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Irina18 [472]
3 years ago
14

What is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?

Physics
1 answer:
Art [367]3 years ago
3 0

CORRECT ANSWER:

a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.

STEP-BY-STEP EXPLANATION:

The complete question from book is

According to Figure 9.6, what is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?

a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.

b- Signaling molecules that bind to cell-surface receptors lead to cellular responses restricted to the cytoplasm; signaling molecules that bind to intracellular receptors lead to cellular responses restricted to the nucleus.

c- Cell-surface receptors bind to specific signaling molecules; intracellular receptors bind any signaling molecule.

d- Cell-surface receptors typically bind to signaling molecules that are smaller than those bound by intracellular receptors.

e- None of the other answer options is correct.

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Shtirlitz [24]

question one b

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3  d

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8 idk

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3 0
3 years ago
Scientists study how the continents move. Why might scientists use a model
tigry1 [53]

Answer:

B. It is too slow to observe directly

Explanation:

They move too slow to be able to observe how they move.


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2 years ago
Form conise note on heat energy specific heat application evaporation boiling sublimation relative humidity and dew point
weqwewe [10]
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8 0
3 years ago
A shopper does 157 J of work pushing a cart with 10.9 N force
Tanzania [10]

The cart travelled a distance of 14.4 m

Explanation:

The work done by a force when pushing an object is given by:

W=Fd cos \theta

where:

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

In this problem we have:

W = 157 J is the work done on the cart

F = 10.9 N is the magnitude of the force

\theta=0^{\circ}, assuming the force is applied parallel to the motion of the cart

Therefore we can solve for d to find the distance travelled by the cart:

d=\frac{W}{F cos \theta}=\frac{157}{(10.9)(cos 0)}=14.4 m

Learn more about work:

brainly.com/question/6763771  

brainly.com/question/6443626  

#LearnwithBrainly

4 0
2 years ago
A 60 N force is applied over a distance of 15 m. How much work was done?
GuDViN [60]

Answer:

900J

Explanation:

w =f×s

60×15

=900J

thus the k.e of the body is 900j

3 0
2 years ago
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