The food chain and food web are used to transfer energy. Plants, who are the primary energy providers in the ecosystem, use their chloroplasts to collect sunlight, which is then partially converted into chemical energy during photosynthesis.
a)
- When herbivores eat (are the primary consumers of) plants as food, this energy is transferred to the primary consumers in the food chain.
- This energy is stored in various organic products in plants. Then, the chemical energy contained in plant products is converted into kinetic energy, degrading the energy by turning it into heat. The secondary consumers come next.
- Further deterioration will occur when these herbivores are consumed by secondary consumers.
- Finally, energy will once more be destroyed when tertiary consumers eat the carnivores. As a result, the energy flow is only in one direction i.e., unidirectional.
b)
- Additionally, the energy flow in a food chain adheres to the 10% law.
- This law states that only 10% of energy is transferred from one trophic level to the next, with the remaining 90% being lost during the digestion process of the organism itself.
c)
- There are typically fewer organisms at the top of an energy pyramid because It has the least quantity of energy, the top level of an energy pyramid has the fewest organisms.
- Most ecosystems only have four trophic levels because there is eventually insufficient energy to maintain further trophic levels.
d)
- For instance, let us assume that a plant at the producer level produces 1000 Kcal of energy.
- When a primary producer eats this plant, it will only get 10% of the energy produced by the plant i.e., 1000/10 = 100 Kcal. the rest 90% will be used up by the plant itself for its metabolism.
- Further when a secondary consumer eats the primary consumer, it will only get 10% of the energy produced by the plant i.e., 100/10 = 10 Kcal.
- Lastly, as the tertiary consumer eats the secondary consumer, it will only get 10/10 = 1kcal.
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Answer:
S is rotting, on the other hand Q is fine.
Explanation:
S is not watered, and will rot soon, and Q is watered, so the behavior will be different.
Answer:
A. 50% Type A, 50% Type B
Explanation:
To answer this we can draw a punnett square.
Since o if recessive, it can only be expressed if it is not masked by a dominant allele. Therefore the male with type O blood has the genotype oo. A and B are co-dominant, so the woman has genotype AB.
_ | A | B
o |Ao | Bo
o | Ao | Bo
Since A and B are dominant alleles, they will mask the o allele. Therefore genotype Ao will be a Type A phenotype, and genotype Bo will be a Type B phenotype.
Therefore, the phenotypic probabilities for this cross is 50% Type A, 50% Type B. Hence, the answer is A.
Hope this helped!
Answer:
Yes, if there is a mutation in the operator of the lac operon in a cell such that the lac repressor always stays bound to the operator. If lactose is added to the cell, the lac repressor loses its ability to bind DNA. It clear off the operator, clearing the way for RNA polymerase to transcribe the operon.
Answer: Option A - alpha ketoglutarate dehydrogenase
Explanation:
One of the peculiarities of the Citric acid cycle is that it contains two successive decarboxylation steps.
The 1st decarboxylation step is catalyzed by Isocitrate dehydrogenase, while the 2nd step is catalyzed by Alpha ketoglutarate dehydrogenase.
So, the reaction where 4CO2 is observed will be:
Oxidative decarboxylation of α-ketoglutarate to Succinyl-CoA