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Pepsi [2]
3 years ago
6

Select the correct mechanism of stomatal opening and closing.A) The radial orientation of cellulose microfibrils in the cell wal

ls of guard cells causes them to bow outward and open the stomatal pore when turgid.B) The cell walls of guard cells are thickened on the side of the stomatal opening and the thinner walls bow outward when the guard cells become turgid, to close the stomata.C) Light stimulates proton pumps in the plasma membrane of the guard cells, causing them to lose K+ and become flaccid.
Biology
2 answers:
NNADVOKAT [17]3 years ago
5 0

Answer:

A) The radial orientation of cellulose microfibrils in the cell walls of guard cells causes them to bow outward and open when the stomatal pore when turgid.

Explanation:

Stomata are pores on the surface of leaves that work as valves controlling the gas exchange in plants; they are bordered by a pair of curved guarded cells in charge of opening and closing by increasing or decreasing the stomatal pore area.

Guard cells walls structure consist of a combination of stiff cellulose microfibrils oriented along the circumference of the cells and a softer matrix material which makes it possible for stomata pore to open by increasing water pressure (turgor).  

Option B states that when guard cells become turgid, they close the stomata. This is a false statement as we explained before.

Option C states that light (known stimulant for opening stomata) causes guard cells to become flaccid which is also false. Guard cells become turgid in order to open the stomatal pore.

Option A is correct.

Elenna [48]3 years ago
4 0

Answer:

A) The radial orientation of cellulose microfibrils in the cell walls of guard cells causes them to bow outward and open the stomatal pore when turgid.

Explanation:

The opening of the stomata happens when the cells become turgid. This is because the radial orientation of the cellulose microfibrils in the cell walls of the guard cells causes them to curve outward and open the stomatal pore when turgid.

On the other hand, stoma closure occurs when cells become flabby and withered. all of this, through the osmotic movement that determines the need for a cell to become wilted or turgid.

Through the process of opening and closing stomata, plants are able to control the entry and exit of gases in their organisms, without the need to lose a lot of water.

The stoma are formed by the guard cells and the stole.

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Answer:

a. 1/64 AATTDD

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d. 2/64 = 1/32 AATTDd or aattDd or AAttDd or aaTTDd

Explanation:

<u>Available data:</u>

  • Flower position: Axial (A) is dominant to Terminal (a)
  • Stem length: Tall (T) is dominant to dwarf (t)
  • Seed shape: Round (R) is dominant to wrinkled (r)

<u>Cross</u>:

Parental) AaTtDd   x   AaTtDd

Knowing that these four genes assort independently, then we can infer that the F1 will have the next genotypic proportions for each gene:

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       Aa =2/4

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  • DD = 1/4

        Dd = 2/4

        dd = 1/4

<em>This is the same as developing a Punnett square for each gene separately.</em>

Knowing this, to get each of the mentioned genotypes for each trait, we must multiply their respective genotypic proportions, like this:

a. homozygous for the three dominant traits, AATTDD

   AA1/4 x TT 1/4 x DD 1/4 = AATTDD 1/64

b. homozygous for the three recessive traits, aattdd

   aa 1/4 x tt 1/4 x dd 1/4 = aattdd 1/64

c. heterozygous for all three characters, AaTtDd

   Aa 2/4 x Tt 2/4 x Dd 2/4 = AaTtDd 8/16=1/8

d. homozygous for axial and tall, heterozygous for seed shape. There are four possibilities, all of them with the same result:

AATTDd --> 1/4  x 1/4 x 2/4 = 2/64=1/32

aattDd ---> 1/4 x 1/4 x 2/4 = 2/64=1/32

AAttDd ---> 1/4 x 1/4 x 2/4 = 2/64=1/32

aaTTDd ---> 1/4 x 1/4 x 2/4 = 2/64=1/32

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