Question

Quantification of sausage thought to contain wild game: You are provided with several cooked sausages that were seized by a conservation officer suspecting that the restaurant is selling wild game (which is illegal in Ontario as the meat has not been inspected and may contain pathogens).  You take three small biopsies from different parts of two sausages in an attempt to minimize the cost of the analyses and to avoid sampling error from the sausage if more than one species of meat is present but not well mixed in the sausage.  The restaurant was marketing the sausage a moose sausage (hence the investigation), but when interviewed, they claimed it was really just spiced pork sausage.  You use this information to perform the following assessments of the DNA quality and quantify for further analyses subsequent to a basic DNeasy column extraction where the sample was eluted in 200ul of TE_0.1

a) Run and uncut DNA gel to assess the quantity and quality of the cooked sausage.

b) Picogreen of both 2ul of stock and 2ul of 1/10 stock DNA

c) Nanodrop of both 2ul of stock and 2ul of 1/10 stock DNA

d) qPCR of 2ul of 1ng/ul DNA, a concentration based on your picogreen results.

Table 1.

Sample (1/10 dilution)	Uncut (ng/ul)	Nanodrop (ng/ul)	Picogreen (ng/ul)	qPCR - Moose (ng/ul)	qPCR-Pork (ng/ul)
S1a	20	15	10	0.51	0.39
S1b		14	9	0.39	0.50
S1c		13	8	0.02	0.89
S2a		18	7	0.46	0.47
S2b		14	8	0.98	0.05
S2c		13	9	0.45	0.45
Average

Explain the differences in the results between each quantification assay in table 1 and specifically, From can you explain your qPCR results and what would you say to the conservation officer?

Answer 1

Ans. From the given table it can be observed that the quantity of DNA in S1 is 15+14+13/3 = 14

And mean quanity of DNA from S2 = 18+14+12/3 = 14.6

While by picogreen quantification the mean DNA quantity of S1 = 10+9+8/3 = 9

And for S2 = 7+8+9/3 = 8

When performed qPCR for moose it can be seen that the S1c is having very low DNA

And similarly S2b is having unexpectedly very low DNA. This shows that there is some discrepancy in the sample. But sample 1 is having proportional quantity with respect to the DNA and on the other side sample 2 is having proper result in case of moose q PCR.

This result indicates that the sample 1 is having moose sausage and the sample 2 is having pork sausage. So it can be justified to the conservation officer that we are using moose sausage and pork sausage separately not the wild game .