All categories

4 years ago

How does the nuclear pore restrict the passage of large molecules that do not bear the correct nuclear localization signal?

Physics

1 answer:

frez [133]4 years ago

7 0

Answer:

The pores remain closed until they are stimulated by the binding of protein with the proper localization signal.

Explanation:

The pores are constructed from a class of proteins called “nucleoporins,” a subset of which contains a tandem series of phenylalanine-glycine (FG) repeats that line the central transport channel of the pore . The inbound proteins are captured by the nuclear basket and released by GTP hydrolysis. The cytosolic fibrils obstruct access to the pore and can be parted by nuclear import receptors. Thus, the pores remain closed until they are stimulated by the binding of protein with the proper localization signal.

Allen, T. D., Cronshaw, J. M., Bagley, S., Kiseleva, E., and Goldberg, M. W.(2000) J. Cell Sci. 113, 1651–165

You might be interested in

Recall that the spring constant is inversely proportional to the number of coils in the spring, or that shorter springs equate t

ruslelena [56]

Answer:

$x_1= 0.0425m$

Explanation:

Using the tension in the spring and the force of the tension can by describe by

T = kx

, T = mg

Therefore:

$m*g = k*x$

With two springs, let, T1 be the tension in each spring, x1 be the extension of each spring. The spring constant of each spring is 2k so:

$T_1 = 2k*x_1$

$2T_1 = m*g=4k x_1$

Solve to x1

$x_1=\frac{m*g}{4k}$

$x_1=\frac{k*x}{4*k}$

$x_1=\frac{x}{4}$

$x_1 = 0.170 / 4$

$x_1= 0.0425m$

7 0

3 years ago

_______ are nonmetals that react with metals to form salts.

kari74 [83]

i’m pretty sure it’s , the alkali metals

5 0

3 years ago

A flywheel of mass 182 kg has an effective radius of 0.62 m (assume the mass is concentrated along a circumference located at th

musickatia [10]

Answer:

A)5524J,

B) 29.2Nm

Explanation:

This question can be treated using work- energy theorem

Work= change in Kinectic energy

W= Δ KE

Work= difference between the final Kinectic energy and intial Kinectic energy.

We know that

Kinectic energy= 1/2 mv^2 .............eqn(1)

This can be written in term of angular velocity, as

KE= 1/2 I

4 0

3 years ago

A 24-cm-diameter vertical cylinder is sealed at the top by a frictionless 15 kg piston. The piston is 90 cm above the bottom whe

abruzzese [7]

Answer:

(A) $P_i=3249.41\ Pa$

(B) $V_f=0.3358\ m$

Explanation:

Given:

diameter of the cylinder, $d= 0.24\ m$

mass of piston sealing on the top, $m_p=15\ kg$

initial temperature of the piston, $T_i=315+273= 588\ K$

initial height of piston, $h_i=0.9\ m$

atmospheric pressure on the piston, $p_a=1 atm=101325\ Pa$

(A)

Initial pressure of gas is the pressure balanced by the weight of piston:

$P_i=\frac{m_p.g}{\pi.d^2\div 4}$

$P_i=\frac{15\times 9.8}{\pi\times (0.24^2\div 4)}$

$P_i=3249.41\ Pa$

Which is gauge pressure because it is measured with respect to the atmospheric pressure.

(B)

Given:

Final temperature, $T_f=18+273=291\ K$

Now, volume of air initially in the cylinder:

$V_i=\pi.d.h_i$

$V_i=\pi\times 0.24\times 0.9$

$V_i=0.6786\ m^3$

Using gas law:

$\frac{P_i V_i}{T_i}= \frac{P_f V_f}{T_f}$ ........................................(1)

∵In every condition of equilibrium the gas pressure will be balanced by the weight of the piston so it is an isobaric transition.

∴ $P_i=P_f$

Hence eq. (1) is reduced to:

$\frac{V_i}{T_i}= \frac{V_f}{T_f}$

putting respective values:

$\frac{0.6786}{588}= \frac{V_f}{291}$

$V_f=0.3358\ m$

6 0

4 years ago

Two 22.7 kg ice sleds initially at rest, are placed a short distance apart, one directly behind the other, as shown in Fig. 1. A

boyakko [2]

Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced

(a) The final speeds of the ice sleds is approximately 0.49 m/s each

(b) The impulse on the cat is 11.0715 kg·m/s

The reason for arriving at the above values is as follows:

The given parameters are;

The masses of the two ice sleds, m₁ = m₂ = 22.7 kg

The initial speed of the ice, v₁ = v₂ = 0

The mass of the cat, m₃ = 3.63 kg

The initial speed of the cat, v₃ = 0

The horizontal speed of the cat, v₃ = 3.05 m/s

(a) The required parameter:

The final speed of the two sleds

For the first jump to the right, we have;

By the law of conservation of momentum

Initial momentum = Final momentum

∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'

Where;

v₁' = The final velocity of the ice sled on the left

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05

∴ 22.7 × v₁' = -3.63 × 3.05

v₁' = -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

For the second jump to the left, we have;

By conservation of momentum law, m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'

Where;

v₂' = The final velocity of the ice sled on the right

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05

∴ 22.7 × v₂' = -3.63 × 3.05

v₂' = -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

(b) The required parameter;

The impulse of the force

The impulse on the cat = Mass of the cat × Change in velocity

The change in velocity, Δv = Initial velocity - Final velocity

Where;

The initial velocity = The velocity of the cat before it lands = 3.05 m/s

The final velocity = The velocity of the cat after coming to rest =

∴ Δv = 3.05 m/s - 0 = 3.05 m/s

The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s

(c) The required information

The average velocity

Impulse = $F_{average}$ × Δt

Where;

Δt = The time of collision = The time it takes the cat to finish landing = 12 ms

12 ms = 12/1000 s = 0.012 s

We get;

$F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}$

∴ $F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s} = 922.625 \ kg\cdot m/s^2 = 922.625 \ N$

The average force on the right sled applied by the cat while landing, $\mathbf{F_{average}}$ = 922.625 N

Learn more about conservation of momentum here:

brainly.com/question/7538238

brainly.com/question/20568685

brainly.com/question/22257327

8 0

3 years ago