Answer:
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Glucose is the preferable source of energy for E. coli cells. By adding small amounts of glucose, 0.05% in this case, it would enable cell growth to proceed as usual until it has reached mid log growth phase. At this point, the small amount of glucose added should have been depleted and the E.coli cell will now be dependent on lactose present as its source of nutrients. Lactose will also inactivate lac repressors and given the lack of glucose, CAP binding to high amounts of cAMP will be activated and increase expression of lac operon genes.
It should be noted that this system of expression may not tightly regulate the expression of lac operon before the E.Coli reaches exponential phase. As lactose is still present, lac repressors will be inactivated throughout the entire experiment and hence small amounts of proteins might be produced even when not auto-induced. After auto-induction, CAP-cAMP protein complex will simply upregulate expression of lac operon genes.
<span>Firstly, producers get their energy energy from the sun through photosynthesis, creating or producing nutrients in the plant. A herbivore consumer can then eat it getting it's nutrients so, consumers rely on producers for their food but decomposers rely on a carnivore or omnivore (consumer) to eat some species of a consumer. The decomposer, lets say a earth worm eats and consumes the rest of the animal breaking it down to it's core nutrients. So I guess you could say decomposers rely on carnivores/omnivores to kill each other and herbivores, and herbivores/omnivores rely on producers for nutrients to eat</span>
Tia would be classified as obese
Answer:
Explanation:
The first one is Prostista: Since they are mostly unicellular and are classified as eukaryotic.
The second one is Eubacteria: Since all Eubacteria has peptidoglycan in their cell walls.
The Last One is Achaebacteria: They tend to help in digestion which is a very acidic environment and fit the other two criteria
