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marusya05 [52]
1 year ago
6

The majority of water molecules moving across plasma membranes by osmosis do so via a process that is most similar to ____. hint

s the majority of water molecules moving across plasma membranes by osmosis do so via a process that is most similar to ____. active transport a process that requires energy from the cell facilitated diffusion cotransport simple diffusion
Biology
2 answers:
Brilliant_brown [7]1 year ago
8 0

Answer:

Simple diffusion

Explanation:

Water  moves from one side to the other in a membrane by osmosis, because it is a small particle, so doesn´t need a carrier to pass trough the membrane, and also, the water moves in favor of the gradient, that means, it moves from one side to another to equilibrate it concentration.

Simple diffusion is a similar process, the difference is that this is used by bigger molecules but not by water.

Hope this info is useful .

goldfiish [28.3K]1 year ago
6 0

The majority of water molecules moving across plasma membranes by osmosis do so via a process that is most similar to simple diffusion. The process of osmosis involves the moment of water molecules across a semipermeable membrane like the cell membrane from a region of high water concentration to a region of low water concentration. This process is similar to the process of simple diffusion where the solute particles move across in a similar fashion.

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In humans, there are four blood types: A, B, AB and O. Each parent can give their offspring one of three blood type alleles, IA
Delvig [45]

Answer: One parent is IAi and the other parent is IBi

Explanation: As this trait is codominant, the child can inherited IA, IB or i.

Tina has type O, which means she is ii and her sister is AB, so her genotype is IAIB. Now, to have a child with a recessive trait both parents has to carry at least one allele for the recessive, i. Rosa is type AB which means she had to have inherited one allele IA from one of her parents and the other IB from the other parent. Thus, one parent is IAi and the other is IBi.

5 0
2 years ago
Black mice and tan mice are native species that are observed in a particular region. Human habitation in the region caused an in
4vir4ik [10]

Answer:

A

Explanation:

The correct answer would be that <u>the availability of food resources for black mice and brown mice will decrease.</u>

<em>Since the food requirements of the black mice are the same as that of the invasive brown mice, the available food supply that used to be only for the black mice would now be shared by the two strains of mice. Hence, the available food for the two groups of mice will naturally decrease.</em>

There is no sufficient information to conclude that the population of tan mice will decrease, hence, option B is incorrect.

The black mice and tan mice have different food requirements going by the information available in the illustration, hence, both cannot compete for food resources. Option C is, therefore, incorrect. In the same vein, option D is incorrect because the tan mice have different food requirements from the brown mice.

<u>The only correct option is A.</u>

3 0
1 year 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
2 years ago
Where is the greatest volume of blood found in the body?
bagirrra123 [75]
The greatest volume of blood is found in veins
5 0
2 years ago
Pseudomonas sp. has a mass doubling time of 2.4 h when grown on acetate. The saturation constant using this substrate is 1.3 g/l
erastova [34]

Answer:

a)  Cell concentration when the dilution rate is one-half of the maximum is  0.598g cell/L

b) the substrate concentration when the dilution rate is 0.8 D_{max}   is 5.2g/l

c) the maximum dilution rate is : 0.41 h⁻¹

d)  the cell productivity at 0.8  D_{max}   is 2.40g cell/L

Explanation:

Given data :

Mass doubling time of Pseudomonas sp. = 2.4 hr

Saturation constant = 1.3 g/L

Cell yield  on acetate = 0.46g cell/g acetate

We are to find;

a. Cell concentration when the dilution rate is one-half of the maximum.

Here, cell yield =amount of cell produced / amount of substrate consumed.

[S] at 0.5D max is determined using the Monod's equation.

Using the formula :

D = \frac {D_{max}[S] }{ks+[S]}

, where D is the dilution rate,

[S] is substrate concentration; &

Ks is the saturation constant.

By replacing the values, we get :

0.5 = \frac{S}{1.3+[S]}

\\\0.65=0.5[S]

[S]=1.3g/L

The cell concentration at 0.5Dmax= cell yield x substrate consumed at 0.5Dmax.

=0.46×1.3

= 0.598g cell/L

b)

Substrate concentration when the dilution rate is 0.8 D_{max}    is calculated as:

D = \frac {D_{max}[S] }{ks+[S]}

0.8=[S]/1.3+[S]

1.04+0.8[S]=[S]

[S]= 5.2g/L

Therefore ,  the substrate concentration when the dilution rate is 0.8 D_{max}   is 5.2g/l

c)

Maximum dilution rate is calculated using the expression D_{max} = \frac{1}{time}

=1/2.4

=0.41 h⁻¹

So, the maximum dilution rate is : 0.41 h⁻¹

d)

The cell productivity at 0.8 D_{max} can be calculated by multiplying the amount  of the cell yield with the amount of the substrate consumed at 0.8D_{max}  

Cell yield = \frac {cell \ productivity \ at \  0.8Dmax}{amount \ of \ substrate\ consumed \at\ 0.8 \D_{max}}

Cell productivity at 0.8 D_{max}    = 0.46 × 5.2

=2.40g cell/L

Therefore, the cell productivity at 0.8  D_{max}   is 2.40g cell/L

5 0
2 years ago
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