Chapter 6, Problem 20EQ

In Drosophila, an allele causing vestigial wings is $\text{12}\text{.5 mu}$ away from another allele that causes purple eyes. A third gene that affects body color has an allele that causes black body color. This third gene is $\text{18}\text{.5 mu}$ away from the vestigial wings allele and $\text{6 mu}$ away from the allele causing purple eyes. The alleles causing vestigial wings, purple eyes, and black body are all recessive. The dominant (wild-type) traits are long wings, red eyes, and gray body. A researcher crossed wild-type flies to flies with vestigial wings, purple eyes, and black bodies. All ${\text{F}}_{\text{1}}$ flies were wild type. ${\text{F}}_{\text{1}}$ female flies were then crossed to male flies with vestigial wings, purple eyes, and black bodies. If 1000 offspring were observed, what are the expected numbers of the following types of flies?  

long wings, red eyes, gray body  

long wings, purple eyes, gray body  

long wings, red eyes, black body  

long wings, purple eyes, black body  

short wings, red eyes, gray body  

short wings, purple eyes, gray body  

short wings, red eyes, black body  

short wings, purple eyes, black body

Which types of flies can be produced only by a double crossover event?

Summary Introduction

To review:

The type of flies, which can be produced by a double crossover event.

Introduction:

When the phenomenon of crossing over take place between two genes, it is known as a two-point cross. Likewise, when three genes are involved in acrossing over, it is known as a three-point cross. Further, this cross helps in determining the actual orders of the genes considered in the chromosome and also, it is used for determining the distances between the genes by calculating the map distance between them. Map distance expresses the distance between the genes in the unit mu (that stands for map unit). This is defined by the division of all the recombinants by the total of all the offspring (that includes both recombinant and non-recombinant offspring), which is finally multiplied by 100.

Explanation of Solution

The exchange of genes within two chromosomes, which results in the production of non-identical chromatids which further holds the genetic components of the gametes is known as crossing over. The mechanism results in the formation of countless gametes, each different from the other and thus constitute a unique character upon fusion to form the embryo. For finding which type of flies can be produced only by a double cross over, one can start by considering the genes in pairs.

Firstly, considering the genes, which are very near/close to each other. The genes which are close are the ones that are responsible for the eye color and the length of the wings. These two genes are separated by 12.5 mu. So, with the help of this information, one can predict that the remaining 87.5% of the offspring will have eitherred eyes (that are dominant over the purple eyes) and long wings (this is dominant over the vestigial one)or the recessive phenotypes (short wings and red eyes; long wings and purple eye).

Further, 12.5% is the recombinationfrequency that has resulted from the crossing over of the genes (genes responsible for wing length and eye color).Also, as all the phenotypes have an equal chance of occurring, therefore the two parental phenotypes (both the dominant pairs and the recessive pairs) have an equal (43.75%) chances of occurrence. Similarly, the two recombinants will also have equal (6.25%) chances of occurrence.

Therefore, if there are a total of 1000 individuals resulting after a cross, then the number of offspring expected for different phenotypes can be calculated as follows:

$\begin{array}{c}\text{Red eyes and long wings}=\frac{\text{43}\text{.75}}{\text{100}}\times \text{1000 }\\ \text{Red eyes and long wings}=\text{437}\text{.5 }\\ \text{Red eyes and long wings}\simeq \text{438}\end{array}$

$\begin{array}{c}\text{Purple eyes and short wings}=\frac{\text{43}\text{.75}}{\text{100}}\times \text{1000 }\\ \text{Purple eyes and short wings}=\text{437}\text{.5 }\\ \text{Purple eyes and short wings}\simeq \text{438}\end{array}$

$\begin{array}{c}\text{Red eyes and short wings}=\frac{\text{6}\text{.25}}{\text{100}}\times \text{1000 }\\ \text{Red eyes and short wings}=\text{62}\text{.5 }\\ \text{Red eyes and short wings}\simeq \text{63}\end{array}$

$\begin{array}{c}\text{Purple eyes and long wings}=\frac{\text{6}\text{.25}}{\text{100}}\times \text{1000 }\\ \text{Purple eyes and long wings}=\text{62}\text{.5 }\\ \text{Purple eyes and long wings}\simeq \text{63}\end{array}$

Similarly, in the second gene pair, genes responsible for the eye color and the color of the body are considered. These two genes are separated by 6 mu. Thus, with the help of this information, one can predict that the remaining 94% of the offspring will have either red eyes (that are dominant over the purple eyes) and grey body (thatis dominant over the black body) or the recessive phenotypes (red eyes and black body; purple eyes and grey body). Further, 6% is the recombination frequency that has resulted from the crossing over of the genes (genes responsible for the body color and eye color).

Among the 438 flies having the long wings and the red eyes, are expected to have the gray body gene along with them, whose chances of occurrence is about 94% which is about:

$\begin{array}{c}\frac{\text{94}}{\text{100}}\times \text{438}=\text{411}\text{.72 }\\ \frac{\text{94}}{\text{100}}\times \text{438}\simeq \text{412}\end{array}$

Similarly, the occurrence of the three genes short wings, purple eyes, and the black bodies also have the same chances of occurrence as their dominant forms have, so they also have $\text{94\%}$ the chance of occurrence which is about:

$\begin{array}{c}\frac{\text{94}}{\text{100}}\times \text{438}=\text{411}\text{.72 }\\ \frac{\text{94}}{\text{100}}\times \text{438}\simeq \text{412}\end{array}$

Further, the recombinants having phenotypes short wings, purple eyes, grey body and the long wing, red eyes, and black bodies, each have about $\text{6\%}$ chances of occurrence, thus, it can be said that the expected offspring of this phenotypes would be around 26

$\begin{array}{c}\frac{6}{100}\times \text{438}=\text{26}\text{.28}\\ \frac{6}{100}\times \text{438}\simeq \text{26 }\end{array}$

Also, among the 62 flies, which possess long wings and purple eyes, there are about 94%, which is:

$\begin{array}{c}\frac{\text{94}}{\text{100}}\times \text{62}=\text{58}\text{.28 }\\ \frac{\text{94}}{\text{100}}\times \text{62}\simeq \text{58}\end{array}$

This shows thechances of occurrence of the phenotype long wings, purple eyes, black bodies and also there will be 6% which is:

$\begin{array}{c}\frac{\text{6}}{\text{100}}\times \text{62}=\text{3}\text{.72 }\\ \frac{\text{6}}{\text{100}}\times \text{62}\simeq \text{4}\end{array}$

Chances of occurrence of the phenotypes long wings, purple eyes, and gray bodies.

Likewise, the phenotype short wings, red eyes, and gray bodies will also have 94% of chances of occurrence which is:

$\begin{array}{c}\frac{\text{94}}{\text{100}}\times \text{62}=\text{58}\text{.28 }\\ \frac{\text{94}}{\text{100}}\times \text{62}\simeq \text{58}\end{array}$

Among the other 62 flies, the phenotypes having short wings, red eyes and gray bodies will be expected in 58 offspring which is:

$\begin{array}{c}\frac{\text{94}}{\text{100}}\times \text{62}=\text{58}\text{.28 }\\ \frac{\text{94}}{\text{100}}\times \text{62}\simeq \text{58}\end{array}$

The other phenotype that is, short wings, red eyes, and black bodies will be expected to have 4 offspring which is:

$\begin{array}{c}\frac{\text{6}}{\text{100}}\times \text{62}=\text{3}\text{.72 }\\ \frac{\text{6}}{\text{100}}\times \text{62}\simeq \text{4}\end{array}$

Conclusion

Therefore, it can be concluded that the offspring produced as a result of double cross over will have the least number of expected offspring. The offspring with long wings, purple eyes, and gray body or the short wings, red eyes, and black body will be obtained with a double crossover event.