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Daniel [21]
3 years ago
7

If 14c-labeled uracil is added to the growth medium of cells, what macromolecules will be labeled?

Chemistry
2 answers:
deff fn [24]3 years ago
8 0

If 14c-labeled uracil is added to the growth medium of cells, what macromolecules will be labeled? RNA

<h3>Further explanation </h3>

14C-labeled compounds

Specifications:

  • Specific Activity: 50-60 mCi/mmol 1.85-2.22 GBq/mmol
  • Synonym: Uracil [6-14C]
  • Formula: Not available
  • Molecular Weight: 112.1
  • Solvent: Sterile water
  • Concentration: 0.1 mCi/ml
  • CAS Number: Not available
  • Shipped in dry ice: No
  • Exclusive: Yes
  • Reference: Not available

If 14c-labeled uracil is added to the growth medium of cells, the macromolecules will be labeled Ribonucleic acid (RNA).

Ribonucleic acid (RNA) is a polymeric molecule essential in various biological roles in coding, decoding, regulation and expression of genes. RNA (ribonucleic acid) is made up of four chemical bases: adenine (A), cytosine (C), guanine (G) and uracil (U) connected by a ribose backbone. RNA is a single-stranded molecule and it is not stable under alkaline conditions. RNA directly codes for amino acids and as acts as a messenger between DNA and ribosomes to make proteins.

When there are 14c-lable uracil that are added to the growth medium of cells, the macromolecules that will be labled are RNA. Uracil is a nucleobase that make up the DNA or the RNA. In RNA, uracil binds with other nucleobase (adenine) through hydrogen bonds.

<h3>Learn more</h3>
  1. Learn more about RNA brainly.com/question/9091941

<h3>Answer details</h3>

Grade:  9

Subject:  chemistry

Chapter:  RNA

Keywords:  RNA

ziro4ka [17]3 years ago
6 0
When there are 14c-lable uracil that are added to the growth medium of cells, the macromolecules that will be labled are RNA. Uracil is a nucleobase that make up the DNA or the RNA. In RNA, uracil binds with other nucleobase (adenine) through hydrogen bonds.
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Explanation:

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7 0
3 years ago
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The most common source of copper (cu) is the mineral chalcopyrite (cufes2). how many kilograms of chalcopyrite must be mined to
tigry1 [53]

Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

Solution : Given,

Mass of Cu = 300 g

Molar mass of Cu = 63.546 g/mole

Molar mass of CuFeS_2 = 183.511 g/mole

  • First we have to calculate the moles of Cu.

\text{ Moles of Cu}=\frac{\text{ Given mass of Cu}}{\text{ Molar mass of Cu}}= \frac{300g}{63.546g/mole}=4.7209moles

The moles of Cu = 4.7209 moles

From the given chemical formula, CuFeS_2 we conclude that the each mole of compound contain one mole of Cu.

So, The moles of Cu = Moles of CuFeS_2 = 4.4209 moles

  • Now we have to calculate the mass of CuFeS_2.

Mass of CuFeS_2 = Moles of CuFeS_2 × Molar mass of CuFeS_2 = 4.4209 moles × 183.511 g/mole = 866.337 g

Mass of CuFeS_2 = 866.337 g = 0.8663 Kg         (1 Kg = 1000 g)

Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.


3 0
3 years ago
What is the density of a liquid if it's volume is 125 mL and it's mass is 50g?
masha68 [24]

Answer:

0.4g/ml

Explanation:

density= mass/volume

density=50g/125ml

density=0.4g/ml

6 0
2 years ago
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Ivan
Distance and period of time is the correct answer

Hope this helps!
7 0
3 years ago
Read 2 more answers
Consider the reaction given below.
Drupady [299]

Answer:

  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

  • r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

6 0
3 years ago
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