Answer:
A) That resistance in bacteria is produced due to inactivation of ampicillin by the beta lactamase enzyme. This enzyme is expressed by the bla gene found in the plasmid. This enzyme is secreted into the culture medium, thereby inactivating ampicillin. Thanks to this inactivation, the bacteria colonies will be able to develop. After a day of incubation, only those bacteria that took the plasmid that gives them resistance to ampicillin will grow after transformation. After prolonged incubation, two types of colonies can be observed in the culture medium. One large colony with ampicillin resistance, and another small colony, both of which are sensitive to ampicillin.
B) Large colonies are characterized by being resistant to ampicillin. When Ramón isolates the plasmid, he will have the gene that provides resistance to antibiotics. Said plasmid can be used again on those bacteria that are sensitive to ampicillin.
On the other hand, satellite colonies are sensitive to ampicillin. These types of colonies do not have the plasmid that contains the gene that gives ampicillin resistance. It is not possible to isolate any plasmids from these satellite colonies. These satellite bacteria will not be able to grow if they are transferred to a plate containing fresh ampicillin, while large colonies, which possess the plasmid that gives them resistance to ampicillin, will be able to grow on that plate.
Explanation:
Blue litmus paper turns red in the presence of an acid. Therefore, it can be assumed that the substance in the beaker is an acid.
Acids have a pH level of less than 7. Consequently, it can be assumed that the substance has a pH level less than 7.
Answer:
An Alkaline solution should be the answer to that.
The reaction described above is the formation of an acetal. The initial starting material has a central carbonyl and two terminal alcohol functional groups. In the presence of acid, the carbonyl will become protonated, making the carbon of the carbonyl susceptible to nucleophilic attack from one of the alcohols. The alcohol substitutes onto the carbon of the carbonyl to provide us with the intermediate shown.
The intermediate will continue to react in the presence of acid and the -OH that was once the carbonyl will become protonated, turning it into a good leaving group. The protonated alcohol leaves and is substituted by the other terminal alcohol to give the final acetal product. The end result of the overall reaction is the loss of water from the original molecule to give the spiroacetal shown in the image provided.
Answer:
392.97 litres
Explanation:
From the equation of reaction, we can see that 1 mole of methane yielded 1 mole of carbon iv oxide. Hence, 15.9 moles of methane will yield 15.9 moles of carbon iv oxide.
At s.t.p one mole of a gas occupies a volume of 22.4L ,hence 15.9 moles of a gas will occupy a volume of 22.4 × 15.9 which equals
356.16L.
Now, we can use the general gas equation to get the volume produced at the values given.
We have the following values;
V1 = 356.16L P1= 1 atm ( standard pressure) T1 = 273K ( standard temperature) V2 = ? T2 = 23.7 + 273 = 296.7K P2 = 0.985 atm
The general form of the general gas equation is given as :
(P1V1)T1 = (P2V2)/T2
After substituting the values , we get V2 to be 392.97Litres
