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Rudik [331]
3 years ago
8

Isoleucine, leucine, and lysine are three examples of amino acids that the body can not make in sufficient quantity. How are the

se compounds classified
Chemistry
2 answers:
Natalija [7]3 years ago
8 0

Answer:

Leucine, isoleucine and lysine are classified as <u>essential amino acids.</u>

Explanation:

Amino acids are the important biomolecules in which an alpha-carbon is attached to a side chain R, two functional groups and one hydrogen.

The two functional groups are: amine group (-NH₂) and carboxyl group (-COOH).

There are total <u>21 amino acids in a human body</u>.<em><u> They can be divided into two types: Essential amino acids and Nonessential amino acids</u></em>

Essential amino acids are the 9 amino acids that can not be produced by the human body and thus has to be taken through diet.

<u>Example: leucine, isoleucine, lysine.</u>

<u>Therefore, leucine, isoleucine and lysine are classified as essential amino acids.</u>

kotegsom [21]3 years ago
4 0

Answer:

Isoleucine, leucine and lysine are essential amino acids.

Explanation:

The amino acids are classified based on different criteria like

a) Polarity

b) acidc, basic or neutral behavior

c) aromatic or non aromatic

One important classification is that amino acids can be "essential" or "non essential".

Essential amino acids are those which cannot be made by our body and thus are required in our diet.

Non essential amino acids are those which our body can synthesize. So are not required in our diet.

So  Isoleucine, leucine and lysine are essential amino acids.

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balandron [24]

Answer:

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6 0
3 years ago
Read 2 more answers
Describe, in detail, to a freshman undergraduate how to make 1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /m
mario62 [17]

Answer:

Explanation:

The objective here is to prepare  1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /ml final concentration) liquid media.

Given that :

the Stock concentration of Amp: 100 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml since 100 mg/ml = 100000  μg/ml

However, using formula C₁V₁=C₂V₂ (Ampicilin),

where:

C₁ = 100000 μg/ml,

V₁=?,

C₂= 50  μg/ml,

V₂=1000 ml

100000 μg/ml × V₁ = 50  μg/ml × 1000 ml

V₁ =  50  μg/ml × 1000 ml/100000 μg/ml

V₁ =   0.5 ml

Given that:

the Stock concentration of Kan: 25 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml , 25 mg/ml = 25000 μg/ml

Now by using formula C₁V₁=C₂V₂ (Kanamycin),

C₁ = 25000 μg/ml,

V₁=?,

C₂= 100  μg/ml,

V₂=1000 ml

25000 μg/ml × V₁ = 100  μg/ml × 1000 ml

V₁ =  100  μg/ml × 1000 ml/25000 μg/ml

V₁ =   4 ml

Thus; in 1 lite of Lb+ Kan+Amp preparation;

0.5 ml of Amp & 4 ml of kanamycin is used for their stock preparation.

Finally;

Sterilization step should be carried out in flask (Clean dry glass wares) for media in an autoclave, the container size should be twice the volume of media which is prepared.

5 0
4 years ago
How to draw Hess' Cycle for this question ?
NISA [10]

Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)     \Delta H_1=-1411kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.7kJ/mole

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.9kJ/mole

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ/mole

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.7kJ/mole)=-787.4kJ/mole

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.9kJ/mole)=-571.8kJ/mole

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+1411kJ/mole)+(-787.4kJ/mole)+(-571.8kJ/mole)

\Delta H=51.8kJ/mole

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

7 0
3 years ago
How many moles of aluminum are produced from 4.9 moles of lithium?
Otrada [13]

Answer:

2.95078 × 10^23

hope this helps you ☺️☺️

5 0
3 years ago
How are the troposphere and the thermosphere alike
vazorg [7]
This layer also contains ratios of nitrogen and oxygen similar to the troposphere, except the concentrations are 1000 times less and there is little water vapor there, so the air is too thin for weather to occur. The thermosphere is the uppermost layer of the atmosphere.
4 0
3 years ago
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