16.1-2ITD - Campbell Biology

Question

Using Chargaff’s rules, fill in the table with your predictions of the missing percentages of bases, starting with the wheat genome and proceeding through E. coli, human, and ox. Show how you arrived at your answers.

Step-by-Step Solution

Answer

Step 1: Chargaff’s rule

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.

Step 2: Base percentage for wheat

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.

Step 3: Base percentage for E. coli

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

To find: [C] and [T]

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

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