Chapter 10.3, Problem 29AQP

a.

Summary Introduction

To determine:

The ratio of $\frac{\left(\text{A+G}\right)}{\left(\text{T+C}\right)}$ and ratio of$\frac{\left(\text{A+T}\right)}{\left(\text{C+G}\right)}$for the organisms given in the table.

Introduction: Erwin Chargaff discovered some rules that help to explain the double helix structure of DNA. He is the first to identify that the number of guanine is equal to number of cytosine, and number of adenine is equal to number of thymine. He also stated that the amount of nitrogenous bases alters in different species.

Explanation of Solution

Organism and tissue
Sheep thymus	$\begin{array}{l}\frac{\left(\text{29}\text{.3}+21.4\right)}{\left(\text{28}\text{.3+21}\text{.0}\right)}\\ =\frac{50.7}{49.3}\\ =1.02\end{array}$	$\begin{array}{l}\frac{\left(\text{29}\text{.3}+28.3\right)}{\left(\text{21}\text{.0+21}\text{.4}\right)}\\ =\frac{57.6}{42.4}\\ =1.36\end{array}$
Pig liver	$\begin{array}{l}\frac{\left(\text{29}\text{.4}+20.5\right)}{\left(\text{29}\text{.7+20}\text{.5}\right)}\\ =\frac{49.9}{50.2}\\ =0.99\end{array}$	$\begin{array}{l}\frac{\left(\text{29}\text{.4}+29.7\right)}{\left(\text{20}\text{.5+20}\text{.5}\right)}\\ =\frac{59.1}{41.0}\\ =1.44\end{array}$
Human thymus	$\begin{array}{l}\frac{\left(30.9+19.9\right)}{\left(\text{29}\text{.4+19}\text{.8}\right)}\\ =\frac{50.8}{49.2}\\ =1.03\end{array}$	$\begin{array}{l}\frac{\left(30.9+29.7\right)}{\left(\text{20}\text{.5+20}\text{.5}\right)}\\ =\frac{60.6}{41.0}\\ =1.48\end{array}$
Rat bone marrow	$\begin{array}{l}\frac{\left(\text{28}\text{.6}+21.4\right)}{\left(\text{28}\text{.4+20}\text{.4}\right)}\\ =\frac{50.0}{48.8}\\ =1.02\end{array}$	$\begin{array}{l}\frac{\left(\text{28}\text{.6}+28.4\right)}{\left(\text{20}\text{.4+21}\text{.4}\right)}\\ =\frac{57.0}{41.8}\\ =1.36\end{array}$
Hen erythrocytes	$\begin{array}{c}\frac{\left(\text{28}\text{.8}+20.5\right)}{\left(\text{29}\text{.2+21}\text{.5}\right)}\\ =\frac{49.3}{50.7}\\ =0.97\end{array}$	$\begin{array}{c}\frac{\left(28.8+29.2\right)}{\left(\text{21}\text{.5+20}\text{.5}\right)}\\ =\frac{58.0}{42.0}\\ =1.38\end{array}$
Yeast	$\begin{array}{c}\frac{\left(31.7+18.3\right)}{\left(\text{32}\text{.6+17}\text{.4}\right)}\\ =\frac{50.0}{50.0}\\ =1.00\end{array}$	$\begin{array}{c}\frac{\left(31.7+32.6\right)}{\left(\text{17}\text{.4+18}\text{.3}\right)}\\ =\frac{64.3}{35.7}\\ =1.80\end{array}$
E.coli	$\begin{array}{c}\frac{\left(\text{26}\text{.0}+24.9\right)}{\left(\text{23}\text{.9+25}\text{.2}\right)}\\ =\frac{50.9}{49.1}\\ =1.03\end{array}$	$\begin{array}{c}\frac{\left(\text{26}\text{.0}+23.9\right)}{\left(\text{25}\text{.2+24}\text{.9}\right)}\\ =\frac{49.9}{50.1}\\ =0.99\end{array}$
Human sperm	$\begin{array}{c}\frac{\left(30.9+19.1\right)}{\left(\text{31}\text{.6+18}\text{.4}\right)}\\ =\frac{50.0}{50.0}\\ =1.00\end{array}$	$\begin{array}{c}\frac{\left(30.9+31.6\right)}{\left(\text{18}\text{.4+19}\text{.1}\right)}\\ =\frac{62.5}{37.5}\\ =1.67\end{array}$
Salmon sperm	$\begin{array}{c}\frac{\left(\text{29}\text{.7}+20.8\right)}{\left(\text{29}\text{.1+20}\text{.4}\right)}\\ =\frac{50.5}{49.5}\\ =1.02\end{array}$	$\begin{array}{l}\frac{\left(\text{29}\text{.7}+29.1\right)}{\left(\text{20}\text{.4+20}\text{.8}\right)}\\ =\frac{58.8}{41.2}\\ =1.42\end{array}$
Herring sperm	$\begin{array}{l}\frac{\left(\text{27}\text{.8}+22.1\right)}{\left(\text{27}\text{.5+20}\text{.7}\right)}\\ =\frac{49.9}{48.2}\\ =1.04\end{array}$	$\begin{array}{l}\frac{\left(\text{27}\text{.8}+27.5\right)}{\left(\text{20}\text{.7+22}\text{.1}\right)}\\ =\frac{55.3}{42.8}\\ =1.29\end{array}$

b.

Summary Introduction

To determine:

Whether the ratios are constant among the organisms or they vary.

Explanation of Solution

$\frac{\left(\text{A+G}\right)}{\left(\text{T+C}\right)}$Ratio among all the organisms is almost constant as the value ranges around 1.0, this concludes that the DNA is double stranded in these organisms. $\frac{\left(\text{A+T}\right)}{\left(\text{C+G}\right)}$Ratio is not constant, as the percentage of GC base pair and AT base pair are varying with the organism so it suggests that all the organisms belongs to different species.

c.

Summary Introduction

To explain:

Whether $\frac{\left(\text{A+G}\right)}{\left(\text{T+C}\right)}$ is different for samples of sperm, to explain with reason.

Explanation of Solution

$\frac{\left(\text{A+G}\right)}{\left(\text{T+C}\right)}$Ratio will remain same for the samples of sperms because the percentage of pyrimidine and purines must be equal in the case of double stranded DNA.