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Options (Choose One) - Biochemistry

Grade 12IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Metabolism involves the complex network of chemical reactions in cells, categorized into anabolism (building molecules, requiring energy, ΔG>0\Delta G > 0) and catabolism (breaking down molecules, releasing energy, ΔG<0\Delta G < 0).

Amino acids are the building blocks of proteins and possess both an amino group (NH2-NH_2) and a carboxyl group (COOH-COOH). At specific pHpH, they exist as zwitterions: H3N+CHRCOOH_3N^+-CHR-COO^-.

The isoelectric point (pIpI) is the pHpH at which an amino acid has a net charge of zero. At pH<pIpH < pI, the amino acid is protonated and carries a positive charge; at pH>pIpH > pI, it is deprotonated and carries a negative charge.

Proteins have four levels of structure: Primary (sequence of amino acids), Secondary (α\alpha-helices and β\beta-pleated sheets held by HH-bonds), Tertiary (3D folding due to RR-group interactions), and Quaternary (assembly of multiple polypeptide chains).

Enzymes function as biological catalysts by lowering the activation energy (EaE_a). Enzyme kinetics are often described by the Michaelis-Menten model, where KmK_m represents the substrate concentration at which the reaction rate is 12Vmax\frac{1}{2}V_{max}.

Lipids include triglycerides, which are esters formed from glycerol and three fatty acids. The degree of unsaturation in a lipid can be determined using the Iodine Number, defined as the mass of I2I_2 in grams that reacts with 100g100\,g of the lipid.

Carbohydrates have the general empirical formula Cx(H2O)yC_x(H_2O)_y. Monosaccharides like glucose (C6H12O6C_6H_{12}O_6) undergo condensation reactions to form disaccharides and polysaccharides, linked by glycosidic bonds.

Vitamins are micronutrients. Fat-soluble vitamins (e.g., Vitamin A,D,E,KA, D, E, K) have long hydrocarbon chains or rings, while water-soluble vitamins (e.g., Vitamin CC) have multiple OH-OH groups allowing for hydrogen bonding with water.

📐Formulae

v=Vmax[S]Km+[S]v = \frac{V_{max}[S]}{K_m + [S]}

pI=12(pKa1+pKa2)pI = \frac{1}{2}(pK_{a1} + pK_{a2})

Iodine Number=m(I2) reacting with lipidm(lipid)×100\text{Iodine Number} = \frac{m(I_2) \text{ reacting with lipid}}{m(\text{lipid})} \times 100

Energy Content=q=mcΔT\text{Energy Content} = q = mc\Delta T

pH=pKa+log([A][HA])pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)

💡Examples

Problem 1:

A sample of linoleic acid (C18H32O2C_{18}H_{32}O_2, Mr=280.45gmol1M_r = 280.45\,g\,mol^{-1}) contains two C=CC=C double bonds. Calculate the Iodine Number of linoleic acid.

Solution:

  1. One mole of linoleic acid reacts with 2mol2\,mol of I2I_2 because there are two double bonds.
  2. Mass of 1mol1\,mol of linoleic acid = 280.45g280.45\,g.
  3. Mass of 2mol2\,mol of I2=2×(2×126.90gmol1)=507.6gI_2 = 2 \times (2 \times 126.90\,g\,mol^{-1}) = 507.6\,g.
  4. Iodine Number = 507.6gI2280.45gacid×100=181\frac{507.6\,g\,I_2}{280.45\,g\,acid} \times 100 = 181.

Explanation:

The Iodine Number measures the degree of unsaturation. Since each C=CC=C bond reacts with one molecule of I2I_2 via an addition reaction, the stoichiometry is 1:21:2 for linoleic acid to I2I_2.

Problem 2:

Explain the effect of a competitive inhibitor on the Michaelis-Menten constants VmaxV_{max} and KmK_m.

Solution:

VmaxV_{max} remains unchanged; KmK_m increases.

Explanation:

A competitive inhibitor competes with the substrate for the active site. At very high substrate concentrations ([S][S]), the substrate outcompetes the inhibitor, so the maximum velocity (VmaxV_{max}) can still be reached. However, more substrate is needed to reach 12Vmax\frac{1}{2}V_{max}, thus the Michaelis constant (KmK_m) increases, indicating a lower affinity of the enzyme for the substrate in the presence of the inhibitor.

Problem 3:

Determine the dominant species of Glycine (NH2CH2COOHNH_2CH_2COOH, pKa1=2.3,pKa2=9.6pK_{a1}=2.3, pK_{a2}=9.6) in a solution at pH=1.0pH = 1.0.

Solution:

H3N+CH2COOHH_3N^+-CH_2-COOH

Explanation:

At pH=1.0pH = 1.0, which is below the first pKapK_a (2.32.3), both the amino group and the carboxyl group are fully protonated. The amino group exists as NH3+NH_3^+ and the carboxyl group remains as COOHCOOH, resulting in a net positive charge on the molecule.

Biochemistry - Revision Notes & Key Formulas | IB Grade 12 Chemistry