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3 years ago

which of the following amino acids are used to measure the protein concentration by UV? A. His B. Pro C. Trp D. Phe E. Tyr

Chemistry

1 answer:

Goryan [66]3 years ago

7 0

Answer:

C. Trp D. Phe E. Tyr

Explanation:

The concentration of a protein has a direct relation with absorbance of the protein in a UV spectrophotometer. The formula which relates concentration with absorbance is described as under:

A = ∈ x c x l

where, A = Absorbance

∈ = Molar extinction co-efficient

c = Concentration of absorbing species i.e. protein

l = Path length of light

Tryptophan (Trp), phenylalanine (Phe ) and tyrosine (Tyr) are three aromatic amino acids which are used to measure protein concentration by UV. It is mainly because of tryptophan (Trp), protein absorbs at 280 nm which gives us an idea of protein concentration during UV spectroscopy.

The table depicting the wavelength at which these amino acids absorb and their respective molar extinction coefficient is as under:

Amino acid Wavelength Molar extinction co-efficient (∈)

Tryptophan 282 nm 5690

Tyrosine 274 nm 1280

Phenylalanine 257 nm 570

In view of table above, we can easily see that Molar extinction co-efficient (∈) of Tryptophan is highest amongst all these 3 amino acids that is why it dominates while measuring concentration.

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3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0

4 years ago

An unknown compound displays singlets at δ 2.1 ppm and 2.56 ppm in the ratio of 3:2. what is the structure of the compound?

olasank [31]

1) As can be seen from any 1H NMR chemical shift ppm tables, hydrogens which have δ values from 2ppm to 2.3ppm are hydrogens from carbon which is bonded to a carbonyl group. From this, we can conclude that our hydrogens belong to the type, but from 2 different alkyl groups because of 2 different signals.

2) So, one alkyl group is CH3 and second one can be CH or CH2.

3) If we know that ratio between two types of hydrogens is 3:2, it can be concluded that second alkyl group is CH2.

4) Finally, we don't have any other signals and it indicates that part of the compound which continues on CH2 is exactly the same as the first part.

The ratio remains the same, 3:2 ie 6:4

7 0

3 years ago

Which part of the crust is the thickest

Sergio039 [100]

Answer:

I'll take a gander at your question

Explanation:

If you mean the Earth's crust, your answer is The Mantle.

5 0

3 years ago

The two boron atoms listed in the table are isotopes of the boron. The two carbon atoms and the two oxygen atoms are also called

Ivan

Answer:

A type of an atom which has a different number of neutrons but the same atomic number, therefore making it the same element. This atom would still have the same properties as well. (Ex: Vanadium-51 is an isotope of Vanadium that has 51 neutrons but still has 23 protons, as its atomic number is 23.)

3 0

2 years ago

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Trava [24]

Answer:

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4 0

3 years ago

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