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neonofarm [45]
3 years ago
11

Which of the following reactions takes place in red blood cells in the capillaries surrounding the cells of active tissues?

Biology
1 answer:
Irina-Kira [14]3 years ago
5 0

Answer:B

Explanation:

This reaction demonstrated Bohr effect. it takes place in the  cytoplasm  of RBC , because this is the location of the enzyme carbonic anhydrase which catalyzed it .  

The hydrogen ion and bicarbonate  are transferred out  of the  RBC into plasma. Chloride ion diffused into the RBC to neutralize drop in pH(chloride shift).

 The hydrogen ion is mobbed by haemoglobin in its buffer effect, while the carbonate is transferred to the lungs  where it is broken down, to release C02 .85% of blood C02 is transported in this form .

The CO2 released out of  the lungs , created a concentration gradient for  influx of Oxygen into the lungs. This is the significance of this reaction.

In addition, Option D which involved direct reaction of C02 with Haemoglobin to form carbaminoheamoglobin  is another reaction which takes place in the RBC. it is a reversible reaction where the C02 is transported to the lungs.10% C02 is tranported in this form,

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I think it was Thomas Hunt Morgan

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The number of individuals that a particular habitat can support with no degradation of that habitat is called _____.
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carrying capacity (carrying capacity is the total population size that a particular environment can comfortably support with little increase or decrease over a relatively long period of time) given the food, habitat, water e.t.c. that is available in the environment.

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Velocity calculation
svetoff [14.1K]
Answer choice: [C]:
_________________________________________

V = d / t  = 1800 mi / 40 min = 180 mi / 40 min = 45 mi / min .

Which very well could be "Answer choice: [C]:  45 m/min" ; except, this answer choice may have been written incorrectly—and should read:
                                                 "45 m/min."
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Now, notice that all the answer choices given are in "meters" per (unit time—seconds or mins.

So, we need to convert:

45 mi.  to "meters", or "m".
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Note the following conversions:
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 1 mi. = 1750 yd
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3 ft. = 1 yd.
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1 m = 3.2 ft.
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So, 1 mi. = ? m ;
_____________________
 1 mi * ( 1760 yd/ mi) * (3 ft / 1 yd) * ( 1m / 3.2 ft) = _?_  m ?
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On the "left hand side" , the units of  "mi", "yd", and "ft" ; cancel to "1" ;
  ____________________________________________________
    and we are left with:
______________________________________
[(1760 * 3)]  /  [(3)*(3.2)]   m  = (5280 / 9.6) m  =  550 m .
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So, we have:  550 m / min ; which match NONE of the answer choices.

However, it is possible that choices A or B, which are given in "m/sec" (meters per second); are among the correct choices.
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So, we convert:   550 m / min to:   "m / sec " ;
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( 550 m/ min)*(1 min/ 60 sec) =  (550/60) m/sec = (55/6) m/ sec
                                                                                    = 9 1/6 m / sec .
_____________________________________________________
                                                                 which is:  None of the above.
______________________________________________________
                     Unless the answer choices were written incorrectly.




4 0
3 years ago
Which of these is the system of classifying living things into categories, based on physical and chemical features?
scoundrel [369]
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4 0
4 years ago
1) A class collected water samples from a nearby pond and wants to observe the samples for
spayn [35]

Answer:

Compound light microscope

Explanation:

This is the ideal microscope because it requires the passage of  harmless beam of light through the specimen.Since the pond contain living creatures this is best to use.

C and D both require the transmission of beam of electrons through a sample. D depends on the reflection of the electron  after scanning the specimen surfaces and sent this to the lenses of the observer.(C)involved the  passing of the beam of electrons through the specimen the interaction of the electrons with the atoms in the specimen(,therefore in both cases  these specimens can not be wet and living.

The specimen should be dry in both cases and therefore non -living ,because the chamber for the specimen must be a vacuum,and living organism can not survive in the vacuum. Thus pond water can not be used as specimen because the specimen in the water but me be living.

4 0
3 years ago
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