MATH DNA Length

Surface Area to Volume Ratio - back

Assumption 1 AMU = 1 g/mol

Assumption 34Angstroms/Base Pair OR 3.4 Angstroms/Base Pair

Nitrogenous Bases:

Wikipedia weights for the nitrogenous bases:

• Adenine = 111.10 g/mol
• Guanine = 151.13 g/mol
• Cytosine = 135.13 g/mol
• Tyrosine = 181.19 g/mol

Remember that when this is bonded to the sugar that it loses 1 Hydrogen atom so it loses about 1 g/mol
Rough estimation for the deoxyribose and phosphate 6O, 5C, 6H, and 1P atom = 187.2 g/mol
Remember that when this is phosphodiester bonded to the C3 of the next sugar that it loses 1 Hydrogen atom so it loses another 1 g/mol
Don't worry there is a way to compensate the two 3' and 5' ends of the DNA by taking the masses of the TOTAL and adding 4 g/mol

• The terminal phosphate has a OH and the terminal 3' ends have a protonated alcohol in the C3

• This is a bit OCD because when compared to the initial number of 5E10 AMUs, 4 AMUs is simply negligible, especially when scientific notation is used

Nucleotide Monophospahte Weights:

So first thing is first, lets build a Nucleotide (such as Adenine --> Adenosine) Monophosphate (the deoxyribose sugar and phosphate)

• Adeonosine monophospahte (AMP) = 298.3 g/mol
• Guanosine monophosphate (GMP) = 338.33 g/mol
• Cytosine monophosphate (CMP) = 322.33 g/mol
• Thymidine monophosphate (TMP) = 368.39 g/mol

Base Pair Weights:

• C has three hydrogen bonds to G
  • CG with the parallel sugars and phosphate attached = 660.66 g/mol = 660.66 AMUs
• A has two hydrogen bonds to T
  • AT with the parallel sugars and phosphate attached = 666.69 g/mol = 666.69 AMUs

Refering Back To Total Weight:

• In terms of dissimilar CG and AT bonding ratios we set two different values of the amount of CG bonding and AT bondings using (b) and (a) repsectively
  • We were given that 5E10 AMUs and a CG:AT bonding of 1:1.56 (for every one CG there is a 1.56 ATs or better yet, for every two CGs there is roughly 3 ATs) - that means that multiplying 1.56 to the number of CGs would balance it back to the number of ATs
  • 5E10 = (AT)a + (CG)b + 4
  • Where a=1.56b

• 5E10=1.70E3b
• 2.94E7=b
• 5E10=1.09E3a
• 4.58E7=a

Sum up a and b to find the number of total base pairs in its entirety using weight as a guide

• 7.52E7 a+b

Times this by the lenght in angstroms per base pair: 3.4 Ang

• 7.52E7 X 3.4 = 2.56E8 Ang = 0.0256 meters = 2.56 centimeters

• In terms of similar CG and AT bonding
  • We were given that 5E10 AMUs and a CG:AT bonding of 1:1
  • 5E10 = (AT)a + (CG)b - 2
  • Where a=b

• 5E10=1.33E3a (which would be the same for b)
  • 3.76E7=a (which would be the same for b)

Sum up a and b to find the number of total base pairs in its entirety using weight as a guide

• 7.52E7 a+b

Times this by the lenght in angstroms per base pair: 3.4 Ang

• 7.52E7 X 3.4 = 2.56E8 Ang = 0.0256 meters = 2.56 centimeters