Summary of Biochemistry and Analytical Chemistry Calculations

Biochemistry & Analytical Chemistry Calculations: Oxidative Stress

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Introduction

Clinical biochemistry calculations are essential tools for laboratory medicine and physiological understanding. They allow clinicians and laboratorians to convert between concentrations, volumes, activities, and to assess properties such as osmolarity, ionic strength, and buffer pH. This guide explains common calculation types, shows step-by-step examples, and highlights how results apply in clinical contexts.

Definition: Clinical biochemistry calculations are quantitative methods used to determine concentrations, activities, osmotic/ionic properties, and acid–base characteristics of biological and clinical solutions.

Contents

  1. Concentration and dilution calculations
  2. Osmolarity and ionic strength
  3. pH and buffer calculations
  4. Enzyme activity and inhibition calculations
  5. Worked examples from clinical problems

1. Concentration and dilution calculations

Key concepts

  • Amount of substance concentration (molarity): number of moles of solute per liter of solution, unit mol·L^{-1} (often written mmol·L^{-1}). Use $n = cV$ where $n$ is amount in moles, $c$ is concentration, $V$ is volume in liters.
  • Dilution and mixing: when mixing solutions, total amount of solute is conserved. For two solutions: $$n_{total} = c_1V_1 + c_2V_2$$ Final concentration: $$c_{final} = \frac{c_1V_1 + c_2V_2}{V_1 + V_2}$$

Definition: Molarity $c$ is $n/V$, where $n$ is moles and $V$ is volume in liters.

Example: Mixing glucose solutions

Problem: Mix $15\ \text{mL}$ of glucose solution at $5\ \text{mmol·L^{-1}}$ with $25\ \text{mL}$ at $4\ \text{mmol·L^{-1}}$. Calculate final concentration.

Steps:

  • Convert volumes to liters: $V_1 = 0.015\ \text{L}$, $V_2 = 0.025\ \text{L}$.
  • Amounts: $n_1 = c_1V_1 = 5\times10^{-3}\times0.015$, $n_2 = 4\times10^{-3}\times0.025$.
  • Final concentration: $$c_{final} = \frac{5\times10^{-3}\times0.015 + 4\times10^{-3}\times0.025}{0.015 + 0.025} = 4.375\ \text{mmol·L^{-1}}$$

Example: Adding water to adjust concentration

Problem: What volume of physiological solution (starting volume unknown) is needed to have final volume $200\ \text{mL}$ and final NaCl concentration $90\ \text{mmol·L^{-1}}$, given an initial NaCl solution? (Interpretation: if you have a stock physiological solution at some concentration, often the question asks for how much of that stock plus water gives final; in the provided problems the answer corresponds to mixing 120\ \text{mL} of stock with water to 200\ \text{mL}.)

General approach: use $c_1V_1 = c_{final}V_{final}$ when diluting a single stock.

2. Osmolarity and ionic strength

Key concepts

  • Osmolarity (osmotic concentration) counts total solute particles per liter. For an electrolyte that dissociates into $i$ particles, osmolarity contribution is $i,c$ where $c$ is molar concentration.
  • Ionic strength $I$ measures electrostatic interactions among ions: $$I = \tfrac{1}{2}\sum_i c_i z_i^2$$ where $c_i$ is molar concentration of ion $i$ and $z_i$ is its charge.

Definition: Ionic strength quantifies the combined effect of ion concentrations and their charges on solution behavior.

Practical rules

  • For strong electrolytes assume full dissociation. Example: \ce{NaCl -> Na+ + Cl-} gives $i=2$.
  • For salts with multiple charges, include stoichiometry: \ce{Fe2(SO4)3} produces 2 Fe^{3+} and 3 SO4^{2-} per formula unit, total particles $i=5$.

Examples

  1. Isoosmotic K3PO4 with blood serum
  • To be isoosmotic with plasma (assume plasma osmolarity ~0.3\ \text{mol·L^{-1}}), account dissociation: \ce{K3PO4 -> 3K+ + PO4^{3-}} gives $i=4$. For target osmolarity $0.3$ mol·L^{-1}, concentration $c = 0.3/4 = 0.075\ \text{mol·L^{-1}}$.
  1. Ionic strength of \ce{FeSO4} at $0.02\ \text{mol·L^{-1}}$
  • Dissociation: \ce{FeSO4 -> Fe^{2+} + SO4^{2-}}. Ionic strength: $$I = \tfrac{1}{2}\left(0.02\times 2^2 + 0.02\times (-2)^2\right)=\tfrac{1}{2}\left(0.08 + 0.08\right)=0.08\ \text{mol·L^{-1}}$$
  1. Osmolarity and io
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Clinical Biochemistry Calculations

Klíčové pojmy: Molarity $c$ equals $n/V$, use liters for volume, When mixing solutions, conserve total moles: $c_{final}=(c_1V_1+c_2V_2)/(V_1+V_2)$, Osmolarity equals sum of particle contributions $\sum i\,c$ where $i$ is dissociation particles, Ionic strength $I=\tfrac{1}{2}\sum c_i z_i^2$ using ion charges $z_i$, Use Henderson–Hasselbalch $\mathrm{pH}=\mathrm{p}K_a+\log_{10}([A^-]/[HA])$ for buffer design, Convert enzyme units: $1\ \text{U}=1\ \mu\text{mol·min^{-1}}=1.667\times10^{-8}\ \text{katal}$, Percent inhibition $\%I=(A_0-A_s)/A_0\times100$, For weak acids, approximate $[H^+]\approx\sqrt{K_a c}$ when $x\ll c$, Compute ionic contributions for multi-ion salts by stoichiometry (e.g., \ce{Fe2(SO4)3} gives 5 particles), Always convert volumes to liters and masses to moles before final concentration calculations

## Introduction Clinical biochemistry calculations are essential tools for laboratory medicine and physiological understanding. They allow clinicians and laboratorians to convert between concentrations, volumes, activities, and to assess properties such as osmolarity, ionic strength, and buffer pH. This guide explains common calculation types, shows step-by-step examples, and highlights how results apply in clinical contexts. > Definition: Clinical biochemistry calculations are quantitative methods used to determine concentrations, activities, osmotic/ionic properties, and acid–base characteristics of biological and clinical solutions. ## Contents 1. Concentration and dilution calculations 2. Osmolarity and ionic strength 3. pH and buffer calculations 4. Enzyme activity and inhibition calculations 5. Worked examples from clinical problems --- ## 1. Concentration and dilution calculations ### Key concepts - **Amount of substance concentration (molarity)**: number of moles of solute per liter of solution, unit mol·L^{-1} (often written mmol·L^{-1}). Use $n = cV$ where $n$ is amount in moles, $c$ is concentration, $V$ is volume in liters. - **Dilution and mixing**: when mixing solutions, total amount of solute is conserved. For two solutions: $$n_{total} = c_1V_1 + c_2V_2$$ Final concentration: $$c_{final} = \frac{c_1V_1 + c_2V_2}{V_1 + V_2}$$ > Definition: Molarity $c$ is $n/V$, where $n$ is moles and $V$ is volume in liters. ### Example: Mixing glucose solutions Problem: Mix $15\ \text{mL}$ of glucose solution at $5\ \text{mmol·L^{-1}}$ with $25\ \text{mL}$ at $4\ \text{mmol·L^{-1}}$. Calculate final concentration. Steps: - Convert volumes to liters: $V_1 = 0.015\ \text{L}$, $V_2 = 0.025\ \text{L}$. - Amounts: $n_1 = c_1V_1 = 5\times10^{-3}\times0.015$, $n_2 = 4\times10^{-3}\times0.025$. - Final concentration: $$c_{final} = \frac{5\times10^{-3}\times0.015 + 4\times10^{-3}\times0.025}{0.015 + 0.025} = 4.375\ \text{mmol·L^{-1}}$$ ### Example: Adding water to adjust concentration Problem: What volume of physiological solution (starting volume unknown) is needed to have final volume $200\ \text{mL}$ and final NaCl concentration $90\ \text{mmol·L^{-1}}$, given an initial NaCl solution? (Interpretation: if you have a stock physiological solution at some concentration, often the question asks for how much of that stock plus water gives final; in the provided problems the answer corresponds to mixing 120\ \text{mL} of stock with water to 200\ \text{mL}.) General approach: use $c_1V_1 = c_{final}V_{final}$ when diluting a single stock. --- ## 2. Osmolarity and ionic strength ### Key concepts - **Osmolarity (osmotic concentration)** counts total solute particles per liter. For an electrolyte that dissociates into $i$ particles, osmolarity contribution is $i\,c$ where $c$ is molar concentration. - **Ionic strength $I$** measures electrostatic interactions among ions: $$I = \tfrac{1}{2}\sum_i c_i z_i^2$$ where $c_i$ is molar concentration of ion $i$ and $z_i$ is its charge. > Definition: Ionic strength quantifies the combined effect of ion concentrations and their charges on solution behavior. ### Practical rules - For strong electrolytes assume full dissociation. Example: \ce{NaCl -> Na+ + Cl-} gives $i=2$. - For salts with multiple charges, include stoichiometry: \ce{Fe2(SO4)3} produces 2 Fe^{3+} and 3 SO4^{2-} per formula unit, total particles $i=5$. ### Examples 1) Isoosmotic K3PO4 with blood serum - To be isoosmotic with plasma (assume plasma osmolarity ~0.3\ \text{mol·L^{-1}}), account dissociation: \ce{K3PO4 -> 3K+ + PO4^{3-}} gives $i=4$. For target osmolarity $0.3$ mol·L^{-1}, concentration $c = 0.3/4 = 0.075\ \text{mol·L^{-1}}$. 2) Ionic strength of \ce{FeSO4} at $0.02\ \text{mol·L^{-1}}$ - Dissociation: \ce{FeSO4 -> Fe^{2+} + SO4^{2-}}. Ionic strength: $$I = \tfrac{1}{2}\left(0.02\times 2^2 + 0.02\times (-2)^2\right)=\tfrac{1}{2}\left(0.08 + 0.08\right)=0.08\ \text{mol·L^{-1}}$$ 3) Osmolarity and io