# User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/10/17

(Difference between revisions)
 Revision as of 13:45, 27 October 2012 (view source) (→Preparation of Lysozyme stock)← Previous diff Revision as of 13:51, 27 October 2012 (view source) (→Preparation of Lysozyme stock)Next diff → Line 29: Line 29: * Lysozyme (14,307 Da) appeared as a white granular solid provided by Sigma. Knowing that 1 Da = 1 g/ mol, the following calculations were made to obtain a stock solution of 15.24 μM. * Lysozyme (14,307 Da) appeared as a white granular solid provided by Sigma. Knowing that 1 Da = 1 g/ mol, the following calculations were made to obtain a stock solution of 15.24 μM. - .012 L of water × $\frac{15E(-6)mol}{1 L}$ = $\frac{1.80E(-7)mol}{L}$ × $\frac{14,307 g}{mol}$ = 0.00258 g of lysozyme + .012 L of water × $\frac{15E(-6)mol}{1 L}$ = $\frac{1.80E(-7)mol}{L}$ × $\frac{14,307 g}{mol}$ = 0.00258 g of lysozyme to obtain 15 μM + + + * The amount to be measured is too small to be certain that the entire solid would be dissolved in water. As a result, a multiple of the calculated amount was chosen to be dissolved in water; the amount chosen was 0.0109 g. Once dissolved, this will be diluted to 15 μM. + + 0.0109 g of lysozyme × $\frac{1 mol}{14,307 g}$ = 7.62E(-7) mol ÷ .050 L of water = 1.524E(-5) M = 15.24 μM + +

## Revision as of 13:51, 27 October 2012

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## Electrophoresis of the mutated DNA plasmids

• Transferred 5 μL of the mutated plasmid into a new sterilized tube; leaving out one tube containing 45 μL.
• Simultaneously, added 1 μL of the gel loading dye 6x blue into the 5 μL plasmid and added 1 μL of DpnI into the 45 μL plasmid.
• The sample containing the DpnI was placed in a VWR analog heatblock at 37°C.
• The agarose gel was prepared by dissolving 1.23 g of agarose in 35 mL of TAE buffer. The agarose was heated in a microwave for 35s.
• When the solution is clear, the agarose was poured into the gel electrophoresis tray with the comb in place.
• A wait period of 30 min. was allotted for the gel to solidify. Upon observing the gel is already in its solidified state, the comb was gently removed to avoid breakage.
• The chamber containing the gel tray was filled with TAE buffer to the point where the gel is completely submerged of the buffer.
• The plasmid with the loading dye were inserted to the wells 3 and 4 using an automatic delivery pipet.
• The Horton 5B Electrophoresis system (chamber containing the tray) was connected to the BioRad power supply powerpac 200 at 85V.
• Completion of the electrophoresis was followed by the addition of ethidium bromide (EtBr). Great caution was done in handling EtBr since this agent is carcinogenic by binding to any DNA in contact.
• 20 μL of the 12.3 g/mL of EtBR was added to the gel suspended in TAE placed on a microwaveable plastic container. This was stirred in a VWR analog shaker at a slow speed.
• Assuming that the gel has absorbed the EtBr, the gel was placed on top of a UV lamp. Only the DNA ladded inserted on the left most part of the gel exhibited orange fluorescence.
• The orange fluorescence is attributed to the EtBr binding to the DNA.

## UV-Vis for Chemiluminescence

• An attempt was made to use the UV-Vis equipment for chemiluminescence by blocking the source lamp (D-2). The measuring mode was set on energy setting. The widest slid width available was 5.0 nm.
• Data collection was not possible. No signal was collected.

## Preparation of Lysozyme stock

• Michael Nagle prepared a new set of Au/BSA solutions. However, calculations made are not available in his notebook.
• Lysozyme (14,307 Da) appeared as a white granular solid provided by Sigma. Knowing that 1 Da = 1 g/ mol, the following calculations were made to obtain a stock solution of 15.24 μM.

.012 L of water × $\frac{15E(-6)mol}{1 L}$ = $\frac{1.80E(-7)mol}{L}$ × $\frac{14,307 g}{mol}$ = 0.00258 g of lysozyme to obtain 15 μM

• The amount to be measured is too small to be certain that the entire solid would be dissolved in water. As a result, a multiple of the calculated amount was chosen to be dissolved in water; the amount chosen was 0.0109 g. Once dissolved, this will be diluted to 15 μM.

0.0109 g of lysozyme × $\frac{1 mol}{14,307 g}$ = 7.62E(-7) mol ÷ .050 L of water = 1.524E(-5) M = 15.24 μM