SCGI Practical Rev Flashcards
Explain the purpose of each of the following components in the isolation of genomic DNA: a) Cell lysis b) Salt solution c) Protease K d) RNase
a) Cell Lysis: break open the cells to release the contents of the cell, including the genomic DNA. The goal is to access the DNA within the cell while minimizing damage to the DNA.
b) Salt Solution:neutralizing the charges on the DNA molecule and the proteins. It helps to prevent the DNA from being soluble in water
c) Protease K: degrade proteins that are bound to the DNA or are present in the solution after lysis. This includes histone proteins that package the DNA in the cell.
d) RNase: degrades RNA. In a DNA isolation procedure, it’s important to get rid of RNA so it doesn’t contaminate the DNA sample
What is the difference between genomic DNA (gDNA) and complementary DNA (cDNA)? How would this difference affect the interpretation of data following a PCR experiment?
Genomic DNA (gDNA):
- Contains the complete genetic information of an organism, including coding (exons) and non-coding regions (introns, regulatory sequences, and intergenic spaces).
- isolated directly from the nucleus of a cell.
Complementary DNA (cDNA):
- synthesized from messenger RNA (mRNA) using reverse transcriptase, and thus contains only the exonic (coding) sequences.
- represents only the genes that were actively being transcribed into mRNA at the time of RNA extraction.
Effects on PCR Data Interpretation:
gDNA as a template:
- Reflect the entire genomic sequence, including both introns and exons
- The amplicon size might be larger due to the presence of introns.
cDNA as a template:
- Only reflects the exonic sequences, equivalent to the processed mRNA.
- The amplicon size would be smaller as it doesn’t include introns.
If investigating genetic mutations across entire gene sequences, including introns, gDNA would be more suitable.
What is the function of trypsin during the harvesting of embryonic stem cells, and why is it necessary to inactivate the enzyme?
- protease enzyme used to cleave proteins and peptides.
- trypsin is primarily used to break down the proteins that hold the cells together and attach them to the culture dish.
trypsinization, effectively detaches the cells from the dish and each other, allowing them to be easily harvested.
Explain the principle behind using a spectrophotometer to measure the concentration of nucleic acids (DNA and RNA). What are the specific wavelengths used, and why are they important?
Used to measure the absorbance of light by a solution at specific wavelengths.
Nucleic acids (DNA and RNA) absorb ultraviolet light, with the absorbance being proportional to the concentration of the nucleic acids in the solution.
By comparing the absorbance of the unknown sample to standards with known concentrations, the concentration of nucleic acids in the unknown sample can be calculated.
260 nm: Both DNA and RNA absorb strongly at this wavelength. It’s used to measure the concentration of nucleic acids in the sample.
280 nm: Proteins, particularly those with aromatic rings in their amino acids (like tryptophan and tyrosine), absorb light at this wavelength. This is used to assess protein contamination in the sample.
The 260/280 nm absorbance ratio is often calculated to check the purity of the nucleic acid sample. Pure DNA should have a ratio of ~1.8, while pure RNA should have a ratio of ~2.0. A lower ratio indicates protein contamination.
discuss the significance of the A260/A280 ratio.
Used to assess the purity of nucleic acid samples (DNA and RNA) in relation to contaminating proteins.
Nucleic acids absorb maximally at a wavelength of 260 nm, while proteins (especially those with aromatic amino acids) absorb maximally at 280 nm.
An A260/A280 ratio of ~1.8 is generally accepted as “pure” for DNA; a ratio of ~2.0 is accepted as “pure” for RNA.
A lower ratio may indicate the presence of protein, phenol, or other contaminants that absorb at or near 280 nm, implying that the nucleic acid sample is not pure.
Accurate determination of this ratio is crucial for downstream applications that require pure nucleic acid samples, such as sequencing, cloning, or PCR.
Explain how you would ensure accurate readings when using a spectrophotometer for DNA quantitation. Discuss the importance of blanking the machine and any other steps you would take.
Blanking the Machine: Always start by blanking the spectrophotometer with the solvent used to dissolve the DNA (typically water or buffer). This sets a baseline reading to eliminate any absorbance by the solvent.
DNA Sample Preparation: Make sure the DNA sample is properly dissolved and free of particulate matter which might interfere with the light path.
Replicate Measurements: Replicate measurements can improve accuracy by averaging any random errors.
All these steps are essential to ensure the accuracy of the readings. Inaccurate readings could lead to errors in downstream applications such as PCR, cloning, or sequencing that require precise amounts of DNA.
If the DNA sample was found to be impure, what steps could be taken to improve the quality and purity of the DNA for future experiments?
- Ensure you are following the extraction protocol correctly and carefully. Avoid overusing mechanical disruption methods (like vortexing or pipetting) that can shear the DNA.
- Use ethanol or isopropanol precipitation to further purify the DNA. This can help remove contaminants and concentrate the DNA.
- In case of severe contamination, you may need to re-extract the DNA from the source material.
Proteinase K Digestion Buffer (0.2 M NaOH, 1 % (w/v) SDS (sodium dodecyl sulphate)) require 25mls
Sodium Hydroxide (NaOH):
Assume you have a stock solution of 1 M NaOH.
You need a final concentration of 0.2 M in 25 ml.
Using the formula (C1)(V1) = (C2)(V2), where C1 is the initial concentration, V1 is the volume of the initial solution required, C2 is the final concentration, and V2 is the final volume.
Plug in the values: (1 M)(V1) = (0.2 M)(0.025 L). Solving for V1 gives 5 ml.
Therefore, you need to add 5 ml of the 1 M NaOH solution.
Sodium Dodecyl Sulphate (SDS):
You need a final concentration of 1% (w/v) SDS. This means 1 g of SDS in 100 ml of solution.
For a 25 ml solution, you need 0.25 g of SDS. (25 ml is 1/4 of 100 ml, so you need 1/4 of 1 g)
Measure out 0.25 g of SDS and add to the solution.
Lysis Buffer (50 mM Glucose, 25 mM Tris/HCl (pH 8.0), 10 mM EDTA) require 100mls
Glucose:
Desired final concentration (C2) = 50 mM = 0.05 M
Final volume (V2) = 100 ml = 0.1 L
Available stock solution concentration (C1) = 1 M
Using the formula (C1)(V1) = (C2)(V2), solve for V1 (volume of glucose to add):
(1 M)(V1) = (0.05 M)(0.1 L)
V1 = 0.005 L = 5 ml
Tris/HCl:
Desired final concentration (C2) = 25 mM = 0.025 M
Final volume (V2) = 100 ml = 0.1 L
Available stock solution concentration (C1) = 1 M
Again, using the formula (C1)(V1) = (C2)(V2), solve for V1 (volume of Tris/HCl to add):
(1 M)(V1) = (0.025 M)(0.1 L)
V1 = 0.0025 L = 2.5 ml
EDTA:
Desired final concentration (C2) = 10 mM = 0.01 M
Final volume (V2) = 100 ml = 0.1 L
Available stock solution concentration (C1) = 0.5 M
Again, using the formula (C1)(V1) = (C2)(V2), solve for V1 (volume of EDTA to add):
(0.5 M)(V1) = (0.01 M)(0.1 L)
V1 = 0.002 L = 2 ml
TE Buffer (10 mM Tris/HCl, 1 mM EDTA (pH 8.0)) require 5mls
Tris/HCl:
You want a final concentration of 10 mM, or 0.01 M, in a final volume of 5 ml or 0.005 L.
Use the formula (C1)(V1) = (C2)(V2), where C1 is the initial concentration, V1 is the volume of the initial solution required, C2 is the final concentration, and V2 is the final volume.
Plugging in the numbers: (1 M)(V1) = (0.01 M)(0.005 L).
Solving for V1 gives 0.00005 L or 50 µl.
Therefore, add 50 µl of the 1 M Tris/HCl stock solution.
EDTA:
You want a final concentration of 1 mM, or 0.001 M, in a final volume of 5 ml or 0.005 L.
Again, use the formula (C1)(V1) = (C2)(V2).
Plugging in the numbers: (0.5 M)(V1) = (0.001 M)(0.005 L).
Solving for V1 gives 0.00001 L or 10 µl.
Therefore, add 10 µl of the 0.5 M EDTA stock solution.
Water:
Subtract the volumes of Tris/HCl and EDTA you added from the total volume.
5 ml - 50 µl - 10 µl = 4940 µl.
Add 4940 µl of water to reach the final 5 ml volume.
6M NaCl – require 250mls
The molecular weight of NaCl is approximately 58.44 g/mol.
Using the formula for molarity:
M = moles/Liter
You can rearrange this to solve for moles:
moles = M x Liter
So, to find the number of moles of NaCl needed for a 6 M solution:
moles = 6 M x 0.25 L = 1.5 moles
Next, convert moles to grams using the molecular weight:
grams = moles x molecular weight
grams = 1.5 moles x 58.44 g/mol = 87.66 g
So, to prepare 250 mL of a 6 M NaCl solution, you would need to dissolve 87.66 g of NaCl in enough water to make 250 mL of solution.
Band intensities of PCR for undifferentiated and differentiated
The intensity or brightness of these bands is often related to the quantity of the PCR product, and therefore, the amount of the original template DNA.
Brighter Band for Undifferentiated Cells:
If the PCR product band is brighter for the undifferentiated ES cells, it would suggest that the target gene was more highly expressed in these cells.
Brighter Band for Differentiated Cells:
Conversely, if the PCR product band is brighter for the differentiated ES cells, it would imply that the target gene was more highly expressed in the differentiated cells.
Equal Band Intensity:
If both bands are of approximately equal brightness, it would suggest that the gene is expressed at similar levels in both cell types.
If there is no band in the No Template Control (NTC) lanes, what does this indicate about the PCR reaction and possible contamination?
No Contamination: The absence of a band indicates that there was no DNA contamination in your PCR reagents or setup.
Specificity of PCR: The lack of a band in the NTC also suggests that your primers are specific to your target sequence and are not binding and amplifying any unintended sequences.
Reliable Results: Thus, the lack of a band in the NTC lane indicates that any bands seen in your sample lanes are likely due to the amplification of your target sequence from your DNA template and not due to contamination.
If there is a band in the NTC lanes, what could be the possible reasons for this observation? solutions?
- contamination. The contaminants could be template DNA from other reactions, PCR products from previous experiments (carry-over contamination), or microbial contamination.
- primer-dimer formation, which happens when primers self-anneal and extend, forming a product in the absence of template DNA. This typically results in a smaller, faint band, often different in size from the expected product.
solutions:
- Make sure all reagents are fresh and stored appropriately. Use new aliquots of reagents if possible.
- Follow good molecular biology lab practices, such as using separate areas for setup and analysis, regularly cleaning workspaces, and using filter pipette tips.
- Check your primer design and consider redesigning to prevent primer-dimer formation.
what is a housekeeping gene
A housekeeping gene is a gene that is constantly expressed at a high rate in virtually all types of cells because it provides basic functions needed for the survival of the cell. Housekeeping genes are often used as reference genes in experiments like Polymerase Chain Reaction (PCR) and Reverse Transcription PCR (RT-PCR) because their constant level of expression allows for normalization of gene expression levels.
e.g Beta-actin
