Answer:
Such transport mechanism is called<u> Antiport.</u>
Explanation:
The production of hydrochloric acid (HCl) by the parietal cells requires the activity of the
enzyme, carbonic anhydrase.In the presence of carbonic anhydrase CO₂ and H₂O are converted to HCO⁻ and H⁻.
The newly generated proton (H+
)s transported into the lumen of the stomach via the e H
₊₋K
+ATPase (called the proton pump)Concurrently, HCO₃
⁻ exits from the basal surface (blood side) of the parietal cell via the HCO₃⁻
₋ Cl antiporter.
The chloride ionneeded to form HCl, enters the parietal cell from the blood via the HCO₃⁻₋Cl⁻ antiporter and exits at the luminal side by the Cl⁻ channel. Once within the lumen of the stomach, cl⁻ combines with H⁺to form HCl.
The luminal K⁺ that is needed to maintain the activity of the proton pump(H⁺₋K⁻ATPase)enters the parietal cells from the stomach lumen by the H⁺₋K⁺ + ATPase and is then recycled back into the lumen of the stomach by the K⁺ channel.
Answer:
Moderate aerobic physical activity involves the physical activities that will result to increase heart rate and break a sweat.
Depending on the severity of the operation moderate aerobic esercise is recommended for three (intense) to five days per week.
For older adults, moderate aerobic exercise accumulate 30 minutes a day for low- pressure, it means about 150 minutes a week. Moderate aerobic exercise for adults includes brisk walking, dancing or household chores.
Hence, the intensity level of moderate aerobic exercise for older adults should be 30 minutes a day.
No the right answer is vary with the taxonomist.
Answer:
This cycle takes approximately 100,000 years to complete.
Explanation:
PLATO
Answer:
a. 29, 29, 27, 27.
Explanation:
Separation of homologous chromosomes to the opposite poles during anaphase-I would have produced the haploid chromosome number of the daughter cells. This means that the parent cell with 56 chromosomes would produce four daughter cells each of which would have 56/2= 28 chromosomes. If one of the chromosome pairs did not segregate during anaphase-I, one daughter cells formed by the end of meiosis-I would have "n+1" chromosomes while the other would have "n-1" chromosomes. This chromosome number is maintained by meiosis-II.
Therefore, nondisjunction at anaphase-I in the parent cell with 56 chromosomes would produce a total of four daughter cells. The two daughter cells would have "n+1= 29" chromosomes and rest two would have "n-1=27" chromosomes.
