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nexus9112 [7]
3 years ago
10

A client who has been on a high-protein diet comes to the emergency department with respiratory symptoms. Upon analysis of arter

ial blood gases (ABGs), the client is diagnosed with hypercapnia. Which ABG results confirm this diagnosis?
Biology
1 answer:
HACTEHA [7]3 years ago
7 0

Answer:

PCO2 of 51 mm Hg

Explanation:

ABG is a test that is use to measure or determine the level of carbon dioxide and some other gases from an artery.

Hypercapnia is a condition which is considered abnormal because it involves abnormal increase in carbon dioxide level in the blood. This poise a problem, because normally, carbon dioxide , which is a gaseous product of metabolism, needed to be expelled through breathing sothat the oxygen will be able to reach the lung.

Hypercapnia affects a number of body functions, including acid-base balance and renal, neurological, and CV functions. Elevated levels of PCO2 (38 to 42) produce a decrease in pH (7.35 to 7.45) and respiratory acidosis. Compensatory mechanisms result in an increase in serum HCO3 (22 to 28). In this example, the PO2 level is within normal range. Serum K+ is not part of the ABG analysis.

In summary, the ABG results given as answer confirm the diagnosis.

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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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