Answer:
a. Oxygen performs an essential role in the mentioned microbial cell in a manner that it takes part in the procedure of glycolysis, Krebs cycle, and electron transport chain, which eventually assists in the production of energy from food substrates and this generation of energy helps the cell to survive.
In the existence of oxygen, sugar gets dissociated through glycolysis to generate pyruvate, which again in the existence of oxygen is transformed into acetyl CoA. This moves into the Krebs cycle and gets dissociated to water and carbon dioxide generating ATP through ETC. This generation of ATP helps the cell to survive.
In low oxygen surrounding or in the absence of oxygen, some of the aerobic microbes can switch their respiratory pathway and carry on the process of fermentation and anaerobic respiration to produce energy and thrive. However, the mentioned microbial cell, which when it comes in contact with the low oxygen environment cannot carry out fermentation process and would die eventually.
b. This organism can be classified as obligate aerobes as they always need oxygen and do not possess the tendency to carry out the process of anaerobic respiration or fermentation under the absence of oxygenic environment.
The factor that poses the greatest threat is loss of habitat. Have a nice day!
Answer: B
Explanation:
Ammonia is so toxic that it can be transported and excreted only in large volumes of very dilute solution. As a result most terrestrial animals simply don't have access to sufficient water to routine excrete ammonia.
The main advantage of urea is its very low toxicity. Animals can transport urea in the circulatory system and store it safely in high concentrations. Also much less water is loss when a given quantity of nitrogen is excreted in a concentrated solution of urea than would be in a dilute solution of ammonia.
Answer:
The correct answer is b.Amplify the gene using PCR. Insert the gene into a plasmid vector. Transform the vector into the bacteria.
Explanation:
If I have a very small amount of gene for a fluorescent protein than the first step is to amplify the gene so that appropriate protein can be produced. PCR is the instrument that is used to amplify the protein.
So after amplification of the gene, the plasmid vector will be used in which the gene will be inserted because the plasmid vector is used to insert this gene in host cells where protein will be expressed.
The final step will be to transform bacteria with recombinant plasmid so that plasmid can make its copy and express a fluorescent protein in bulk.
Answer:
According to the image, the approximate QT interval is 0.4 seconds.
Explanation:
The QT interval is the space between the beginning of the QRS complex and the end of the T wave, which represents from the beginning of depolarization to ventricular repolarization.
Considering that the extandardized measurements of an EKG, where the paper circulates at a speed of 25 mm/s, 1 mm horizontal —measuring time— has an equivalence of 0.04 s.
In the image, there is 10 mm between the beginning of the QRS and the end of the T wave, so:
0.04 seconds X 10 mm = 0.4 seconds.
Then, the estimated QT interval is 0.4 seconds.
