Biol2 Lecture 1 V1
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This document summarizes key points from a lecture on molecular biology and DNA quantification: 1. It introduces molecular biology and why it is being studied, then outlines what will be covered in Biol2 including nucleic acid quantification. 2. It explains how to quantify DNA and RNA concentrations using absorbance measurements at 260nm and calculation formulas involving dilution factors and extinction coefficients. 3. It discusses potential contaminants like protein and phenol that can interfere with absorbance readings and describes using A260/A280 ratios to assess DNA purity. Examples are provided to demonstrate these quantification calculations.
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3. Preserve the integrity of the DNA (prevent degradation and shearing)
Step 1 can be accomplished in a number of ways, such as mechanical disruption (grinding, mincing), protein denaturation (detergents), and protein degradation (via proteases). These can be used singly or in combination depending on the type of biological sample you are starting with. Grinding the samples provides more surface area for the denaturants/proteases to interact with the cellular proteins, thus speeding up the denaturation process. We used liquid nitrogen (N2) and protein degradation (Proteinase K) in lab 2. Various salts are included in a cell lysis solution to stabilize the DNA by providing positive ions which insert between the negatively charged phosphates in the DNA backbone (creating a “salt bridge”). Buffers (such as Tris) also help to preserve DNA integrity by maintaining a neutral pH.
Once the cells have been lysed, contaminating proteins, lipids, etc. must be separated from the DNA. A widely used and efficient way to remove proteins from nucleic acids solutions is to extract with a 1:1 mixture of phenol and chloroform (CHCl3). Phenol and CHCl3 are both hydrophobic organic solvents that unfold proteins. When mixed with an aqueous DNA/protein solution and then centrifuged, the denatured proteins are selectively partitioned into the denser organic phase, while the DNA (plus RNA and salt) remains in the aqueous phase. This procedure takes advantage of the fact that deproteinization is more efficient when two different organic solvents are used instead of one. Additionally, chloroform removes any lingering traces of phenol from the nucleic acid preparation (which would interfere with later applications). Since the aqueous phase contains RNA and salt in addition to the DNA, phenol:CHCl3 extraction is followed by ethanol (EtOH) precipitation. DNA (a polar molecule) is soluble in water (also polar) because the water molecules intercalate into the phosphate backbone of the DNA and thus maintain it in a soluble state, but DNA is insoluble in 95% EtOH (nonpolar). Water molecules have a higher affinity for the EtOH than the DNA, so when you add EtOH and salt [10 M ammonium acetate (NH4Ac); pH 5.2], Na+ ions replace water in the DNA backbone, essentially removing the water molecules, and the DNA is forced out of solution (precipitates). After precipitating with 95% EtOH, the DNA is “washed” in 70% EtOH to remove the salt. Since 70% EtOH contains 30% water, the salt, having a greater affinity for the water than the DNA, remains in the EtOH, and the DNA is forced out.
The final step in the purification process is to preserve the DNA in a sta ...
DNA Preparation for sequencing
Preparing DNA for sequencing requires careful purification of the template DNA to ensure high quality results. Key steps include extracting DNA, quantifying the DNA concentration and quality, purifying DNA or PCR products to remove contaminants like primers, salts and proteins, and checking for purity and fragment size on a gel. Using purified template DNA free of impurities is crucial for accurate sequencing reads.
Preparing Genomic DNA for Sequencing
DNA sequencing is the process of determining the nucleic acid sequence – the order of nucleotides in DNA.
DNA extraction.pptx
DNA extraction is a method used to purify DNA from a sample by separating it from other cellular components using physical and chemical techniques. It involves lysing cells and solubilizing the DNA using detergents and enzymes. Common extraction methods include organic extraction, non-organic extraction, and adsorption. DNA can then be extracted from various samples for purposes like genetic testing, forensic analysis, and body identification.
12021947.ppt
This document provides instructions for extracting genomic DNA from human blood. The objectives are to isolate DNA and analyze it using agarose gel electrophoresis. The extraction process uses EDTA to inhibit DNase, Tris as a buffer, SDS as a detergent to disrupt cell membranes and release DNA, and NaCl and alcohol for precipitation. Blood is collected and red blood cells are lysed using lysis buffer. White blood cells are then lysed and DNA is precipitated using NaCl and ethanol before being collected and analyzed.
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みなさんこんにちはこれ何文字まで入るの？40文字以下不可とか本当に意味わからないけどこれ限界文字数書いてないからマジでやばい文字数いけるんじゃないの？えこ...
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みなさんこんにちはこれ何文字まで入るの？40文字以下不可とか本当に意味わからないけどこれ限界文字数書いてないからマジでやばい文字数いけるんじゃないの？えこ...
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Biol2 Lecture 1 V1
- 1. Biol 2 – Lecture 1 1. Intro to Molecular Biology 2. DNA Quantification 3. Standard Curves Image credit: http://www.flickr.com/photos/86712369@N00/438500895
- 2. What is Molecular Biology?
- 3. Why are we studying Molecular Biology? Image credit: http://farm4.static.flickr.com/3209/2831443228_131e565e91.jpg
- 4. What is Biol2? Image Credit: http://www.flickr.com/photos/37101220@N00/2604395220
- 5. What Biol2 isn’t? Image Credits: http://farm3.static.flickr.com/2160/2446536264_822f20b28c.jpg http://www.flickr.com/photos/29302171@N00/15411429
- 11. Nucleic Acid Quantification NAME DATE
- 13. Nucleic Acid Quantification At 260 nm, 50 µ g/ml dsDNA = 1 absorbance unit Therefore, [dsDNA] = A 260 x dilution-factor x 50 µ g/ml 1 abs unit Notation Note: OD 260 = A 260
- 14. What about ssDNA or RNA? At 260 nm, 40 µ g/ml RNA = 1 absorbance unit Therefore, [RNA] = A 260 x dilution-factor x 40 µ g/ml 1 abs unit 33 µg/ml ssDNA = 1 absorbance unit
- 19. [dsDNA] = A 260 x dilution-factor x 50 µ g/ml 1 abs unit [dsDNA] = 0.138 abs units x 1000 µ l x 50 µ g/ml 1 abs unit 2 µl [dsDNA] = 0.138 x 500 x 50 µg/ml [dsDNA] = 3450 µg/ml (or 3.45 µg/µl)
- 21. A 260 A 280 = 0.138 = 1.51 0.091 Not quite pure enough (ratio < 1.6), probably contaminated with protein.
- 22. Eric’s quote of the week! Image credit: http://www.flickr.com/photos/32434850@N00/139184796 “ Always read the label on the stock bottle to make sure you have the right reagent ”