How many DNA bases?

DNA ( Deoxyribonucleic acid ) is the genetic material for life and is made of $4$ Nitrogenous bases, which are Adenine, Guanine, Cytosine, and Thymine. According to Scientist Erwin Chargaff, Adenine always bonds with thymine $(A-T)$ and Cytosine always bonds with Guanine $(G-C)$ . A Nucleotide is the basic unit of DNA, which consists of a base, Deoxyribose sugar, and a Phosphate $(PO_{4}^{3-})$ . Because of Chargaff's rules , the number of Adenine nucleotides is equal to the number of Thymine nucleotides, and the number of Guanine nucleotides is equal to the number of Cytosine nucleotides.

We are given the formulas for Adenine, Guanine, Cytosine, and Thymine. Their respective molar masses are $135$ grams/mol, $126$ grams/mol, $111$ grams/mol, and $151$ grams/mol. In this problem, it is useful to find the molar mass of an Adenine-Thymine $(A-T)$ base pair and the molar mass of a Guanine-Cytosine $(G-C)$ base pair. This is because the DNA is made up of only $(A-T)$ and $(G-C)$ since it is unmutated. If $x$ is the moles of $(A-T)$ and $y$ is the moles of $(G-C)$ in the $500$ -gram sample, we can use a System of Equations to solve for $x$ and $y$ .

A $(A-T)$ base pair is made of $1$ Adenine, $1$ Thymine, $2$ Deoxyribose sugars ( $1$ on each side chain), and $2$ Phosphates ( $1$ on each side chain). We can add the molar masses, but we have to subtract out the molar mass of $4$ water molecules since there were $4$ Hydrolysis bonds ( $2$ for Base-Sugar Glycosidic, $2$ for Phosphate-Sugar Phosphodiester) when producing the base pair. Thus, the molar mass for an $(A-T)$ base pair is $(135+126+2(134)+2(95)-4(18))=647$ grams/mol. Similarly, the molar mass of a $(G-C)$ base pair is $(111+151+2(134)+2(95)-4(18))=648$ grams/mol. Thus, $500=647x+648y$ .

In the problem, it states that there are $22\%$ more Guanine bases than Adenine bases. Since Guanine and Adenine bases make up $50\%$ of DNA, then there would be $14\%$ Adenine bases and $(14+22)\%=36\%$ Guanine bases. In other words, there are $\frac{36}{14}$ times as many Guanine as Adenine, so $y=2.57x$ . Substituting this back into $500=647x+648y$ yields $x=0.216$ moles of $(A-T)$ nucleotides and $y=0.555$ moles of $(G-C)$ nucleotides.

In the problem, we considered the molar masses of $(A-T)$ and $(G-C)$ base pairs, and we solved the system based on those. Since there are $6.02*10^{23}$ molecules in a mole, and each molecule is an $(A-T)$ or $(G-C)$ base pair, there would be $0.216*6.02*10^{23}=1.3*10^{23}$ $(A-T)$ base pairs and $2.57(0.216)*6.02*10^{23}=3.341*10^{23} (G-C)$ base pairs. As a result, the final answer is $3.341*10^{23}-1.3*10^{23}=2.04*10^{23}$ more $(G-C)$ base pairs and $(A-T)$ base pairs.