91Ó°ÊÓ

Chargaff’s rule states that the base percentage of adenine is equal to thymine, and the same goes for guanine and cytosine. Also, the sum of all base percentages for adenine, guanine, cytosine, and thymine equals 100.

For wheat, given that$\left[A\right]=28.1$, $\left[G\right]=28.1$,$\left[C\right]=22.7$.

Consequently, the base percentage of thymine must be approximately 28 as adenine.

To find: [T]

$$\begin{gathered} \left[A\right]+\left[G\right]+\left[C\right]+\left[T\right]=100 \\ 28.1+21.8+22.7+\left[T\right]=100 \\ 72.6+\left[T\right]=100 \end{gathered}$$

$$\begin{gathered} \left[T\right]=100-72.6 \\ \left[T\right]=27.4 \end{gathered}$$

Therefore, the base percentage for thymine is 27.4 for wheat.

For E. coli, given that $\left[A\right]=24.7$and $\left[G\right]=26.0$.

To find: [C] and [T]

role="math" localid="1643686737583" $$\begin{gathered} \left[A\right]+\left[G\right]=24.7+26.0 \\ \left[A\right]+\left[G\right]=50.7 \\ \left[A\right]+\left[G\right]+\left[C\right]+\left[T\right]=100 \\ \left[C\right]+\left[T\right]=100-50.7 \\ \left[C\right]+\left[T\right]=49.3 \end{gathered}$$

We know that [A]=[T] and [G]=[C], and considering, A+G=C+T, we have 0.7% extra in purine and 0.7% less in pyrimidine. In purine, adenine is less than guanine. So, thymine percentage should be less in proportion than cytosine.

Dividing $\left[C\right]+\left[T\right]=49.3$equally between [C] and [T], we have 24.65 for each.

Adjusting 0.7 between the pyrimidine bases to achieve [A]=[T], and [G]=[C] we have