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Kipish [7]
3 years ago
8

The _____________ cells in the _________________ of a nephron can cause and increase or decrease in body pH. A. intercalated; co

llecting duct B. principal; proximal convoluted tubule C. macula densa; distal convoluted tubule D. glomerular; nephron loop

Biology
2 answers:
gayaneshka [121]3 years ago
7 0

Answer:

The intercalated cells in the distal convoluted tubule of a nephron can cause and increase or decrease in body pH.

Explanation:

The renal collecting duct is the nephron segment where the final urine content of acid equivalents and inorganic ions are determined.

Two types of cells regulate the acid-base and volume homeostasis.

Intercalated cells, which express acid-base transporters and vacuolar H+-ATPase, maintain an apropriate acid-base balance and principal cells that express the epithelial sodium channel ENaC and aquaporin 2, regulate electrolyte reabsorption.

Filtered bicarbonate is predominantly reabsorbed from the proximal convoluted tubule about 80% and the remaining bicarbonate is reabsorbed from the thick ascending limb of the loop of Henle, the distal nephron, specifically the distal convoluted tubule & the connecting tubule, and finally the cortical and both the outer and inner medullary collecting duct.

Intercalated cells are enriched in mitochondria, and express proteins involved in transport of proton equivalents such as vacuolar H+-ATPase, carbonic anhydrase II and bicarbonate transporters. There are two types of intercalated cells Type-A and Type-B.

anastassius [24]3 years ago
7 0

Answer:

hmmm

Explanation:

hmmmmm

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Which term best describes the change from more light-colored moths to more dark-colored moth:
maria [59]

Answer:

The correct option is B) adaptation

Explanation:

The features and properties of an organism that make it favourable to live in an environment are termed as the adaptations of that organism. Over time, evolution occurs and organisms with better adaptations survive as they are better suited to live in a particular environment. The light-coloured moths were adapted to live on Earth when the colour of the trees had not gone dark. With time, as the colour of trees got darker, the moth had to change colour to be better adapted to live on Earth.

6 0
3 years ago
A solution of acetic acid is prepared in water by adding 11.1 g of sodium acetate to a volumetric flask and brining in the volum
velikii [3]

Answer:

The concentration of

the Acetic Acid is 0.0325M

the Sodium Acetate is 0.1028M

Explanation:

Given

The dissociation of acetic acid is

CH3COOH + CH3COO- + H+ + HA -> A- + H+

We calculate the molarity of sodium acetate by using

Molarity = Moles/Volume

The formula weight of sodium acetate is 82.03 g/mol.

Moles = 11.1g/82.03g/mol

= 0.1353M

Since the volume is 1L, the molarity of sodium acetate will still be 0.1353M

The starting concentration of sodium acetate is 0.1353M

The final concentration of sodium acetate is the starting amount of sodium acetate, minus the amount that was protonated.

Assume that, x = concentration of acetic acid, which is the amount of sodium acetate that was protonated.

From the Henderson-Hasselbalch equation

PH = pKa + log[A-]/[HA]

[A-] = 0.1353 - x

[HA] = x

pKa = 4.75

PH = 5.25

Substitute in these values

5.25 = 4.75 + log(0.1353-x)/x

log(0.1353 - x)/x = 5.25 - 4.75

log(0.1353-x)/x = 0.5

log(0.1353 - x)/x = ½

(0.1353-x)/x = 10^½

(0.1353 - x)/x = 3.1623

0.1353 - x = 3.1623x

0.1353 = x + 3.1623x

0.1353 = 4.1623x

x = 0.1353/4.1623

x = 0.0325M

Therefore [HA] = x = 0.0325M

[A-] = 0.1353 - x = 0.1353 - 0.0325 =

[A-] = 0.1028M

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solmaris [256]

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Explanation:

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Learn more about Amantadine to visit this link

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