Answer:
I believe the answer is C.
Explanation:
Hope my answer has helped you!
55 is the answer
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Answer:
This is due to the event of Speciation that happened for the rodents in Island B but not for the rodents in Island C.
Explanation:
- Due to splitting of the population,
- The sub-population of rodents formed in Island B are B1 and B2.
- The sub-population of rodents formed in Island C are C1 and C2.
- In case of Island B, each of the B1 and B2 sub-populations that got split from each other developed certain mutations that were necessary for them to adapt to the particular diverse environment each of them were exposed to, through the period of 50,000 years. These mutations were so varied that reproductive isolation was generated between them that resulted in each of them to develop into different species.Hence, speciation happens here and B1 and B2 are incapable of inter-breeding.
- In case of Island C, each of the C1 and C2 sub-populations that got split might have got exposed to similar environmental change or no environmental change or the environmental change might have been too small to cause drastic change in each of the sub-populations. As a result of this the two sub-populations might have acquired certain mutations to adapt to the environment each of them were exposed to, through a period of 100,000 years. These mutations might not have been too variable or contrasting to cause reproductive isolation between C1 and C2. Hence, no new speciation happens here and C1 and C2 are capable of inter-breeding.
Answer: and Explanation:
A.)The reason for the different products of glycogen breakdown in the two tissues is that glucose 6-phosphotase which is
a known enzyme that brings about hydrolysis of glucose 6-phosphate as a result of the creation of a phosphate group and free glucose is not available in heart and skeletal muscle, therefore,any glucose 6-phosphotase that is produced will just enters the glycolytic pathway and get converted to lactate through pyruvate, in the absence of Oxygen O2.
B) Whenever a situation involving fight or flight arises, the concentration of glycolytic precursors becomes high in order to prepare for muscular activity. Since the membrane is impermeable to any charged species, and at the same time glucose 6-phosphotase enzyme cannot be moved through the glucose transporter, then there cannot be a release of Phosphorylated intermediates from the cell. The blood glucose level must be maintained by the liver by releasing of glucose.
glucose that is later formed from glucose 6-phosphotase then enters the bloodstream.
