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ohaa [14]
3 years ago
15

Blood ph and cerebrospinal fluid ph are affected by carbon dioxide content. how does this enable the organism to sense a disturb

ance in gas levels? question 2 options:
a.the brain alters the ph of the cerebrospinal fluid to force the animal to retain more or less carbon dioxide
b.the brain directly measures and monitors carbon dioxide and causes breathing changes accordingly
c.the medulla oblongata is able to control the concentration of bicarbonate ions in the blood
d.the medulla oblongata, which is in contact with cerebrospinal fluid, monitors ph and uses this measure to control breathing
e.the lungs sense changes in oxygen concentration, which causes the medulla oblongata to speed up or slow breathing
Biology
1 answer:
frutty [35]3 years ago
3 0

The answer is B. The cardiac, respiratory, and vasomotor control centers are in the medulla. This means that the medulla controls heart rate, blood pressure, breathing, and the size of the body's blood vessels. They have chemoreceptors, pH sensing structures that detect levels of carbon dioxide in the blood and adjust the breathing rate accordingly.






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natima [27]

The correct answer is option A, that is, a landfill in the region of nonporous bedrock.  

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6 0
3 years ago
How does leaf structure best represent the relationship between structure and function? It indicates the maximum height an organ
vova2212 [387]

Answer:

The correct answer is - *It relates directly to the characteristics of the plant's environment*

Explanation:

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6 0
3 years ago
NO NEED TO ANSWER ALL! WOULD BE HELPFUL THOUGH :)
frutty [35]

Answer: question 6

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8 0
3 years ago
What is the role of RNA polymerase?
juin [17]
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6 0
3 years ago
Read 2 more answers
You dilute your bacteriophage for plating 10^-6. You mix 0.1ml of the bacteriophage dilution with 0.3ml of bacteria and plate wi
Blizzard [7]

Answer:

46 * 10^7 pfu/ml

Explanation:

Given -

Bacteriophage dilution = 0.1 ml

Volume of bacteria and plate with top agar  = 0.3 ml

Number of plaques = 46 pfu

Dilution (d) = 10^{-6}

Final concentration of phage stock is equal to

= \frac{pfu}{d*v}

Where pfu is the count of plaques

d is the dilution and v is the volume

Substituting the given values, we get -

\frac{46}{10^{-6} * 10^{-1}} \\= 46 * 10^7 pfu/ml

7 0
3 years ago
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