Question

Select the diagram that represents the shape of a red blood cell when placed in each of the following a to $e:(9.6)$

a. $0.9\%(\mathrm{m}/\mathrm{v})\mathrm{NaCl}$solution

b. localid="1654008537048" $10\%(m/v)$glucose solution

c. localid="1654008233523" $0.01\%(m/v)NaCl$solution

d. localid="1654008237506" $5\%(m/v)$glucose solution

e. localid="1654008241312" $1\%(m/v)$glucose solution

Short answer

Part a) As a result, a red blood cell will not alter.

Part b) A red blood cell will crenate as a result.

Part c) As a reaction, hemolysis occurs in a red blood cell.

Part d) As a corollary, a red blood cell will remain unchanged.

Part e) As an outcome, hemolysis occurs in a red blood cell.

Step-by-step solution

Introduction.(Part a)

Depending on the osmotic pressure of the medium in which it is placed, a red blood cell will experience crenation, hemolysis, or no change.

Because there is an equal flow of water in and out of a red blood cell in an isotonic solution, the cell preserves its volume. A solution and a $5\%(m/v)$glucose solution serve as isotonic solutions.

Given data (Part a).

Water rushes into red blood cells when they are placed in a solution, causing them to inflate and burst. Hemolysis is the name for this process. The amount of solute in a hypotonic solution is lower.

Water leaves a red blood cell when it is added to the solution, leading it to shrink. Crenation is the term for this procedure. The amount of solute in a hypertonic solution is higher.

Explanation (Part a). 

The form of a red blood cell in isotonic, hypotonic, and hypertonic solutions is depicted in the diagram below:

Isotonicity is defined as a $0.9\%(m/v)NaCl$solution. As a result, a red blood cell will remain unchanged. As a result, the diagram depicting the form of a red blood cell in a $0.9\%(m/v)NaCl$solution is

Given data (Part b).

(b) In comparison to a $5\%(m/v)$glucose solution, a $10\%(m/v)$glucose solution has a larger solute concentration. As a result, a $10\%(m/v)$glucose solution is hypertonic. A red blood cell will crenate as a result.

Explanation (Part b). 

As a result, the diagram depicting the form of a red blood cell in a$10\%(m/v)$glucose solution is.

Given data (Part c).

(c) When contrasted to a $0.9\%(m/v)NaCl$solution, a $0.01\%(m/v)NaCl$answer has a lower osmotic pressure. As a result, a $NaCl$solution of $0.01\%(m/v)$is hypotonic. As a result, hemolysis occurs in a red blood cell.

Explanation (Part c).

Therefore, the diagram that represents the shape of a red blood cell when placed in a $0.01\%(m/v)NaCl$solution is

Given data (Part d).

(d) A glucose solution containing $5\%(m/v)$glucose is isotonic. As a result, a red blood cell will stay unchanged.

Explanation (Part d).

As a corollary, the image showing the size of a red blood cell in a $5\%(m/v)$glucose solution is as follows:

Given data (part e).

(e) When opposed to a $5\%(m/v)$glucose solution, a $1\%(m/v)$sucrose solution has a lower osmotic pressure. A $1\%(m/v)$glucose solution is therefore hypotonic. As a result, hemolysis develops in a red blood cell.

Explanation (Part e).

As a result, the diagram illustrating the form of a red blood cell in a $1\%(m/v)$glucose solution is.