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2 years ago

Which reaction connects the citric acid cycle to the electron‑transport chain and amino acid synthesis?

1 answer:

7 0

Conversion of succinylcoA to succinate is the reaction that link acid cycle to the electron‑transport chain and amino acid synthesis.

<h3>Conversion of succinylcoA to succinate</h3>

SuccinylCoA is converted into succinate through the hydrolytic release of coenzyme A by succinyl-CoA synthetase Also succinyl-CoA is porphyrin synthesis, where it combine with glycine by ALA synthase to form δ-aminolevulinic acid.

Succinyl CoA is catalytically converted to succinate. The enzyme stimulates the hydrolysis of succinyl CoA into succinate and ATP and this represents a remarkable point of substrate level phosphorylation. After this, it link up to the electron transport chain via citric acid cycle.

Therefore, Conversion of succinylcoA to succinate is the reaction that link citric acid cycle to the electron‑transport chain and amino acid synthesis.

Learn more about Citric acid cycle conversion here.

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Define: atomic mass unit(amu)

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Answer:

The atomic mass is the mass of an atom. It is approximately equivalent to the number of protons and neutrons in the atom

Explanation:

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3 years ago

The term wavelength refers to which of the following?

svlad2 [7]

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3 years ago

Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

alukav5142 [94]

Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

For aluminium:

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

For copper:

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

Aluminium wire:

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

Copper wire:

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

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3 years ago

A scientist designed a foam container to help keep frozen foods from melting. Which best explains how the foam works?

FrozenT [24]

I think the answer should be D.It reduces the amount of thermal energy that is transferred from outside to inside the container.

6 0

4 years ago

Read 2 more answers

An 888.0 kg elevator is moving downward with a velocity of 0.800 m/s. It decelerates uniformly and comes to a stop in a distance

bagirrra123 [75]

Answer:

The value of tension on the cable T = 1065.6 N

Explanation:

Mass = 888 kg

Initial velocity ( u )= 0.8 $\frac{m}{sec}$

Final velocity ( V ) = 0

Distance traveled before come to rest = 0.2667 m

Now use third law of motion $V^{2}$ = $u^{2}$ - 2 a s

Put all the values in above formula we get,

⇒ 0 = $0.8^{2}$ - 2 × a ×0.2667

⇒ a = 1.2 $\frac{m}{sec^{2} }$

This is the deceleration of the box.

Tension in the cable is given by T = F = m × a

Put all the values in above formula we get,

T = 888 × 1.2

T = 1065.6 N

This is the value of tension on the cable.

5 0

3 years ago