Question

Product Concentrations Using the Handbook of Biochemistry and Molecular Biology (G. Fasman, ed., CRC Press, Cleveland, 1976) or a similar source or a suitable biochemistry textbook (in other words, looking up information), find the necessary information to determine the amount of material required to make \(1 \mathrm{ml}\) of each of the following aqueous solutions: (a) \(0.01 \mathrm{M}\) cytochrome \(\mathrm{c}\) (b) \(1 \times 10^{-7} \mathrm{M} \beta\)-galactosidase (c) \(0.01 \mathrm{M}\) porcine insulin (d) \(0.01 \mathrm{M}\) human hemoglobin (e) \(0.1 \mathrm{M}\) streptomycin (f) \(1 \times 10^{-6} \mathrm{M}\) oligonucleotide with 10 nucleotides Also calculate the concentrations in terms of the following additional standard means of expressing bioproduct concentrations: percent (weight per volume) and milligrams per milliliter. Assuming that the solutions are in pure water, also express the concentrations as mole fractions. Discuss the feasibility of making each one of these solutions.

Short answer

The steps provide a comprehensive method for calculating the amount of material required and the concentrations in different units. Feasibility depends on factors such as the solubility of the substance, availability of the bio-product, and cost.

Step-by-step solution

Locate Needed Information

The first step is to find the molecular weights (MW) of the bio-products. These can be found in a suitable biochemistry textbook, biochemical handbook or a reliable online source.

Calculate Amount of Material Required

The amount (in grams) of each substance required to make 1 ml of a solution with the specified molarity can be calculated using the formula: \( \text{{mass}} (g) = \text{{Molarity}} (M) \times \text{{Volume}} (L) \times \text{{Molecular weight}} (g/mol) \) Remember to convert volume from ml to L by dividing by 1000. Use the molecular weights obtained in Step 1 for the calculations.

Convert Concentrations into Different Units

For percent weight/volume concentration, use the formula: \( \text{{Percent weight/volume}} = \frac{{\text{{mass}} (g)}}{{\text{{volume}} (ml)}} \times 100 \) For milligrams per millilitre, convert grams into milligrams by multiplying the mass by 1000. For mole fractions, assuming the solutions are in pure water, calculate the moles of water in 1 ml, use the density of water (1g/ml), and the molar mass of water (18.01528 g/mol), then calculate the mole fraction using the formula: \( \text{{Mole fraction}} = \frac{{\text{{moles of the solute}}}}{{\text{{moles of the solute}} + \text{{moles of water}}}} \)

Discuss the Feasibility

Discuss the feasibility of making each one of these solutions based on the calculated material required. Consider solubility limits, purification processes, and cost implications.

Key concepts

Molarity Calculations

Understanding molarity calculations is pivotal for biochemists, especially when preparing precise solutions for experiments. Molarity, denoted as 'M', represents the number of moles of a solute dissolved per liter of solution. It's a measure of concentration that signifies how strong or weak a solution is.

When a student is tasked with creating a solution of a certain molarity, calculating the amount of substance required is straightforward using the formula:
\[ \text{mass} (g) = \text{Molarity} (M) \times \text{Volume} (L) \times \text{Molecular weight} (g/mol) \]
For example, to prepare 1 ml (or 0.001 liters) of a 0.01 M solution, and if the molecular weight of the substance is say, 123.456 g/mol, the calculation would be:
\( 0.01 \times 0.001 \times 123.456 \) grams.
Remember to always ensure your units are consistent; here, we converted ml to liters by dividing by 1000. Knowing this will help not only in your chemistry classes but in any field where solution preparation is necessary.

Bioseparations

Bioseparations refer to the process of purifying biological products such as proteins, nucleic acids, and other biomolecules from a complex mixture. Bioproducts like enzymes, antibiotics, and hormones need to be isolated for various applications in research and medicine. The feasibility of preparing certain concentrations of these bioproducts is dependent on their solubility and the effectiveness of the bioseparation techniques.

Techniques like centrifugation, filtration, chromatography, and precipitation are often used to separate the desired product from impurities. For instance, some proteins may precipitate (solidify) out of solution when reaching a certain concentration, thus knowing the proper preparation method is crucial.

When solving problems involving bioproduct solution preparation, always consider if the product can be effectively separated and if it remains stable in the desired concentration. This practical knowledge accompanies the theoretical understanding of molarity and concentration conversions in real-world applications.

Concentration Units Conversion

Converting between different concentration units is a common task in the laboratory. Typically, scientists may work with percent solutions (weight/volume), milligrams per milliliter (mg/ml), and mole fractions, among others. Knowing how to interconvert these units is essential for accurately reporting and understanding experimental data.

To convert the mass of a substance in grams to a percent solution, you would use the formula:
\[ \text{Percent weight/volume} = \frac{\text{mass} (g)}{\text{volume} (ml)} \times 100 \]
This is very useful for applications where dosage or concentration needs to be communicated in a clear and standardized manner.

Similarly, to convert grams to milligrams, simply multiply by 1000, since there are 1000 milligrams in a gram. Mole fraction, a less common but equally important concentration unit, particularly in physical chemistry, refers to the ratio of the number of moles of solute to the total number of moles in the solution.

Understanding these conversions allows for flexibility when measuring and preparing solutions across different scientific disciplines, ensuring precise communication and experimentation.