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3 years ago

The ideal temperature at which to study the effect of proteases on protein is 37°C. Why are higher temperatures not advisable?

Physics

2 answers:

elena-14-01-66 [18.8K]3 years ago

7 0

Answer:

The higher temperatures are not advisable because

A) This could contribute to irreversible denaturing of protein.

D) Only at the specified temperature, the protein interacts with the proteases.

Explanation:

The higher temperatures are not advisable because

A) This could contribute to irreversible denaturing of protein.

D) Only at the specified temperature, the protein interacts with the proteases.

1. The protease at a temperature higher than 37 degree is not active

2) The protein you are researching to be destroyed by protease may be denatured and the protease may not function because the protein(s) that it targets are messed up by the structure.

loris [4]3 years ago

3 0

Answer:a

Explanation:

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2 years ago

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Answer:

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3 years ago

A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.

Nesterboy [21]

Answer:

The velocity after the collision is 2.82 m/s

Explanation:

Law Of Conservation Of Linear Momentum

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2$

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

The common velocity after this situation is:

$\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

$\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}$

$\displaystyle v'=\frac{22.287}{7.91}$

$\boxed{v' = 2.82\ m/s}$

The velocity after the collision is 2.82 m/s

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2 years ago

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2 years ago

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2 years ago