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Arte-miy333 [17]
2 years ago
14

Chloride ions, Cl-, follow actively transported Na+ ions from the nephrons into the blood. Would you not expect the Cl- concentr

ation to decrease as fluids are extracted along the nephron?
Biology
1 answer:
DerKrebs [107]2 years ago
5 0

Answer:

Bicarbonate ion, HCO3- (which has a similar charge to chloride ions) also follow sodium ions into the blood. Also, potassium ions, K+ are transported into the nephron so some chloride ions and bicarbonate ions remains in the nephron to balance the charge.

Explanation:

Sodium is the primary positively charged electrolyte in extracellular fluid. Most of the solute reabsorbed in the proximal tubule is in the form of sodium bicarbonate and sodium chloride. Water is also reabsorbed in order to balance osmotic pressure

When sodium ions are reabsorbed into the blood, few of the substances that are transported with Na+ on the membrane facing the lumen of the tubules include Cl- ions, Ca2+ ions, amino acids, and glucose. Sodium is actively exchanged for K+ using ATP on the basal membrane.

In the distal convoluted tubule, K+ and H+ ions are selectively secreted into the filtrate, while Na+, Cl-, and HCO3- ions are reabsorbed to maintain pH and electrolyte balance in the blood.

Some chloride ions remains in the nephron to balance the charge of the secreted K+ ions and also due to the bicarbonate ions that are removed.

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What is the equation used to measure liquid pressure?​
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In a unique species of plants, flowers may be yellow, blue, red, or mauve. All colors may be true-breeding. If plants with blue
Zolol [24]

Answer:

F1) 100% bbRr, red flowered plants.

Explanation:

<u>Available data:</u>

  • flowers may be yellow, blue, red, or mauve
  • colors may be true-breeding
  • the cross of blue-flowered plants with red-flowered plants, produce plants that have yellow flowers
  • F2 generation: 9/16 yellow, 3/16 blue, 3/16 red, and 1/16 mauve.

Knowing that the phenotypic ratio is 9:3:3:1, we can assume that there are two genes involved in the flower color expression. We can name these genes B and R with a dominant and a recessive allele each (B, b and R, r respectively).

According to the first cross, we might establish the following genotypes:

1st Cross: blue-flowered plant     x     red-flowered plant

Parentals)         BBrr                     x                bbRR

Gametes)     Br, Br, Br, Br                         bR, bR, bR, bR

F1) 100% BbRr, yellow plants

Parentals)   BbRr     x     BbRr

Gametes) BR, Br, bR, br

                BR, Br, bR, br

Punnett square)     BR       Br        bR          br

                  BR     BBRR    BBRr   BbRR     BbRr

                  Br      BBRr     BBrr     BbRr      Bbrr

                  bR     BbRR    BbRr    bbRR      bbRr                    

                  br      BbRr     Bbrr      bbRr      bbrr

F2)  9/16 yellow ---> 1/16 BBRR + 2/16 BBRr + 4/16 BbRr + 2/16 BbRR  

       3/16 blue ------> 1/16 BBrr + 2/16 Bbrr

       3/16 red---------> 1/16 bbRR + 2/16 bbRr

       1/16 mauve ----> 1/16 bbrr    

So,

  • Yellow-flowered plants: BBRR, BBRr, BbRR, BbRr
  • Red-flowered plants: bbRR, bbRr
  • Blue-flowered plants: BBrr, Bbrr
  • Mauve-flowered plants: bbrr

According to these genotypes, the second cross would be like following,

2nd Cross: true-breeding red-flowered plants with true-breeding mauve flowered plants.                

Parentals)          bbRR       x        bbrr

Phenotype)       Red                   Mauve

Gametes)   bR, bR, bR, bR      br, br, br, br

Punnett square)    bR       bR        bR        bR                    

                    br    bbRr     bbRr    bbRr     bbRr

                    br    bbRr     bbRr    bbRr     bbRr

                    br    bbRr     bbRr    bbRr     bbRr

                    br    bbRr     bbRr    bbRr     bbRr

F1) 100% bbRr, red flowered plants.

6 0
3 years ago
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