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alexdok [17]
2 years ago
6

Osmosis causes water to move from areas of low salt concentration to areas of high salt concentration. The internal NaCl concent

ration of red blood cells is 0.9 M. The diagram below shows red blood cells submerged in solutions of four different NaCl concentrations—0 M, 0.65 M, 0.9 M, and 2 M. Which of the diagrams shows red blood cells in the 2 M solution?

Biology
2 answers:
WITCHER [35]2 years ago
4 0
<span>Red blood cells normally have a biconcave shape (picture a donut, but instead of a hole through the centre there's just a downward slope to the centre of the cell). If the cell takes on water, it will increase in size and the sloped areas will start to flatten out. If the cell loses water, it will begin to shrivel. Use these observations and your understanding of osmosis to answer the question!</span>
Nezavi [6.7K]2 years ago
3 0
You haven't given us the diagrams, but in case you're asking about the shape, imagine this: a disc, with a sunken centre, starting to resemble a donut.

In a low salt solution, the cells would start to absorb water, thus, inflating, up to the point of resembling spheres, and even bursting in the end. In case of a salty solution like the one you proposed, blood cells would lose water and would become a bit amorphous. In the attached pic you can see a standard blood cell (on the left), and a cell submerged in a salty solution (on the right).

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11) In C3 photosynthesis, the reactions that require ATP take place in A) the light reactions alone. B) the Calvin cycle alone.
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Answer:

In C3 photosynthesis, the reactions that require ATP take place in

B) the Calvin cycle alone.

Explanation:

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Alkaptonuria is an infrequent autosomal recessive condition. It is first noticed in newborns when the urine in their diapers tur
Arisa [49]

Answer:

See the answer below

Explanation:

Let the disorder be represented by the allele a.

Since the disease is an autosomal recessive one, affected individuals will have the genotype aa and normal individuals will have the genotype Aa or AA.

Since the four adults are carriers, their genotypes would be Aa.

                    Aa     x     Aa

Progeny:    AA    2Aa    aa

Probability of being affected = 1/4

Probability of being a carrier = 1/2

Probability of not being affected = 3/4

(a) The chance that the child second child of Mary and Frank will have alkaptonuria = 1/2

(b) The chance that  the third child of Sara and James will be free of the condition = 3/4

(c)

(d) If someone has no family history of the disorder, their genotype would be AA.

                 AA     x     aa

                        4 Aa

<em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history </em>= 0

(e)

(f) <em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history</em> = 0

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Explanation:

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