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3 years ago

What is the temperature used for the extension step?

Physics

1 answer:

lisabon 2012 [21]3 years ago

8 0

Explanation:

The polymerase chain response (PCR) could be a strategy to quickly open up groupings of DNA.
Amid a normal PCR, format DNA (containing the locale of intrigued) is blended with deoxynucleotides (dNTPs), a DNA polymerase and small segments of DNA called primers.
PCR involves a series of temperature cycles.
The temperature used for the extension step is 72°C.
The temperature that is used during the extension phase is dependent on the DNA polymerase that is used.
Amid the expansion step (ordinarily 68°C-72°C) the polymerase amplifies the groundwork to create an early DNA strand. This prepare is rehashed different times
Generally at this stage, the reaction mixture is heated to a temperature intermediate between the denaturation and annealing temperatures. The optimal temperature for Taq polymerase is 72°C.

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A turbine blade rotates with angular velocity ω(t) = 2.00 rad/s- 2.1.00 rad/s3 t2. What is the angular acceleration of the blade

STatiana [176]

Answer:

$\alpha =-38.22\ rad/s^2$

Explanation:

given,

ω(t) = 2.00 rad/s- 2.1.00 rad/s³ t²

angular acceleration of the blade at t= 9.10 s

$\alpha =\dfrac{d\omega}{dt}$

$\alpha =\dfrac{d}{dt}(2 - 2.1 t^2)$

$\alpha =-2 \times 2.1 t$

$\alpha =-4.2 t$

angular acceleration of blade at t = 9.10

$\alpha =-4.2\times 9.10$

$\alpha =-38.22\ rad/s^2$

4 0

3 years ago

AP Physics I, shouldn't be too hard.

Nana76 [90]

Answer:

The correct option is;

D. The kinetic energy decreases by 3·m₀·v₀²

Explanation:

The given parameters are;

The mass of object X = m₀

The initial velocity of object X = v₀

The mass of object Y = 2·m₀

The initial velocity of object Y = -2·v₀

By conservation of linear momentum, we have;

The total initial momentum = The total final momentum

Therefore, we have;

The total initial momentum = m₀·v₀ - 2·m₀·2·v₀ = The total final momentum

∴ The total final momentum = -3·m₀·v₀

The total mass of the two object after sticking together = 2·m₀ + m₀ = 3·m₀

Therefore, the velocity of the two objects after collision = (The total final momentum)/(Total mass) = -3·m₀·v₀/(3·m₀) = -v₀

The kinetic energy = 1/2 × Mass × (Velocity)²

Therefore, the kinetic energy after collision = 1/2 × (3·m₀) × v₀² = 3·m₀·v₀²/2

The kinetic energy before collision = 1/2 × m₀ × v₀² + 1/2 × (2·m₀) × (2·v₀)² = (1/2 + 4) × (m₀·v₀²)

∴ The kinetic energy before collision = 9·(m₀·v₀²)/2

The change in kinetic energy = The kinetic energy after collision - The kinetic energy before collision = 3·m₀·v₀²/2 - 9·(m₀·v₀²)/2 = -3·m₀·v₀²

Therefore, the kinetic energy decreases by 3·m₀·v₀².

5 0

3 years ago

A box with mass m = 8 kg is pushed x = 10 m across a level floor by a constant applied force F P = 16.27 N. The coefficient of k

vitfil [10]

Answer:

Final speed of the box after it has moved to x = 10 is given as

v = 3.35 m/s

Explanation:

As we know by work energy theorem that work done by all the forces is equal to the change in its kinetic energy

Here work done by external force + work done by friction = change in kinetic energy of the box

so we have

$W_{ex} + W_{fric} = \frac{1}{2}mv^2$

$F x - \mu mg x = \frac{1]{2}mv^2$

$16.27 (10) - (0.15)(8)(9.8) (10) = \frac{1}{2}(8) v^2$

$162.7 - 117.6 = 4 v^2$

$v = 3.35 m/s$

4 0

3 years ago

The maximum number of electrons in the second energy level of an atom is ____.

ehidna [41]

Your answer will be 8.

6 0

3 years ago

Suppose the earth suddenly came to halt and ceased revolving around the sun. The gravitational force would then pull it directly

AleksAgata [21]

Answer:

613373.65233 m/s

Explanation:

M = Mass of Sun = $1.989\times 10^{30}\ kg$

m = Mass of Earth

v = Velocity of Earth

r = Distance between Earth and Sun = $147.12\times 10^{9}\ m$

$r_e$ = Radius of Earth = $6.371\times 10^6\ m$

$r_s$ = Radius of Sun = $695.51\times 10^6\ m$

In this system it is assumed that the potential and kinetic energies are conserved

$\dfrac{1}{2}Mv_2-\dfrac{GMm}{r_e+r_s}=0-\dfrac{GMm}{r}\\\Rightarrow v=\sqrt{2GM(\dfrac{1}{r_e+r_s}-\dfrac{1}{r})}\\\Rightarrow v=\sqrt{2\times 6.67\times 10^{-11}\times 1.989\times 10^{30}(\dfrac{1}{6.371\times 10^6+695.51\times 10^6}-\dfrac{1}{147.12\times 10^{9}})}\\\Rightarrow v=613373.65233\ m/s$

The velocity of Earth would be 613373.65233 m/s

3 0

3 years ago