Answer:
a. Sucrose is present in the plant cells that provide them energy and helps in the metabolic process of the plants. The pH of the plant cells will increase and their cellular environment becomes basic in nature. The uptake of sucrose is pH specific and the acidic condition in the environment allows the uptake of sucrose. The decrease in pH concentration in the environment increases the pH inside the cells.
b. The inhibitor of ATP inhibits the production and functioning of the ATP molecule. This effects the sucrose transport in the plant cells. As the sucrose movement requires the ATP and it is a active transport. The ATP inhibition decreases the sucrose uptake in the plant cells and the sucrose concentration decreases inside the plant cells.
Sucrose uptake in plant cells seems to require an acidic environment, brought about by the active transport of protons which requires ATP. An inhibitor of ATP regeneration would likely slow or stop this transport and, in turn, sucrose absorption.
The reported results suggest that the process of sucrose uptake in plant cells involves acidification of the surrounding medium prior to sucrose absorption. This can be explained by the proton-sucrose symport mechanism, in which protons (H+ ions) are actively pumped out of the cell in a process that requires ATP energy. When these protons combine with water (H2O) in the cell's environment, they form hydronium ions (H3O+), resulting in a lower pH or more acidic environment. Only after this acidic environment is established does sucrose uptake begin.
Based on this mechanism, introducing an inhibitor of ATP regeneration would be expected to decrease or halt this process, since ATP is required for the active transportation of protons. With less ATP, fewer protons will be pumped out, leading to a less acidic environment and, thus, lower sucrose uptake. This hypothesis is supported by how phosphofructokinase, a key enzyme in glycolysis (ATP production), is affected by low pH levels.
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Answer:
4. Glucose, which reduces cAMP, is an example of positive control, and lactase, which breaks down lactose, is an example of negative control.
Explanation:
During positive control, the presence of Glucose results in the repression of expression of lac operon. This concept is known as catabolite repression. During negative control, the lac genes are switched off by repressor when the inducer is absent (indicating the unavailability of lactose). But when lactose is present, lac binds the repressor protein and modifies it in order to dissociate it from the operator. The removal of the combination of repressor and inducer helps lac to be transcribed and expressed.
passed to offspring must occur
Answer:
Germ cell DNA
Explanation:
mutations that are present in other body cells do not get passed onto offspring. if the mutations are present in a germ cell it does get passed onto offspring
Answer:
We can see those planets because there in between the sun and the earth, that's why we can see them, also we can see mars because it is close to earth.
Explanation:
A. making an observation
B. developing a hypothesis
C. creating an experiment
D. writing a theory
Answer:
b. developing a hypothesis
Answer: Hello!
Explanation: Your answer is Diverse Organisms in a Large Region
was added to 9.9 ml of sterile buffer. After thorough mixing, this
suspension was further diluted by a 1/100 dilution followed by a
1/10 dilution. One-tenth of a ml of this final dilution was plated
on agar plates. After incubation, 52 colonies were present. How
many colony-forming units were present in the total 10 gram sample
of hamburger?
Answer:
5.2 × 10 ⁹ cfus
Explanation:
Using the dilution factors
0.1 ml of the final dilution has 52 colonies
1 ml will have approximately 520 colonies
10 ml of the final sample will have 5200 colonies
at 1 / 100 dilution
1 ml of the sample will have 5200 colonies
100 ml of the sample will have 520000 colonies
1 ml of the 0.1 ml + 9.9 ml has 520000 colonies
10 ml will have 5200000
at the second stage of the dilution
0.1 ml of the slurry had 5200000 colonies
1 ml will have 52000000 colonies
10 ml will have 520000000 colonies
100 ml of the initial sample ( 10 grams + 90 ml ) = 5200000000 colonies =
5.2 × 10 ⁹ cfu