(50 Calculation-Based Questions and Answers Related to Proteins)
![](https://i0.wp.com/biochemz.com/wp-content/uploads/2024/07/img-z6woL33bHT30hVApwS0fOeD7.png?resize=1024%2C1024&ssl=1)
50 CALCULATION-BASED QUESTIONS AND ANSWERS RELATED TO PROTEINS
Protein being a major course in biochemistry is filled with calculations which are not so complex, just the specific formulas need to be noted as shown below:
- Q: What is the molecular weight of a protein composed of 100 amino acids, each with a molecular weight of 110 Da?
A: Molecular weight = 100 amino acids * 110 Da = 11,000 Da.
- Q: Given a protein’s molecular weight of 64,500 Da, how many amino acids are present if each has a molecular weight of 125 Da?
A: Number of amino acids = Molecular weight / Molecular weight of one amino acid = 64,500 Da / 125 Da ≈ 516 amino acids.
- Q: Calculate the isoelectric point (pI) of a protein with pKa values of 2.34 (N-terminus), 9.60 (C-terminus), and 5.00 (amino acid side chains).
A: pI = (pKa1 + pKa2) / 2 = (2.34 + 9.60) / 2 ≈ 5.97.
- Q: A protein is composed of 40% hydrophobic amino acids, 30% polar amino acids, and 30% charged amino acids. If it has 300 amino acids, how many are charged?
A: Number of charged amino acids = 0.30 * 300 = 90 amino acids.
- Q: A protein contains 15 cysteine residues. How many disulfide bonds can potentially form within this protein?
A: Number of disulfide bonds = Number of cysteines / 2 = 15 / 2 = 7.5. Since disulfide bonds form between pairs of cysteine residues, it will be 7 disulfide bonds.
- Q: If a protein has an extinction coefficient of 32,500 M^(-1)cm^(-1) at 280 nm, what is its concentration in mg/mL if the absorbance at 280 nm is 0.6?
A: Concentration (mg/mL) = Absorbance / Extinction coefficient = 0.6 / 32,500 ≈ 0.018 mg/mL.
- Q: A protein solution has an optical density of 0.5 at 450 nm in a 1 cm cuvette. If the molecular weight of the protein is 60,000 Da, what is its concentration in μM?
A: Concentration (μM) = (Absorbance * Molecular weight) / (Extinction coefficient * Path length) = (0.5 * 60,000) / (1 * Extinction coefficient).
- Q: A protein is 70% folded. If its molar ellipticity at 220 nm is -8000 deg cm^2 dmol^(-1), what is the molar ellipticity of the fully folded protein?
A: Molar ellipticity of fully folded protein = Molar ellipticity at 220 nm / Fraction folded = -8000 deg cm^2 dmol^(-1) / 0.70 ≈ -11429 deg cm^2 dmol^(-1).
- Q: Given a protein’s initial concentration of 1 mg/mL, if its absorbance at 280 nm is 0.8, and the extinction coefficient is 20,000 M^(-1)cm^(-1), what is its final concentration after a 1:10 dilution?
A: Final concentration (mg/mL) = (Initial concentration * Initial absorbance) / Final absorbance = (1 * 0.8) / 0.1 = 8 mg/mL.
- Q: A protein has a specific activity of 2500 units/mg. If the enzyme reaction produced 0.2 moles of product, what mass of the protein was present?
A: Mass of protein (mg) = Moles of product / Specific activity = 0.2 moles / 2500 units/mg = 0.00008 mg.
- Q: A protein is 20% alpha-helical and 30% beta-sheet. What percentage of the protein is in random coil conformation?
A: Percentage of random coil = 100% – (Percentage of alpha-helix + Percentage of beta-sheet) = 100% – (20% + 30%) = 50%.
- Q: Calculate the turnover number (kcat) of an enzyme if it converted 2000 moles of substrate into product in 10 seconds with an enzyme concentration of 2 μM.
A: kcat = Moles of product formed / (Enzyme concentration * Time) = 2000 moles / (2 μM * 10 seconds) = 100,000 s^(-1).
- Q: What is the fraction of ionized carboxyl groups (COOH) in a protein at pH 7.5? Given pKa of COOH is 2.4.
A: Fraction ionized = 1 / (1 + 10^(pKa – pH)) = 1 / (1 + 10^(2.4 – 7.5)) ≈ 0.994.
- Q: A protein has a Kd of 2 x 10^(-8) M for its ligand. If the concentration of the protein is 100 nM, what is the fraction of bound protein?
A: Fraction of bound protein = [Ligand] / (Kd + [Ligand]) = 100 nM / (2 x 10^(-8) M + 100 nM) ≈ 0.998.
- Q: How many peptide bonds are present in a protein with 300 amino acids?
A: Number of peptide bonds = Number of amino acids – 1 = 300 – 1 = 299.
- Q: If a protein has a specific rotation of +40°, what is the observed rotation if the concentration is 2 mg/mL in a 1 dm tube?
A: Observed rotation = Specific rotation * Concentration * Path length = +40° * 2 mg/mL * 1 dm = +80°.
- Q: A protein has an absorption coefficient of 0.5 mL mg^(-1) cm^(-1) at 280 nm. What is the concentration of the protein if its absorbance is 1.2 in a 1 cm cuvette?
A: Concentration (mg/mL) = Absorbance / Absorption coefficient = 1.2 / 0.5 = 2.4 mg/mL.
- Q: Given a protein’s extinction coefficient at 280 nm of 5500 M^(-1)cm^(-1) and a path length of 0.5 cm, what is the absorbance if the protein concentration is 0.1 mg/mL?
A: Absorbance = (Extinction coefficient * Concentration * Path length) = 5500 M^(-1)cm^(-1) * 0.1 mg/mL * 0.5 cm = 275.
- Q: A protein is 80% pure, and its total mass is 50 mg. What is the mass of the pure protein in the sample?
A: Mass of pure protein = Total mass * Percentage purity = 50 mg * 0.80 = 40 mg.
- Q: Calculate the Vmax of an enzyme if it catalyzes the conversion of 2 μmol of substrate in 1 minute at an enzyme concentration of 0.1 μM.
A: Vmax = Moles of product formed / Time / Enzyme concentration = 2 μmol / 1 minute / 0.1 μM = 20 μmol/min/μM.
- Q: If a protein has a solubility of 25 mg/mL in water at 25°C, how much of the protein can be dissolved in 500 mL of water?
A: Mass of dissolved protein = Solubility * Volume of solvent = 25 mg/mL * 500 mL = 12,500 mg.
- Q: A protein has an initial velocity (V0) of 2.5 μmol/min at a substrate concentration of 0.5 mM. If the Km is 0.1 mM, calculate the reaction rate at 1 mM substrate concentration.
A: Reaction rate at 1 mM substrate concentration = Vmax * [S] / (Km + [S]) = 2.5 μmol/min * 1 mM / (0.1 mM + 1 mM) = 2.27 μmol/min.
- Q: What is the net charge of a protein with 12 positively charged amino acids (pKa ≈ 9.0) and 7 negatively charged amino acids (pKa ≈ 4.0) at pH 7.4?
A: Net charge = (Number of positive charges – Number of negative charges) = (12 – 7) = +5.
- Q: Given a protein’s UV absorbance at 280 nm of 1.8 and a molar extinction coefficient of 16,000 M^(-1)cm^(-1), what is the concentration of the protein in μM?
A: Concentration (μM) = Absorbance / (Extinction coefficient * Path length) = 1.8 / (16,000 M^(-1)cm^(-1) * 1 cm) ≈ 0.0001125 μM.
- Q: A protein is 60% folded at a denaturant concentration of 0.5 M. What is the fraction of folded protein if the denaturant concentration is reduced to 0.1 M?
A: Fraction of folded protein = 1 / (1 + (Denaturant concentration / IC50)^n), where IC50 is the concentration at which 50% unfolding occurs. Let’s assume n = 1 for simplicity.
Fraction of folded protein = 1 / (1 + (0.1 M / 0.5 M)^1) = 1 / (1 + 0.2) = 1 / 1.2 ≈ 0.833.
- Q: Calculate the specific activity of an enzyme if 0.05 moles of substrate were converted to product in 2 minutes using 10 μg of enzyme.
A: Specific activity = Moles of product formed / (Enzyme amount * Time) = 0.05 moles / (10 μg * 2 minutes) ≈ 2.5 μmol/min/μg.
- Q: A protein has an absorbance of 0.4 at 280 nm in a 1 cm cuvette. If the protein concentration is 2 mg/mL, what is the extinction coefficient at 280 nm?
A: Extinction coefficient (M^(-1)cm^(-1)) = Absorbance / (Concentration * Path length) = 0.4 / (2 mg/mL * 1 cm) = 0.2 M^(-1)cm^(-1).
- Q: If a protein has a half-life of 24 hours, what is its decay rate constant (k)?
A: k = 0.693 / Half-life = 0.693 / 24 hours ≈ 0.0289 hours^(-1).
- Q: A protein requires 1 mole of ATP to fold properly. If 90% of the protein is correctly folded, how many moles of ATP are required to fold 1 mg of the protein?
A: Moles of ATP required = (Total moles of protein * ATP per mole) * Fraction folded = (1 mg * 1 mole/1,000 mg) * 0.90 = 0.0009 moles.
- Q: Calculate the percentage of hydrophobic amino acids in a protein with 250 amino acids if it contains 50 leucine residues.
A: Percentage of hydrophobic amino acids = (Number of hydrophobic residues / Total number of residues) * 100% = (50 / 250) * 100% = 20%.
- Q: A protein is 40% active after a 10-minute incubation at 70°C. What is the protein’s half-life at this temperature?
A: Half-life (t1/2) = ln(2) / Decay rate constant (k) = ln(2) / (ln(Initial activity) – ln(Final activity)) * Incubation time = ln(2) / (ln(1) – ln(0.40)) * 10 minutes ≈ 17.3 minutes.
- Q: Given a protein’s turnover number (kcat) of 100 s^(-1) and a substrate concentration of 0.2 mM, calculate the initial velocity (V0) of the enzyme-catalyzed reaction.
A: V0 = kcat * [S] = 100 s^(-1) * 0.2 mM = 20 μM/s.
- Q: A protein with a molecular weight of 80,000 Da has a diffusion coefficient of 2.1 x 10^(-7) cm^2/s. Calculate its Stokes radius (Rs).
A: Rs = (k * T) / (6 * π * η * D), where k is the Boltzmann constant, T is the absolute temperature, η is the solvent viscosity, and D is the diffusion coefficient.
Assume T = 298 K, k = 1.38 x 10^(-23) J/K, and η = 0.01 poise = 0.001 g/(cm·s).
Rs = (1.38 x 10^(-23) J/K * 298 K) / (6 * π * 0.001 g/(cm·s) * 2.1 x 10^(-7) cm^2/s) ≈ 1.41 nm.
- Q: A protein has an enzymatic activity of 100 units/mL and a molecular weight of 40,000 Da. Calculate the specific activity in units/mg.
A: Specific activity (units/mg) = Enzymatic activity (units/mL) / Protein concentration (mg/mL) = 100 units/mL / (40,000 Da * 1 mg/mL) = 2.5 units/mg.
35.
Q: Calculate the Michaelis constant (Km) for an enzyme with an initial velocity (V0) of 1.5 μmol/min and a substrate concentration of 0.3 mM.
A: Km = (Vmax / 2) / (V0 / [S]) – [S] = (1.5 μmol/min / 2) / (1.5 μmol/min / 0.3 mM) – 0.3 mM = 0.1 mM.
- Q: If a protein has a molecular weight of 45 kDa, how many amino acids are present in the polypeptide chain?
A: Number of amino acids = Molecular weight (Da) / Average molecular weight of an amino acid (Da) ≈ 45,000 Da / 110 Da ≈ 409 amino acids.
- Q: A protein is 20% glycosylated by mass. If the glycosylated form has a molecular weight of 60 kDa, what is the molecular weight of the unglycosylated form?
A: Mass of glycosylated protein = Fraction glycosylated * Total mass = 0.20 * Total mass.
Molecular weight of glycosylated form = 60 kDa.
Therefore, Total mass = (Molecular weight of glycosylated form) / Fraction glycosylated = 60 kDa / 0.20 ≈ 300 kDa.
Molecular weight of unglycosylated form = Total mass – Mass of glycosylation = 300 kDa – (0.20 * 300 kDa) ≈ 240 kDa.
- Q: Given a protein’s turnover number (kcat) of 1500 s^(-1) and a substrate concentration of 0.05 mM, calculate the initial velocity (V0) of the enzyme-catalyzed reaction.
A: V0 = kcat * [S] = 1500 s^(-1) * 0.05 mM = 75 μM/s.
- Q: If a protein has a specific rotation of -30°, what is the observed rotation if the concentration is 1.5 mg/mL in a 2 dm tube?
A: Observed rotation = Specific rotation * Concentration * Path length = -30° * 1.5 mg/mL * 2 dm = -90°.
- Q: A protein has a fluorescence quantum yield of 0.6 and absorbs light at 340 nm. If the incident light intensity is 2 x 10^5 photons/s, how many emitted photons are produced?
A: Number of emitted photons = Fluorescence quantum yield * Number of absorbed photons = 0.6 * 2 x 10^5 photons/s = 1.2 x 10^5 photons/s.
- Q: Calculate the molar absorptivity (ε) of a protein if its absorbance is 0.5 at 280 nm, and its concentration is 0.1 mg/mL in a 1 cm cuvette.
A: ε = Absorbance / (Concentration * Path length) = 0.5 / (0.1 mg/mL * 1 cm) = 5.0 cm^2/(mg).
- Q: A protein solution has a concentration of 1 μM, and its absorbance at 280 nm is 0.3 in a 1 cm cuvette. Calculate the molar absorptivity (ε) at 280 nm.
A: ε = Absorbance / (Concentration * Path length) = 0.3 / (1 μM * 1 cm) = 0.3 L/(mol·cm) = 3 x 10^5 M^(-1)cm^(-1).
- Q: If a protein has a turnover number (kcat) of 500 s^(-1) and an enzyme concentration of 0.5 μM, calculate the initial velocity (V0) of the enzyme-catalyzed reaction.
A: V0 = kcat * [E] = 500 s^(-1) * 0.5 μM = 250 μM/s.
- Q: Given a protein’s extinction coefficient at 280 nm of 12,000 M^(-1)cm^(-1) and a path length of 1 cm, what is the concentration of the protein in mg/mL if its absorbance is 0.6?
A: Concentration (mg/mL) = Absorbance / (Extinction coefficient * Path length) = 0.6 / (12,000 M^(-1)cm^(-1) * 1 cm) ≈ 0.00005 mg/mL.
- Q: A protein with a molecular weight of 36 kDa has a diffusion coefficient of 1.8 x 10^(-7) cm^2/s. Calculate its Stokes radius (Rs).
A: Rs = (k * T) / (6 * π * η * D), where k is the Boltzmann constant, T is the absolute temperature, η is the solvent viscosity, and D is the diffusion coefficient.
Assume T = 298 K, k = 1.38 x 10^(-23) J/K, and η = 0.01 poise = 0.001 g/(cm·s).
Rs = (1.38 x 10^(-23) J/K * 298 K) / (6 * π * 0.001 g/(cm·s) * 1.8 x 10^(-7) cm^2/s) ≈ 1.78 nm.
- Q: A protein requires 2 moles of ATP to fold properly. If 80% of the protein is correctly folded, how many moles of ATP are required to fold 0.5 mg of the protein?
A: Moles of ATP required = (Total moles of protein * ATP per mole) * Fraction folded = (0.5 mg * 1 mole/40,000 mg) * 0.80 = 0.00002 moles.
- Q: A protein with a molecular weight of 60 kDa has a diffusion coefficient of 2.4 x 10^(-7) cm^2/s. Calculate its Stokes radius (Rs).
A: Rs = (k * T) / (6 * π * η * D), where k is the Boltzmann constant, T is the absolute temperature, η is the solvent viscosity, and D is the diffusion coefficient.
Assume T = 298 K, k = 1.38 x 10^(-23) J/K, and η = 0.01 poise = 0.001 g/(cm·s).
Rs = (1.38 x 10^(-23) J/K * 298 K) / (6 * π * 0.001 g/(cm·s) * 2.4 x 10^(-7) cm^2/s) ≈ 1.69 nm.
- Q: Calculate the Michaelis constant (Km) for an enzyme with an initial velocity (V0) of 0.8
μmol/min and a substrate concentration of 0.1 mM.
A: Km = (Vmax / 2) / (V0 / [S]) – [S] = (0.8 μmol/min / 2) / (0.8 μmol/min / 0.1 mM) – 0.1 mM = 0.05 mM.
- Q: Given a protein’s turnover number (kcat) of 300 s^(-1) and a substrate concentration of 0.02 mM, calculate the initial velocity (V0) of the enzyme-catalyzed reaction.
A: V0 = kcat * [S] = 300 s^(-1) * 0.02 mM = 6 μM/s.
- Q: If a protein has a specific rotation of +25°, what is the observed rotation if the concentration is 0.5 mg/mL in a 0.5 dm tube?
A: Observed rotation = Specific rotation * Concentration * Path length = +25° * 0.5 mg/mL * 0.5 dm = +6.25°.