{"id":4070,"date":"2026-10-11T06:33:32","date_gmt":"2026-10-11T06:33:32","guid":{"rendered":"https:\/\/nhdbio.com\/?page_id=4070"},"modified":"2026-10-15T03:21:31","modified_gmt":"2026-10-15T03:21:31","slug":"%e8%82%bd%e5%82%a8%e5%a4%87%e6%b6%b2%e9%85%8d%e5%88%b6%e7%a4%ba%e4%be%8b","status":"publish","type":"page","link":"https:\/\/nhdbio.com\/cn\/analytical-methods\/peptide-stock-preparation-worked-example\/","title":{"rendered":"\u914d\u5236\u80bd\u50a8\u5907\u6db2\uff1a\u8ba1\u7b97\u8d28\u91cf\u548c\u6700\u7ec8\u4f53\u79ef"},"content":{"rendered":"

You want 10 mL of a 100 \u00b5M peptide stock. How much material do you need? The calculation uses three numbers: concentration, final volume and molecular weight. Let us work through them before discussing the powder and solvent.<\/p>\n

Write down the three numbers first<\/h2>\n

A stock solution is a relatively concentrated solution prepared for later dilution. In this example, the target is 100 \u00b5M, the final volume is 10 mL, and the molecular weight is 2,000 g\/mol.<\/p>\n

Convert the concentration to mol\/L and the volume to litres: 100 \u00b5M = 0.0001 mol\/L; 10 mL = 0.010 L.<\/p>\n

\"Figure
Figure 1. Start with concentration, final volume and molecular weight.<\/figcaption><\/figure>\n

Calculate molecular amount, then mass<\/h2>\n

Multiply concentration by volume: 0.0001 \u00d7 0.010 = 0.000001 mol. Multiply that amount by 2,000 g\/mol: the result is 0.002 g, or 2 mg.<\/p>\n

The example therefore needs 2 mg of the specified peptide form, before any correction for other material in the powder.<\/p>\n

0.0001 mol\/L \u00d7 0.010 L \u00d7 2,000 g\/mol = 0.002 g<\/p>\n

\"Figure
Figure 2. The example calculation gives 2 mg of target peptide.<\/figcaption><\/figure>\n

What if only 80% of the powder is the target peptide?<\/h2>\n

Suppose a suitable quantitative test shows that the target peptide is 80% of the powder\u2019s weight. To obtain 2 mg of target peptide, the powder amount is 2 \u00f7 0.80 = 2.5 mg.<\/p>\n

This requires a measured target-peptide mass fraction. An HPLC area-purity result of 98% cannot be substituted for it.<\/p>\n

\"Figure
Figure 3. A measured 80% target-peptide mass fraction would require 2.5 mg of powder.<\/figcaption><\/figure>\n

\u201cMake up to 10 mL\u201d means final solution volume<\/h2>\n

Adding 10 mL of solvent and making a solution up to 10 mL describe different operations. The calculation above uses the final solution volume. Follow an appropriate preparation method that achieves that volume.<\/p>\n

Before preparation, confirm a suitable solvent and handling conditions for the exact material. This arithmetic does not establish solubility.<\/p>\n

Check the calculation with a smaller batch<\/h2>\n
With the same molecular weight, how much target peptide is needed for 5 mL at 100 \u00b5M?<\/strong><\/p>\n
\nSee the answer<\/summary>\n

Half the volume needs half the amount: 1 mg of target peptide. If the measured target mass fraction is 80%, the powder amount would be 1.25 mg.<\/p>\n<\/details>\n<\/div>\n

Record the mass weighed, molecular weight, final volume, solvent and any content correction. Another person should be able to follow the calculation without guessing.<\/p>\n

Before preparing a MOTS-c or SS-31 stock<\/h2>\n

For MOTS-c or SS-31 stock planning, first confirm the exact material form and suitable solvent. Ask what quantity information is available for that material; do not assume the vial label is a measured target-peptide content result.<\/p>\n

MOTS-c<\/a> \/ SS-31<\/a><\/p>\n

Discuss material and document requirements with NHD<\/a> \u00b7 Continue with the analytical guide<\/a><\/p>\n