary structure of the protein. Phenylalanine is a hydrophobic amino acid; its presence at position 508 is critical for correct folding of the protein. Removing it alters the hydrophobic interactions that guide the protein into its correct tertiary shape. The misfolded protein is recognised by cellular quality control mechanisms (e.g., chaperone proteins in the endoplasmic reticulum), and is degraded before it can reach the cell membrane. Even if some protein reaches the membrane, the altered tertiary structure means the chloride channel does not function correctly.

(d) According to the fluid mosaic model, proteins are embedded in the phospholipid bilayer of the cell membrane. Intrinsic (integral) proteins such as CFTR span the entire bilayer; they have hydrophobic regions on their outer surface that interact with the hydrophobic tails of the phospholipids, and hydrophilic regions that protrude into the aqueous environment on either side of the membrane. Extrinsic (peripheral) proteins are attached to the surface of the membrane, typically bound to intrinsic proteins or to phospholipid heads. The phospholipid molecules can move laterally within the bilayer (fluid), giving the membrane flexibility. The proteins form the "mosaic" pattern within this fluid bilayer.

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IT-2: Water Properties and Gas Exchange Surface (with Exchange and Transport)

Question:

The walls of the alveoli in the lungs are lined with a thin layer of fluid. Surfactant, a phospholipid-protein mixture, is secreted by type II alveolar cells and reduces the surface tension of this fluid.

(a) Explain the role of water's cohesion and high specific heat capacity in maintaining mammalian body temperature.

(b) Explain why water has a high surface tension and describe how this would be problematic in the lungs without surfactant.

(c) With reference to the phospholipid structure, explain how surfactant reduces surface tension at the air-water interface in the alveoli.

(d) The thin layer of fluid between the alveolar epithelium and the capillary endothelium facilitates gas exchange. Explain how the properties of water as a solvent contribute to the transport of oxygen and carbon dioxide in the blood.

Solution:

(a) Cohesion (hydrogen bonding between water molecules) means that a significant amount of energy is required to change water from liquid to vapour (high latent heat of vaporisation). When mammals sweat, the evaporation of water from the skin surface removes heat energy from the body, cooling it. Water's high specific heat capacity (due to extensive hydrogen bonding) means that a large amount of energy is required to change its temperature. This buffers the body against rapid temperature fluctuations — blood, which is mostly water, can transport heat from metabolically active organs to the skin for dissipation without large temperature changes in the blood itself.

(b) Water has a high surface tension because the water molecules at the surface experience an unbalanced net inward pull from hydrogen bonds with molecules below them. This creates a strong, cohesive surface film. In the lungs, without surfactant, the high surface tension of the fluid lining the alveoli would cause the alveoli to collapse inward (atelectasis) during exhalation, as the inward pull of surface tension exceeds the outward pressure from the chest wall. This would drastically reduce the surface area available for gas exchange.

(c) Phospholipids have a hydrophilic phosphate head and two hydrophobic fatty acid tails. At the air-water interface in the alveoli, surfactant phospholipids orient themselves with their hydrophilic heads in the water layer and their hydrophobic tails projecting into the air space. This disrupts the hydrogen bonding network between surface water molecules, reducing the cohesive forces between them and therefore lowering the surface tension. The protein component of surfactant also contributes by spreading the phospholipid molecules across the surface.

(d) Water is an excellent solvent for polar molecules and ions because its polar nature allows it to form hydration shells around charged and polar solutes. Carbon dioxide is transported in the blood in three ways: dissolved directly in plasma (a small fraction), bound to haemoglobin as carbaminohaemoglobin, and as hydrogencarbonate ions ($\text{HCO}_3^-$) inside red blood cells (the majority). The conversion of $\text{CO}_2$ to $\text{HCO}_3^-$ requires the enzyme carbonic anhydrase and depends on water as both a solvent and a reactant: $\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$. Oxygen, although non-polar and poorly soluble in water, is carried by haemoglobin (a protein dissolved in the aqueous cytoplasm of red blood cells) — the small amount that dissolves directly in plasma does so in the aqueous portion of blood.

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IT-3: Enzyme Kinetics in Digestion and Absorption (with Exchange and Transport)

Question:

The digestion of starch begins in the mouth with salivary amylase and continues in the small intestine with pancreatic amylase. The resulting maltose is hydrolysed by maltase, an enzyme embedded in the cell surface membrane of epithelial cells lining the small intestine.

(a) Explain why salivary amylase has an optimum pH of approximately 7, whereas pepsin (a protease in the stomach) has an optimum pH of approximately 2.

(b) Describe how the villi and microvilli of the small intestine epithelium are adapted to maximise the rate of absorption of the products of digestion.

(c) Maltase is a membrane-bound enzyme. Explain the advantage of maltase being embedded in the cell surface membrane rather than being secreted into the lumen of the small intestine.

(d) The products of starch digestion are ultimately used in cellular respiration. Explain how the monosaccharides absorbed from the small intestine are transported in the blood and taken up by cells, including the role of a co-transporter protein.

Solution:

(a) The optimum pH of an enzyme reflects the pH of the environment in which it normally functions. The mouth has a pH close to neutral (approximately 6.5-7.5) maintained by saliva, so salivary amylase has evolved an active site with an optimum at pH 7. At this pH, the ionic charges on amino acid residues in the active site are in their correct state for substrate binding and catalysis. The stomach produces hydrochloric acid, creating a pH of approximately 1.5-2.0. Pepsin has an active site with amino acid residues whose charges are optimised at low pH. If pepsin were exposed to neutral pH, the change in protonation of key residues would alter the active site shape and prevent substrate binding. This ensures that protein digestion occurs efficiently in the stomach and that pepsin is not active (and potentially damaging) in other parts of the digestive system.

(b) Villi are finger-like projections of the intestinal wall that increase the surface area for absorption. Each villus has a single layer of epithelial cells (short diffusion distance), a dense capillary network to carry away absorbed molecules and maintain a steep concentration gradient, and a lacteal (lymph vessel) to absorb fatty acids and glycerol. Microvilli are microscopic projections on the apical surface of each epithelial cell, further increasing the surface area (forming the brush border). Together, villi and microvilli increase the effective absorptive surface area of the small intestine by approximately 600-fold compared to a simple tube.

(c) Having maltase embedded in the cell surface membrane (it is a membrane-bound intrinsic protein) means that maltose is hydrolysed to glucose at the site of absorption. The glucose is produced directly adjacent to the transport proteins on the epithelial cell membrane, creating a very high local concentration of glucose at the membrane surface. This maintains a steep concentration gradient for the co-transport of glucose into the cell, increasing the efficiency of absorption. If maltase were secreted into the lumen, glucose would diffuse throughout the lumen and the concentration gradient driving uptake would be reduced.

(d) Monosaccharides (primarily glucose) are absorbed from the small intestine via secondary active transport (co-transport). The sodium-potassium pump ($\text{Na}^+/\text{K}^+$ ATPase) on the basolateral membrane of the epithelial cell actively transports $\text{Na}^+$ out of the cell and $\text{K}^+$ into the cell, using ATP. This maintains a low intracellular $\text{Na}^+$ concentration. $\text{Na}^+$ then diffuses back into the epithelial cell through a $\text{Na}^+$-glucose co-transporter protein (SGLT1) on the apical membrane, coupled with the transport of glucose against its concentration gradient. Glucose exits the epithelial cell via facilitated diffusion through a GLUT2 transporter on the basolateral membrane into the blood. In the blood, glucose dissolves in the plasma (water as solvent) and is transported to body cells. Cells take up glucose via facilitated diffusion through GLUT transporters on their cell surface membranes, driven by the concentration gradient between blood and the cell cytoplasm.