Chapter 7, Problem 12P

In the experiment shown in Fig 7.13 on p. 218, H. J. Muller first performed a control in which the P generation males were not exposed to X-rays. He found that 99.7% of the individual F₁ Bar-eyed females produced some male progeny with Bar eyes and some with wild-type (non-Bar) eyes, but 0.3% of these females produced only wild-type (non-Bar) male progeny.

a.	If the average spontaneous mutation rate for Drosophila genes is 3.5 × 10-6 mutations/gene/gamete, how many genes on the X-chromosome can be mutated to produce a recessive lethal allele?
b.	As of the year 2013, analysis of the Drosophila genome had revealed a total of 2238 genes on the X chromosome is typical of the genome, what is fraction of genes in the fly genome that is essential to survival
c.	Muller now exposed male flies to a specific high dosage of X-rays and found that 12% of F1 Bar-eyed females produced male progeny that were all wild type. What does this new information say?

Summary Introduction

a.

To determine:

The mutation in the number of genes to produce recessive lethal trait.

Introduction:

The genotype of an organism is dependent on its genome and also on the environment of the organism. In some of the animals, it is not as typical as the genotype of the mother plays a crucial role. This type of effect is known as a maternal effect. The genes that participate in this are called maternal effect genes.

Explanation of Solution

The genes that express a lethal recessive phenotype inside the mutant organism are referred to as essential genes. When these become non-functional due to mutation, then the organism cannot survive. Based on the experiment of HJ muller the results show that 99.7% of the F₁ generation females produce males with Bar (mutant) eyes and 0.3% of the F₁ females that produced males had wild (non-Bar) type phenotype.

Hence, the females produced 0.3% wild type progeny with Bar mutation and recessive lethal on X-chromosome. The given average spontaneous mutation rate for Drosophila genes is 3.5×10^-6 mutations/gene/gamete. The rate of mutation on the X-chromosome with respect to the essential gene is 3×10^-3 mutations per gamete. Hence, the number of genes with mutations producing a recessive lethal gene is calculated by formulae:

$\begin{array}{rll}\text{Number}\text{of}\text{genes}\text{with}\text{mutation}& =& \frac{\text{the}\text{rate}\text{of}\text{X-linked}\text{lethal}\text{mutations}\text{per}\text{gamete}}{\text{the}\text{rate}\text{of}\text{mutations}\text{per}\text{gene}\text{per}\text{gamete}}\\ & =& \frac{\text{3}\times {\text{10}}^{\text{-3}}}{\text{3}\text{.5}\times {\text{10}}^{\text{-6}}}\\ & =& 857\end{array}$

Hence, the number of genes that are affected with recessive lethal mutation is 857.

Summary Introduction

To determine:

The fraction of genes present in the Drosophila that are essential for survival.

Introduction:

The mutation is the change in the nucleotide sequence of the gene, which results in either the formation of a defective protein or no protein at all. The mutation can also alter the regulation of certain genes leading to their hyperactivity or hypoactivity. It is different from recombination where gametes from parents are interchanged to produce new zygotes.

Explanation of Solution

The results that are obtained from the experiment in the year 2013 showed that a total of 2238 genes are present on the X-chromosome of Drosophila. The estimated lethal X-linked genes are 857. Thus, the fraction of the genes with affected genotype is calculated as follows:

$\begin{array}{l}\text{The}\text{fraction}\text{of}\text{mutated}\text{genes}=\frac{\text{the}\text{estimated}\text{essential}\text{genes}\text{on}\text{the}\text{X}\text{chromosome}}{\text{total}\text{genes}\text{on}\text{X-chromosome}}\text{×100}\\ =\frac{\text{857}}{\text{2283}}\text{×100}\\ =\text{0}\text{.37×100}\\ =\text{37\%}\end{array}$

Summary Introduction

To determine:

The information conveyed by the change in the experiment where 12% of F₁ Bar-eyed females produced a male with wild type eyes.

Introduction:

The law of independent assortment was given in a dihybrid cross which is the cross between the alleles of two pairs of contrasting characters. The alleles produced by each character are independent to combine with the alleles of the other character. This independent sorting of the alleles irrespective of the dominant or the recessive trait is defined as the law of independent assortment.

Explanation of Solution

The new X-ray induced mutation rate is seen to be 12% of the F₁ females that produced males with wild type eyes. The X-ray induced mutation rate is calculated as:

$\begin{array}{l}=\frac{\text{Mutation}\text{rate}}{\text{Total}\text{genes}\text{on}\text{the}\text{X}\text{chromosome}}\\ =\frac{\text{0}\text{.12}}{\text{857}}\\ =\text{1}{\text{.4×10}}^{\text{-4}}\text{.}\end{array}$

As already proved the given average spontaneous mutation rate for Drosophila genes is 3.5×10^-6 mutations/gene/gamete.

Hence,

$\begin{array}{l}=\frac{\text{the}\text{X-ray}\text{induced}\text{mutation}\text{rate}}{\text{the}\text{rate}\text{of}\text{mutations}\text{per}\text{gene}\text{per}\text{gamete}}\\ ={\frac{\text{1}\text{.4}\times \text{10}}{\text{3}{\text{.5×10}}^{\text{-6}}}}^{-4}\\ =40\end{array}$

Thus, the information conveys that X-rays are mutagenic, because the rate of spontaneous mutation increased with increasing dosage of X-rays up to 40 fold.