Chapter 10, Problem 3P

Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book.

Chargaff’s Rules for the Base Composition of DNA Chargaff’s results (Table 10.1) yielded a molar ratio of 1 1.29 for A to G in ox DNA, 1.43 for T to C, 1.04 for A to T, and 1.00 for G to C. Given these values, what are the approximate mole fractions of A. C, G. and T in ox DNA?

Interpretation Introduction

Interpretation:

The approximate mole fractions of A, C, G, and T in$\text{oxDNA}$are to be calculated.

Concept introduction:

DNA stands for deoxyribonucleic acid, is a biological macromolecule. In organisms, it carries genetic information and is required for protein production, cell reproduction; regulation and metabolism. There are four base types of DNA: adenine (A), cytosine (C), guanine (G) and thymine (T).

The strands of the DNA molecule contain the complementary pairs of bases. One nucleotide sequence of one strand automatically provides the details required to form its partner. The adenine is complementary nucleotide for thymine and cytosine is the complementary nucleotide for guanine.

Answer to Problem 3P

The mole fraction of G in solution is$0.211$.

The mole fraction of T in solution is$0.29$.

The mole fraction of A in solution is$0.28$.

The mole fraction of C in solution is$0.22$.

Explanation of Solution

It is given that the molar ratio for A to G in$\text{oxDNA}$obtained from Chargaff’s results is$1.29$.

The molar ratio for T to C in$\text{oxDNA}$is$1.43$.

The molar ratio for A to T in$\text{oxDNA}$is$1.04$.

The molar ratio for G to C in$\text{oxDNA}$is$1.00$.

The data obtained in the ratios is not completely consistent. Therefore, multiple correct answers may be obtained. The relationship between all nucleotides is one of the ways to calculate the mole ratios. First, the molar ratio for G to C in$\text{oxDNA}$is$1.00$, the amounts of G and C are equivalent to two decimal places. It is expressed as,

$\begin{array}{l}\frac{\text{G}}{\text{C}}=1.00\\ \text{G}=\text{C}\end{array}$

The relationship between A and G is expressed in three ways as follows,

$\begin{array}{l}\frac{\text{A}}{\text{G}}=1.29\\ \text{A}=1.2\text{9 G}\\ \text{G}=\frac{\text{A}}{\text{1}\text{.29}}\end{array}$

Similarly, the relationship between A and T is expressed in three ways as follows,

$\begin{array}{l}\frac{\text{A}}{\text{T}}=1.04\\ \text{A}=1.04\text{ T}\\ \text{T}=\frac{\text{A}}{\text{1}\text{.04}}\end{array}$

The relationship between T and C is expressed in three ways as follows,

$\begin{array}{l}\frac{\text{T}}{\text{C}}=1.43\\ \text{T}=1.43\text{ C}\\ \text{C}=\frac{\text{T}}{\text{1}\text{.43}}\end{array}$

Now, the relationships between C and T and C and G are used to identify the relationship between T and G as shown below.

$\begin{array}{l}\text{T}=1.43\text{ G}\\ \text{G}=\frac{\text{T}}{\text{1}\text{.43}}\end{array}$

The value of mole fraction of any nucleotide is the moles of that nucleotide in solution divided by the total number of moles of all nucleotides in solution. Therefore, the mole fraction of G in solution is calculated by substituting the values of all nucleotides as follows,

${\text{X}}_{\text{G}}=\frac{\text{G}}{\text{A}+\text{C}+\text{G}+\text{T}}$

As it is known that,$\text{G}=\text{C or A}=\text{T}$. Thus, the above expression is expressed as,

$\begin{array}{c}{\text{X}}_{\text{G}}=\frac{\text{G}}{1.2\text{9 G}+\text{G}+\text{G}+1.4\text{3 G}}\\ =\frac{\text{G}}{1.2\text{9 G}+\text{2 G}+1.4\text{3 G}}\\ =\frac{\text{G}}{4.7\text{2 G}}\\ =0.211\end{array}$

The mole fraction of T in solution is calculated by substituting the values of all nucleotides as follows,

${\text{X}}_{\text{T}}=\frac{\text{T}}{\text{A}+\text{C}+\text{G}+\text{T}}$

As it is known that,$\text{G}=\text{C or A}=\text{T}$. Thus, the above expression is expressed as,

$\begin{array}{c}{\text{X}}_{\text{T}}=\frac{\text{T}}{1.04\text{ T}+\raisebox{1ex}{$\text{T}$}\!\left/ \!\raisebox{-1ex}{$\text{1}\text{.43}$}\right.+\raisebox{1ex}{$\text{T}$}\!\left/ \!\raisebox{-1ex}{$\text{1}\text{.43}$}\right.+\text{T}}\\ =\frac{\text{T}}{1.04\text{ T}+\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$1.43$}\right.\text{ T}+\text{T}}\\ =\frac{\text{T}}{\text{3}\text{.44 T}}\\ =0.29\end{array}$

The mole fraction of A in solution is calculated by substituting the values of all nucleotides as follows,

${\text{X}}_{\text{A}}=\frac{\text{A}}{\text{A}+\text{C}+\text{G}+\text{T}}$

As it is known that,$\text{G}=\text{C or A}=\text{T}$. Thus, the above expression is expressed as,

$\begin{array}{c}{\text{X}}_{\text{A}}=\frac{\text{A}}{\text{A}+\text{G}+\text{G}+\raisebox{1ex}{$\text{A}$}\!\left/ \!\raisebox{-1ex}{$\text{1}\text{.04}$}\right.}\\ =\frac{\text{A}}{1.9\text{6 A}+\text{2 G}}\\ =\frac{\text{A}}{1.9\text{6 A}+\text{2}\left(\frac{\text{A}}{\text{1}\text{.29}}\right)}\\ =\frac{\text{A}}{\text{3}\text{.51 A}}\end{array}$

Simplify the above expression,

${\text{X}}_{\text{A}}=0.28$

Now, the mole fraction of C is calculated by the formula,

${\text{X}}_{\text{A}}+{\text{X}}_{\text{C}}+{\text{X}}_{\text{G}}+{\text{X}}_{\text{T}}=1$

Substitute the values of mole fractions of all nucleotides in the above formula to calculate the mole fraction of C.

$\begin{array}{c}{\text{X}}_{\text{A}}+{\text{X}}_{\text{C}}+{\text{X}}_{\text{G}}+{\text{X}}_{\text{T}}=1\\ {\text{X}}_{\text{C}}=1-\left({\text{X}}_{\text{A}}+{\text{X}}_{\text{G}}+{\text{X}}_{\text{T}}\right)\\ {\text{X}}_{\text{C}}=1-\left(0.28+0.211+0.29\right)\\ {\text{X}}_{\text{C}}=1-0.781\end{array}$

Simplify the above expression,

$\begin{array}{c}{\text{X}}_{\text{C}}=0.219\\ \approx 0.22\end{array}$

Conclusion

The mole fraction of G in solution is$0.211$.

The mole fraction of T in solution is$0.29$.

The mole fraction of A in solution is$0.28$.

The mole fraction of C in solution is$0.22$.