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shutvik [7]
3 years ago
6

For each of the following features of nucleic acids, indicate whether it is true of DNA only, of RNA only, of both DNA and RNA,

or of neither. Sort each feature into the appropriate bin.
a. Is involved in the process of protein synthesis in the cytoplasm of the cells in the liver.
b. Contains the base adenine.
c. Contains the nucleotide deoxythymidine monophosphate.
d. Is an inherently directional molecule, with an NN-terminus on one end and a CC-terminus on the other end.
e. Occurs only in single-stranded form.
f. Is synthesized by a process that involves base pairing.


True of DNA only
True of RNA only
Chemistry
1 answer:
11111nata11111 [884]3 years ago
5 0

True for DNA only:

  • Contains the nucleotide deoxythymidine monophosphate.

True for RNA only:

  • Is involved in the process of protein synthesis in the cytoplasm of the cells in the liver.

True for both DNA and RNA:

  • Contains the base adenine.
  • Is synthesized by a process that involves base pairing.

True neither for DNA nor RNA:

  • Is an inherently directional molecule, with an NN-terminus on one end and a CC-terminus on the other end.
  • Occurs only in single-stranded form.

Some facts:

DNA known as Deoxyribonucleic acid acts as the genetic material of all living organism except for some viruses. It stores the genetic information of an organic life form. DNA has deoxyribose sugar and is a double stranded helix.

RNA or Ribonucleic acid  can be genetic or non genetic material. It functions in transferring the genetic information stored in DNA during protein synthesis and in control of gene expression.RNA also have enzyme activity in the form ribozyme and it is the genetic material of some viruses and virioids.

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Follows are the solution:

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Its response decreases over time as well as consumption of a reactants.  

r = -kAB

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\frac{1}{[(100 - 2x)(50 - x)]} dx = -k dt\\\\\frac{1}{2[(50 - x)(50 - x)]} dx = -k dt\\\\\ integral\  \frac{1}{2[(50 - x)^2]} dx =\ integral [-k ] \ dt\\\\\frac{-1}{[100-2x]} = -kt + D \\\\

D is the constant of integration

initial conditions: t = 0, x = 0

\frac{-1}{[100-2x]} = -kt + D   \\\\\frac{ -1}{[100]} = 0 + D\\\\D= \frac{-1}{100}\\\\

hence we get:

\frac{-1}{[100-2x]}= -kt -\frac{1}{100}\\\\or \\\\ \frac{1}{(100-2x)} = kt + \frac{1}{100}

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Insert the above value x into \frac{1}{(100-2x)} equation = kt + \frac{1}{100} to get k.  

\to  \frac{1}{(100-2\times \frac{10}{3})}  = k \times (7) + \frac{1}{100} \\\\ \to  \frac{1}{(100- 2 \times 3.33)} =   \frac{700k + 1}{100} \\\\ \to  \frac{1}{(100-6.66)} = \frac{700k + 1}{100}\\\\ \to \frac{1}{93.34} = \frac{700k + 1}{100} \\\\

\to 100 =  93.34(700k + 1) \\\\ \to 100 =  65,338k + 700 \\\\ \to   65,338k =  -600 \\\\ \to  k =  \frac{-600}{ 65,338} \\\\ \to k= - 0.0091

therefore plugging in the equation the above value of k  

\to \frac{1}{(100-2x)} = kt +\frac{1}{100} \\\\\to \frac{1}{(100-2x)} = -0.0091t + \frac{1}{100}\\\\\to \frac{1}{(100-2x)} =  \frac{1 -0.91t}{100}\\\\\to \frac{1}{2(50-x)} =  \frac{1 -0.91t}{100}\\\\\to \frac{1}{(50-x)} =  \frac{1 -0.91t}{50}\\\\\to 50= (1-0.91t)(50-x)\\\\\to 50 = 50-45.5t-x-0.91tx\\\\\to x+0.91xt= -45.5t\\\\\to x(1+0.91t)= -45.5t\\\\\to x=\frac{-45.5t}{1+0.91t}

Let y = C

, calculate C:

y = 3x

y =3 \times \frac{-45.5t}{1+0.91t}

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x = \frac{-45.5t}{1+0.91t} , plug t = 28

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