Chapter 4 Nucleic Acid Extraction Methods

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kkelley
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Chapter 4 Nucleic Acid Extraction Methods
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2014-04-05 18:24:11
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Chapter Nucleic Acid Extraction Methods
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Buckingham and Flaws Chapter 4 Nucleic Acid Extraction Methods Q&A
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  1. DNA Quantity/Quality

    Calculate the DNA concentration in µg/mL from the following information:

    a. Absorbance reading at 260 nm from a 1:100 dilution = 0.307

    b. Absorbance reading at 260 nm from a 1:50 dilution = 0.307

    c. Absorbance reading at 260 nm from a 1:100 dilution = 0.172

    d. Absorbance reading at 260 nm from a 1:100 dilution = 0.088

    Nucleic acids absorb light at 260 nm through adenine residues.

    Using Beer-Lambert Law, concentrationcan be determined from the absorptivity constants. 

    50 for double-stranded DNA
    40 for RNA

    The relationship of concentration to absorbance is expressed as

    A = €bc

    A = absorbance,
    €= molar absorptivity (L/mol-cm)
    b = path length (cm)
    c = concentration (mg/L).
    • a. 0.307 x 50 ug/mL = 15.35 ug/mL
    •     15.35 ug/mL x 100 = 1535 ug/mL

    • b. 0.307 x 50 ug/mL=15.35 ug/mL
    •     15.35 ug/mL x 50 = 767.5 ug/mL

    • c. 0.172 x 50 ug/mL= 8.60 ug/mL
    •    8.60 ug/mL x 100 = 860 ug/mL

    • d. 0.088 x 50 ug/mL = 4.40 ug/mL
    •    4.40 ug/mL x 100 = 440 ug/mL
  2. a. Absorbance reading at 260 nm from      1:100 dilution = 0.307

    0.307x50 ug/mL = 15.35 ug/mL   15.35 ug/mLx100 = 1535 ug/mL


    b. Absorbance reading at 260 nm from a 1:50 dilution = 0.307

    0.307x50 ug/mL = 15.35 ug/mL   15.35 ug/mLx50 = 767.5 ug/mL


    c. Absorbance reading at 260 nm from a 1:100 dilution = 0.172

    0.172x50 ug/mL = 8.60 ug/mL   8.60 ug/mLx100 = 860 ug/mL


    d. Absorbance reading at 260 nm from a 1:100 dilution = 0.088

    0.088x50 ug/mL = 4.40 ug/mL
    4.40 ug/mLx100 = 440 ug/mL


    If the volume of the above DNA solutions was 0.5 mL, calculate the yield for

    (a) – (d).
    • a. 1535 ug/mL  0.5 mL = 767.5 ug
    • b. 767.5 ug/mL  0.5 mL = 383.8 ug
    • c. 860 ug/mL  0.5 mL = 430 ug
    • d. 440 ug/mL  0.5 mL = 220 ug


  3. Three DNA preparations have the above following A260 and A280 readings:

    Each sample based on the A260/A280 ratio, is each preparation suitable for further use?

    If not, what is contaminating the DNA?
  4. After agarose gel electrophoresis, a 0.5 µg aliquot ofDNA isolated from a bacterial culture produced only a faint smear at the bottom of the gel lane.

    Is this anacceptable DNA sample?
    This amount of bacterial DNA should produce a bright smear near the top of the gel lane.

    This DNA is probagbly degraded and is therefore unacceptable.
  5. Contrast the measurement of DNA concentration by spectrophotometry with analysis by fluorometry with regard to staining requirements and accuracy.
    Spectrophotometry requires no DNA staining. Fluorometry requires staining of DNA togenerate a fluorescent signal. Fluorometry may be more accurate than spectrophotometry, sincedouble-stranded DNA must be intact to stain and generate a signal, where as single nucleotides will absorb light in spectrophotometry.
  6. RNA Quantity/Quality

    Calculate the RNA concentration in µg/mL from the following information:

    a. Absorbance reading at 260 nm from a 1:100 dilution = 0.307.

    b. Absorbance reading at 260 nm from a 1:50 dilution = 0.307.

    c. Absorbance reading at 260 nm from a 1:100 dilution = 0.172.

    d. Absorbance reading at 260 nm from a 1:100 dilution = 0.088.
    • a. 0.307 x 40 ug/mL = 12.28 ug/mL
    •     12.28 ug/mL x 100 = 1228 ug/mL

    • b. 0.307 x 40 ug/mL = 12.28 ug/mL
    •     12.28 ug/mL x 50 = 614 ug/mL

    • c. 0.172 x 40 ug/mL = 6.88 ug/mL
    •     6.88 ug/mL x 100 = 688 ug/mL

    • d. 0.088 x 40 ug/mL = 3.52 ug/mL
    •     3.52 ug/mL x 100 = 352 ug/mL
  7. If the volume of the above RNA solutions was 0.5 mL,calculate the yield for (a)–(d).
    a. 1228 ug/mL x 0.5 mL = 614 ug

    b. 614 ug/mL x 0.5 mL = 308 ug

    c. 688 ug/mL x 0.5 mL = 344 ug

    d. 352 ug/mL x 0.5 mL = 176 ug
  8. An RNA preparation has the following absorbance readings:

    A260= 0.208
    A280= 0.096

    Is this RNA preparation satisfactory for use?
    The A260/A280 ratio is 0.208/0.096 = 2.17


    This RNA preparation is satisfactory for use.
  9. A blood sample was held at room temperature for 5 days before being processed for RNA isolation. Will this sample likely yield optimal RNA?
    This sample will likely not yield optimal RNA due to degradation and changes in gene expression at room temperature.
  10. Name three factors that will affect yield of RNA froma paraffin-embedded tissue sample.
    Isolation of RNA from fixed tissue is especially affected by

    • *type of fixative used
    • *age/length of storage of tissue
    • *preliminary handling of original specimen.

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