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olga_2 [115]
3 years ago
12

A Rac variant, in which the residue at position 61 was replaced with an alanine (Rac61A), was synthesized. Wild-type Rac and Rac

61A were incubated separately with VopC. To obtain data to support that VopC modifies Rac at residue 61, the samples should be analyzed for the presence of which compound?
Physics
1 answer:
ludmilkaskok [199]3 years ago
3 0

Answer:

Ammonia, (NH_{3})

Explanation:

If a Rac variant, in which the residue at position 61 was replaced with an alanine (Rac-61A), was synthesized. Also, Wild-type Rac and Rac-61A were incubated separately with VopC.

In order to obtain data to support that VopC modifies Rac at residue 61, the samples should be analyzed for the presence of Ammonia (NH_{3}[\tex]) compound.In chemistry, the Vanadyl Phthalocyanine (VopC) comprises of a catalytic domain that activates host cells GTPase Rac irreversibly, through the deamide side chain of the residue existing at position 61.Also, we know that a deamidation reaction gives off Ammonia, [tex](NH_{3}) and as such when ammonia is present in the sample containing Wild-type Rac but not in Rac-61A, this simply proves or provide the data to support that VopC modifies Rac at residue 61.

<em>Additionally, deamidation can be defined as the chemical conversion (hydrolysis) of an amide functional group such as glutamine, asparagine, in a polypeptide to another functional group such as glutamic acid or isoaspartic acid respectively by treating it with a strong acid (deamidate, transamidase). </em>

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Two protons in an atomic nucleus are typically separated by a distance of 2 ✕ 10-15 m. The electric repulsion force between the
castortr0y [4]

Answer:

The magnitude of the electric force between the to protons will be 57.536 N.

Explanation:

We can use Coulomb's law to find out the force, in scalar form, will be:

F \ = \ \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{d^2}.

Now, making the substitutions

d \ = \ 2.00 * 10 ^{-15} \ m,

q_1 = q_2 = 1.60 * 10 ^ {-19} \ C,

\frac{1}{4\pi\epsilon_0}=8.99 * 10^9 \frac{Nm^2}{C^2},

we can find:

F \ = \ 8.99 * 10^9 \frac{Nm^2}{C^2} \frac{(1.60 * 10 ^ {-19} \ C)^2}{(2.00 * 10 ^{-15} \ m)^2}.

F \ = 57.536 N.

Not so big for everyday life, but enormous for subatomic particles.

4 0
3 years ago
Read 2 more answers
The diagram shows two forces of equal magnitude acting on an object. If the common magnitude of the forces is 3.6 N and the angl
Nuetrik [128]
<h3>Answer</h3>

6.6 N pointing to the right

<h3>Explanation</h3>

Given that,

two forces acting of magnitude 3.6N

angle between them = 48°

To find,

the third force that will cause the object to be in equilibrium

<h3>1)</h3>

Find the vertical and horizontal components of the two forces

vertical force1 = sin(24)(3.6)

vertical force2= -sin(24)(3.6)

<em>(negative sign since it is acting on opposite direction)</em>

vertical force3 = sin(24)(3.6) - sin(24)(3.6)

                        = 0

<h3>2)</h3>

horizontal force1 = cos(24)(3.6)

horizontal force2= cos(24)(3.6)

horizontal force3 = cos(24)(3.6) + cos(24)(3.6)

                            = 2(cos(24)(3.6))

                            = 6.5775 N

                            ≈ 6.6 N

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4 0
3 years ago
An object weighing 49 N is dropped from a height of 30 m. It is found to be moving with a velocity of 24m/s just before it hits
faltersainse [42]

The  frictional force will be 0.22N.

<h3>What is Frictional force?</h3>

Frictional force is the force generated between two surfaces that are in contact and slide against each other.

Given,

Weight=4N

mass =4.9/9.8=0.5kg

Hieght =30m

velocity=24m/s

Acceration , v²-u²=2as

24²/2×30 =a , u is zero

 a= 1.5m/s²



By Using conservation  of energy ,

30F+1/2mv²=mgh

30F=1150-144

F= 6/30

F=0.2N

The force will be 0.2N

to learn more about Friction clickhttps://brainly.com/question/17608236

#SPJ9

6 0
2 years ago
The beginning of the Phanerozoic is marked by what occurrence
Sergio039 [100]
The beginning of the Phanerozoic is marked by the development of hard body parts, such as shells and bones.
3 0
3 years ago
9. A Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table. How long does it take to hit the floor if no one sneezes? Wh
Paha777 [63]

By Considering the vertical distance and both vertical and horizontal final velocity, the time t = 0.45 s and Velocity V = 6.7 m/s

Given that a Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table.

Height h = 1.0 m

As the ball rolls off the table, it will be fallen under gravity. Where

g = 9.8 m/s^{2}

Initial vertical velocity u_{y} = 0

Initial horizontal velocity u_{x} = 5 m/s

Considering the vertical distance, the formula to use to calculate the time will be;

h = ut + 1/2gt^{2}

1 = 0 + 1/2 x 9.8t^{2}

1 = 4.9t^{2}

t^{2} = 1/4.9

t = \sqrt{0.204}

t = 0.45 seconds

It takes 0.45 seconds to hit the floor if no one sneezes.

To calculate its velocity when it hits the floor, we will need to calculate for both vertical and horizontal final velocity and find the resultant velocity of the two.

Vertical component

V_{y} = U_{y} + gt

V_{y} = 0 + 9.8(0.45)

V_{y} = 4.41 m/s

Horizontal component

V_{x} = u_{x} + at

but a = 0

V_{x} = 5 m/s

Final velocity V = \sqrt{5^{2} + 4.41^{2}  }

V = 6.67 m/s

Therefore, it will hit the floor at a velocity of 6.7 m/s

Learn more here: brainly.com/question/5063616

8 0
3 years ago
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