A rapid lateral flow immunoassay is presented that uses carboxyl-modified superparamagnetic nanoparticles as labels that can be quantified by highly sensitive multi-channel electronic readers. The approach is generic in that it is likely to be applicable to numerous small molecules. The method permits both single- and multiplex assays at a point-of-need without sample pretreatment. It is user-friendly and offers attractive characteristics demonstrated here for detection of morphine, fentanyl and methamphetamine in . The competitive immunoassay uses commercially available reagents that do not require special permissions. After migration of sample, the lateral flow test strips are subjected to an alternating magnetic field at two frequencies. The response from the nanolabels is readout at a combinatorial frequency from the entire volume of a porous immunochromatographic membrane by the magnetic particle quantification technique.
application to the determination of drugs of abuse.
Estimating the detection time of a drug in urine is complex because of many different influencing factors and the lack of experimental data. Detection times vary depending on dose and route of administration, metabolism and characteristics of the screening and confirmation assays. Using a cut-off value of 1000 ng/mL, urinary samples can be positive for amphetamine for up to 5 days after intake of the drug.
Answer:
Option D, a test for the presence of cells that contain DNA
Explanation:
As we know all living organisms are made up of cells and DNA is the basic genetic material that codes for protein and traits with an organism. If even a single cell with DNA is detected on the extraterrestrial planet then it will be clear that it has life on it or the environmental conditions are favorable to support life. Cells are microscopic and hence can be seen through microscope which can be installed easily on a probe.
Hence, option D is correct
Answer:
C if you plug it into the sentance its the only one that makes sense
P: DdCc x DdCc
g: DC Dc dC dc DC Dc dC dc
1.)
DC Dc dC dc
DC DDCC DDCc DdCC DdCc
Dc DDCc DDcc DdCc Ddcc
dC DdCC DdCc ddCC ddCc
dc DdCc Ddcc ddCc ddcc
2.)
G= 1:2:2:1:1:4:2:2:1
3.)
F= 9:3:3:1
Answer:
It occurs in the cytoplasm. Glycolysis also involves two stages which break up glucose. In the 1st stage, Glucose is broken into two phosphorylated 3-carbon compounds through a series of reactions.
