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sveticcg [70]
3 years ago
5

Mr. jones is admitted to your unit from the emergency department with a diagnosis of hypokalemia. his laboratory results show a

serum potassium of 3.2 meq/l (3.2 mmol/l). for what manifestations will you be alert?
Biology
1 answer:
BaLLatris [955]3 years ago
8 0
The answer is muscle weakness, fatigue and dysrhythmias. The distinctive indication of hypokalemia contains muscle faintness, leg spasms, fatigue, paresthesia and dysrhythmias. Indicators of hypercalcemia contain nausea, vomiting, constipation, bone pain, too much urination, dehydration, misperception, weariness and indistinct speech. Reduced cognitive capability and hypertension may outcome from hyperchloremia in which constipation is a indication of hypercalcemia.
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You are researching a cytoplasmic protein associated with a nerve disorder. The native form of the enzyme appears to be globular
kherson [118]

Answer:

This signifies that the protein primarily comprises multiple polypeptide chains connected together with the help of disulfide bonds. The enzymes may be found in the form of dimers, trimers, or tetramers. Various examples of dimers, trimers, and tetramer proteins are known, of them, NEMOs dimers are considered to be held by disulfide bonds.  

Thus, it can be hypothesized that the enzyme under examination is a multimer held in combination by disulfide bonds, with each comprising catalytic sites. On breaking of disulfide bonds, the enzyme dissociates into its many single units.  

This illustrates the reduction in catalytic activity. Each active site in a single unit will work, however, at a gradual rate. This also shows detection of multiple globular proteins after disulfide reduction.  

8 0
3 years ago
An owl has good night vision because its eyes can detect a light intensity as small as 5.0 × 10-13 W/m2. What is the minimum num
Lady bird [3.3K]

Answer:

Thus, the minimum number of photons per second is 77.34

Explanation:

Light intensity, I_{min} =  5\times 10^{-13} W/m^{2}

Pupil has a diameter, d = 8.5 mm

                                      = 8.5 x 10^{-3} m

Radius of the eye, r = 4.25 x 10^{-3} m

∴ Area of the eye, A = \pi .r^{2}

                                 = 3.14\times \left ( 4.25\times 10^{-3} \right )^{2}

                                = 5.6\times 10^{-5} m^{2}

Let P_{min} be the minimum number of photons.

Therefore, P_{min} = I_{min} x A

                                              = 5\times 10^{-13} x 5.6\times 10^{-5}

                                             = 2.8\times 10^{-17} W

Thus the minimum number of photons is given by

N_{min}=P_{min}/E

where E = hc/\lambda

             = \left (6.63\times 10^{-34}\times 3\times 10^{8}  \right )/548\times 10^{-9}

            = 3.62\times 10^{-19} J

Therefore, N_{min} = \frac{2.8\times 10^{-17}}{3.62\times 10^{-19}}

                                              = 77.34 photons per second

Thus, the minimum number of photons per second is 77.34

4 0
3 years ago
In 2003 congress passed a prescrition drug benefit as part of which existing health care law
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What occurs during the G1 and G2 in the cell cycle?
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Answer:

An important cell cycle control mechanism activated during this period (G1 Checkpoint) ensures that everything is ready for DNA synthesis. ... DNA replication occurs during this S (synthesis) phase. Gap 2 (G2): During the gap between DNA synthesis and mitosis, the cell will continue to grow and produce new proteins.

Explanation:

hope this helps you if this is what you needed  plz mark Brainliest

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The correct answer to your question is option B 
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