Answer:
The hydrophilic nature of the stain explains that these molecules are polar and charge molecules. therefore, attracts the water molecule but due to their polar nature, it will avoid lipid or fat molecules.
Adipocytes are the cells that are made up of fat or lipids that are hydrophobic due to their non-polar nature and due to this fact, these hydrophilic stains will not be able to stain adipocytes and will not be colored by the stain.
Answer:
The correct option is : a. Identification of the cell
Explanation:
The cell membrane, also called plasma membrane, is a semi-permeable membrane surrounding the cytoplasm of the cell, that protects and separates the interior of the cells from the external environment.
The carbohydrate layer of the cell membrane is known as the glycocalyx. In eukaryotes, <u>the carbohydrates present on the surface of the cell membrane play a important role in the cell-cell recognition and share information.</u>
Unlike Lipids, the nucleic acids typically do not form polymers with its own kind. The polymers are the substances or molecules composed of large number of repeating units or sub units. Instead of having the same polymers, the nucleic acids are form nucleotides come together through phospodister between carbon atoms.
Answer:
D
Explanation:
D is an example of confirmation bias because confirmation bias involves people looking for information to support their opinions and beliefs, the person confirms their already existing biases and beliefs when they seek out information to support these biases and beliefs whilst ignoring contradicting information. The journalist is confirming his belief that climate is not real when he cites an article discussing how humans cannot be causing climate change and when he ignores several other articles that discuss how humans can cause climate change so as a result he is confirming his own bias and belief.
Answer:
6.75l l/min
Explanation:
Streptomycin is extracted from the fermentation broth using an organic solvent in a counter current staged extraction unit.
for 5 stage counter current staged extraction unit for streptomycin
X5 = 0.2g/l
X0 = 10.0g/l
X5/X0 = 0.02
given
Xn/X0, E, and
n = 5
A estimate of E can be evaluated directly;
E = 1.8
= LKd/H
by substituting the given values; Kd = 40, H= 150l/min
then we have,
L;L = 1.8(150)/40
= 6.75l l/min
