Environmental Drivers of Symbiosis LAB (HELP PLEASE) 1) Approach: Create a summa
ID: 217875 • Letter: E
Question
Environmental Drivers of Symbiosis LAB (HELP PLEASE)
1) Approach: Create a summary of the overall approach you use to answer the question being addressed. Summerize on how the questions below were obtained.
LAB DETAILS:
- If sample A reaches a fluorescence level of "12" after 15 cycles of PCR for a symbiont gene and sample B reaches a fluorescence level of "12" after 18 cycles, which sample has more symbionts? By how much?
- Why was it important to use chelex in the DNA extraction? What is the role of magnesium in terms of DNA?
- Imagine that S. minutum provides its host with more amino acids than S. trenchii, so S. minutum hosting anemones have more protein per cell. How does this affect the validity of symbiont density as measured as symbionts/host protein? How about for symbiont density measured through the symbiont CO1/host 18S approach?
- Why is it important to normalize symbiont values to a similar measurement of the host? Give an example of a scenario where not normalizing symbiont measurements could produce misleading results.
Groups & light/tempature conditions Group 1 (high light/high temp) Group 2 (low light/high temp) Group 3 (low light/high temp) Group 4 (high light/low temp) Group 5 (lhigh light/low temp) Group 6 (low light/high temp) Group 8 (low light/low temp) Group 9 (low light/low temp) Group 10 (high light/high temp) Group 11 (high light/high temp) Group 12 (low light/low temp) Undiluted #1 0.184 0.022 0.020 0.061 0.058 0.153 0.016 0.035 0.109 0.44 0.059 Undiluted #2 0.116 0.034 0.038 0.051 0.042 0.080 0.099 0.061 0.069 0.47 0.047 Dilted 0.085 0.023 0.40 0.037 0.029 0.011 0.092 0.031 0.029 0.62 0.037 Amount of sybionts 12,000 cell/mL 12,500 cell/mL 8,000 cell/mL 12,500 cell/mL 11,240 cell/mL 16,666 cell/mL 8,000 cell/mL 9,530 cell/mL 15,340 cell/mL 16,340 cell/mL 10,380 cell/mLExplanation / Answer
Sample A has more symbionts than sample B that is why it took 3 cycles less than sample B to reach the same intensity. sample A/sample B = 2(delta Ct) = 2 (15-18) = 2-3 = 1/23= 1/ 8= .125. There is .125 times sample A compared to B. [(delta Ct) = difference in number of cycles ]
Chelex disrupts Cell and nuclear membrane, there by releasing the DNA. It binds to divalent metal cations like Mg2+ that are essential co factors of DNases that degrade DNA.
When measured symbionts/ host protein when there is same density of S. trenchii and S. minutum, the latter will have lesser value of symbionts/host protein since host protein will be more in S. minutum hosting anemones. Both types of host cells will express 18S in fairly equal degree so there will not be any discrepancy in symbiont density calculation by symbiont CO1/host 18S method.
inorder to measure two symbiot's gene expression one need to make sure that these symbiots are isolated from same amount of host cells. Since each number of host cells will express the standard gene in same degres, normalizing the two symbiot's gene expression with a similar measurement of their respective host's will give an idea of symbiont's gene expression per unit host gene expression which is now comparable.
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