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9966 [12]
3 years ago
14

H pylori a type of bacteria are one known cause of

Biology
1 answer:
Marina CMI [18]3 years ago
6 0

gastritis

H pylori channels it's way into the cells of the stomach lining. It is the most common cause of gastric ulcer.

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What kind of cell does the virus attack/damage?
yanalaym [24]

Answer:

<u><em>Plant cell, Bacterial cell and Animal cell</em></u>

<u><em>Example: </em></u>Plant viruses:Tobacco Mossiac Virus

                Animal virus: HIV(Human immunodeficeincy virus),Rhino virus      

                Bacteriophage  

Explanation:

Viruses attack living cells on the basis of their host types.

7 0
2 years ago
Which state of matter has the most densely packed particles?<br><br> Gas, solid, liquid, or none?
Leviafan [203]
I think it’s solid. Hope this helps. :)
8 0
3 years ago
Read 2 more answers
Whay can piranhas do in minutes?
Rom4ik [11]
Bonjour!

Piranhas can devour any living thing in minutes, and sometimes even seconds. They are known for their ferocious attitude towards anything living; including others of their kind. They are some of the deadliest fish in the sea - watch out!

Hope this helps you!

~DL

3 0
4 years ago
Part A: If one follows 70 primary oocytes in an animal through their various stages of oogenesis, how many secondary oocytes wou
olga nikolaevna [1]

Answer:

<u>Part A</u> : 70 secondary oocytes will be formed.

<u>Part B</u> : 70 first polar bodies will be formed.

<u>Part C</u> : 70 ootids will be formed.

Explanation:

During oogenesis growth maturation of a single oogonium produces one primary oogonium.

the primary oogonium then undergoes meiosis -1 and produces  one secondary oocyte and first polar body.

The secondary oocyte then undergoes meiosis - 2 and forms an ootid and second polar body.

The ootid then differentiates into the ovum.

As in the above scenario , 70 primary oocytes are present , they undergo meiosis-1 and produces 70 secondary oocytes and 70 first polar bodies. Hence answers of part A and B is 70.

As 70 secondary oocytes are formed , they undergo meiosis -2 and forms 70 ootids which then differentiate in 70 ovums.

8 0
4 years ago
Compare and contrast osmotic challenges faced by animals in freshwater, marine, and terrestrial environments, and the adaptation
Gnesinka [82]

Answer:

  • Fresh water fish have higher salt contents in their bodies than in their environments.
  • Marine fishes have less salt in their bodies than their environment
  • Terrestrial organisms have the challenge of water retention due to atmospheric contact.

Explanation:

FRESH WATER OSMOREGULATION

The salt concentration in salt water fish is higher than the concentration found in its environment (fresh water). This causes water to enter into the body of the fish through osmosis and without regulating processes, the fish is bound to swell and likely burst.To compensate for this challenge, the kidney in fresh water fish produces a large amount of urine, causing them to lose salt. To ensure too salt is not lost beyond the basic requirement, chloride cells in the gills take up ions from the water which are transported into the blood.

MARINE OSMOREGULATION

In marine fishes, the challenge opposes that of fresh water fishes since salt content in this case is lower in their blood than in their environment. To address this challenge, marine fishes lose water constantly while retaining salts to lead to a build up. The water lost, is then made up for and replenished by continual drinking of seawater. The chloride cells in marine fishes works in a manner opposing that of fresh water fish, functioning to compliment the excretion of salts by the kidney.

TERRESTRIAL OSMOREGULATION

The major challenge of osmoregulation in  terrestrial organisms is water regulation in the body owing to their contact with the atmosphere.

Terrestrial organisms possess effective kidneys which enable osmoregulation. A series of processes including filtration, re-absorption and tubular secretion, enable regulation of fluids and water conservation.

Water passes out of the descending limb of the loop of Henle, leaving a more concentrated filtrate inside. Salt diffuses out from the lower, thin part of the ascending limb. In the upper, thick part of the ascending limb, salt is then actively transported into the interstitial fluid. The amount of salt in the interstitial fluid, determines how much water moves out of the descending limb i.e the saltier it gets, the more water moves out of the descending limb. This process leaves a concentrated filtrate inside, so more salt passes out. Water from the collecting ducts moves out by osmosis into this hypertonic interstitial fluid and is carried away by capillaries, achieving osmoregulation.

8 0
3 years ago
Read 2 more answers
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