Numerical on Carbohydrates

1. Find the mass of 1 mole of glucose.

Shortcut method:

The molecular formula of glucose is C6H12O6. The molar mass of carbon (C) is 12 g/mol, hydrogen (H) is 1 g/mol, and oxygen (O) is 16 g/mol.

Mass of 1 mole of glucose = (6 × 12 g/mol) + (12 × 1 g/mol) + (6 × 16 g/mol) Mass of 1 mole of glucose = 72 g/mol + 12 g/mol + 96 g/mol Mass of 1 mole of glucose = 180 g/mol

Therefore, the mass of 1 mole of glucose is 180 g.

2. A sample of glucose contains 12 g of carbon, 2 g of hydrogen, and 16 g of oxygen. Calculate the empirical formula of glucose.

Shortcut method:

The empirical formula represents the simplest whole-number ratio of the elements present in a compound.

• Calculate the number of moles of each element: Number of moles of carbon = 12 g / 12 g/mol = 1 mol Number of moles of hydrogen = 2 g / 1 g/mol = 2 mol Number of moles of oxygen = 16 g / 16 g/mol = 1 mol

• Divide the number of moles of each element by the smallest number of moles: Number of moles of carbon = 1 mol / 1 mol = 1 Number of moles of hydrogen = 2 mol / 1 mol = 2 Number of moles of oxygen = 1 mol / 1 mol = 1

• Write the empirical formula using the simplified mole ratio: Empirical formula of glucose = CH2O

Therefore, the empirical formula of glucose is CH2O.

3. Calculate the percentage composition of carbon, hydrogen, and oxygen in sucrose (C12H22O11).

Shortcut method:

• Calculate the molar mass of sucrose: Molar mass of sucrose = (12 × 12 g/mol) + (22 × 1 g/mol) + (11 × 16 g/mol) Molar mass of sucrose = 144 g/mol + 22 g/mol + 176 g/mol Molar mass of sucrose = 342 g/mol

• Calculate the percentage composition of each element: Percentage of carbon = (12 × 12 g/mol / 342 g/mol) × 100% = 42.43% Percentage of hydrogen = (22 × 1 g/mol / 342 g/mol) × 100% = 6.43% Percentage of oxygen = (11 × 16 g/mol / 342 g/mol) × 100% = 51.14%

Therefore, the percentage composition of sucrose is 42.43% carbon, 6.43% hydrogen, and 51.14% oxygen.

4. What is the molecular weight of starch if it is composed of 2000 glucose units?

Shortcut method:

The molecular weight of starch can be calculated by multiplying the molecular weight of a single glucose unit by the number of glucose units present.

• Molecular weight of glucose = 180 g/mol
• Molecular weight of starch (composed of 2000 glucose units) = 180 g/mol × 2000 Molecular weight of starch = 360,000 g/mol

Therefore, the molecular weight of starch composed of 2000 glucose units is 360,000 g/mol.

Numerical on Proteins

1. Calculate the number of amino acids in a protein with a molecular weight of 10,000 Da.

Shortcut method:

• The molecular weight of an amino acid is approximately 110 Da.
• Number of amino acids in the protein = Molecular weight of protein / Molecular weight of amino acid

Number of amino acids = 10,000 Da / 110 Da Number of amino acids ≈ 91

Therefore, the protein contains approximately 91 amino acids.

2. Determine the isoelectric point (pI) of a protein with the following amino acid composition:

Alanine (Ala): 20 residues (pI = 6.0) Aspartic acid (Asp): 10 residues (pI = 3.0) Glutamic acid (Glu): 15 residues (pI = 4.25) Lysine (Lys): 12 residues (pI = 9.74) Arginine (Arg): 8 residues (pI = 10.76)

Shortcut method:

The isoelectric point (pI) of a protein is the pH at which the net charge of the protein is zero. It can be estimated by calculating the average of the pI values of the individual amino acids, weighted by their respective counts (assuming equal contributions from acidic and basic residues).

• Calculate the total charge of acidic amino acids (Asp + Glu) at pH 7: Aspartic acid (Asp): 10 residues × (-1 charge) = -10 charges Glutamic acid (Glu): 15 residues × (-1 charge) = -15 charges Total negative charges = -25 charges

• Calculate the total charge of basic amino acids (Lys + Arg) at pH 7: Lysine (Lys): 12 residues × (+1 charge) = +12 charges Arginine (Arg): 8 residues × (+1 charge) = +8 charges Total positive charges = +20 charges

• Calculate the net charge of the protein at pH 7: Net charge = Total positive charges + Total negative charges Net charge = +20 charges - 25 charges = -5 charges

Since the net charge is negative at pH 7, the pI of the protein will be lower than 7.

• Calculate the pI by adjusting for the net charge: pI ≈ pH 7 - (Net charge / Total charge of acidic residues) pI ≈ 7 - (-5 charges / 25 charges) pI ≈ 7 + 0.2 ≈ 7.2

Therefore, the isoelectric point (pI) of the protein is approximately 7.2.

Numerical on Lipids

1. Calculate the number of carbon atoms in a saturated fatty acid with 16 carbon atoms.

Shortcut method:

A saturated fatty acid with 16 carbon atoms is typically referred to as palmitic acid.

• The general formula for a saturated fatty acid with ’n’ carbon atoms is CnH2nO2.
• For a saturated fatty acid with 16 carbon atoms (palmitic acid), n = 16.

Substituting the value of n into the formula: Number of carbon atoms in palmitic acid = 16

Therefore, a saturated fatty acid with 16 carbon atoms (palmitic acid) contains 16 carbon atoms.

2. Determine the melting point of a triglyceride composed of one saturated fatty acid (palmitic acid) and two unsaturated fatty acids (oleic acid and linoleic acid).

Shortcut method:

The melting point of a triglyceride depends on the degree of saturation of its fatty acids. Saturated fatty acids have higher melting points than unsaturated fatty acids.

• Palmitic acid (saturated) has a high melting point (around 63°C).
• Oleic acid (monounsaturated) has a lower melting point (around 13°C).
• Linoleic acid (polyunsaturated) has a very low melting point (around -5°C).

Since the triglyceride contains a mixture of saturated and unsaturated fatty acids, its melting point will be intermediate between the melting points of its individual fatty acids.

Melting point of the triglyceride ≈ (Melting point of palmitic acid + Melting points of oleic acid + linoleic acid) / 3 Melting point of the triglyceride ≈ (63°C + 13°C + (-5°C)) / 3 Melting point of the triglyceride ≈ 24°C

Therefore, the melting point of the triglyceride is approximately 24°C.

3. Calculate the iodine number of an oil that contains 50% saturated fatty acids and 50% unsaturated fatty acids.

Shortcut method:

The iodine number is a measure of the degree of unsaturation in an oil or fat. It represents the amount of iodine (in grams) that is absorbed by 100 grams of the sample.

• The higher the iodine number, the higher the degree of unsaturation.

Since the oil contains 50% saturated and 50% unsaturated fatty acids, we can assume average unsaturation.

• A typical iodine number for an oil with 50% unsaturated fatty acids is around 90.

Therefore, the iodine number of the oil is approximately 90.

Numerical on Nucleic Acids

1. Find the number of nucleotides in a DNA molecule with a molecular weight of 6.6 × 106