Question

In fruit flies, black body color (b) is recessive to the normal wild-type body color \(\left(b^{+}\right) .\) Cinnabar eye (cn) is recessive to the normal wild-type eye color \(\left(\mathrm{cn}^{+}\right)\). A homozygous wild-type fly was mated to a fly with black body and cinnabar eyes. The resulting heterozygous \(\mathrm{F}_{1}\) fly was mated to a fly with black body and cinnabar eyes. These were the results in the offspring: 90 wild-type 92 black body and cinnabar eyes 9 black body and wild-type eyes 9 wild-type body and cinnabar eyes What is the map distance between the gene for black body and the gene for cinnabar eyes?

Short answer

The map distance between the gene for black body and the gene for cinnabar eyes in fruit flies is 9 centiMorgans (cM).

Step-by-step solution

Understand basic genetic information

In fruit flies, black body color (b) is recessive to the normal wild-type body color \((b+)\). Cinnabar eye (cn) is recessive to the normal wild-type eye color \((cn+)\). We are given a cross between a homozygous wild-type fly and a fly with black body and cinnabar eyes. We can represent these flies' genotypes as: 1. Homozygous wild-type fly: \((b^+b^+, cn^+cn^+)\) 2. Black body, cinnabar eyes fly: \((bb, cncn)\)

Analyze and represent the F1 generation

When we cross these flies, we get an F1 generation with heterozygous genotypes, represented as: F1 generation: \((b^+b, cn^+cn)\)

Set up the F2 cross

The F1 generation fly is then mated to another fly with black body and cinnabar eyes (\((bb, cncn)\)). This will produce an F2 generation with various combinations of phenotypes.

Analyze the given results

The F2 generation offspring are given as: 1. 90 wild-type: \((b^+b^+, cn^+cn^+)\) or \((b^+b, cn^+cn)\) 2. 92 black body and cinnabar eyes: \((bb, cncn)\) 3. 9 black body and wild-type eyes: \((bb, cn^+cn)\) 4. 9 wild-type body and cinnabar eyes: \((b^+b, cncn)\)

Calculate the recombination frequency

We are interested in the non-parental phenotypes, as these are the result of recombination during meiosis. The non-parental phenotypes are offspring number 3 and 4 (9 black body and wild-type eyes, and 9 wild-type body and cinnabar eyes). Total offspring = 90 + 92 + 9 + 9 = 200 To calculate the recombination frequency, we divide the number of non-parental types (9 + 9) by the total number of offspring (200): Recombination frequency = \(\frac{9+9}{200} = \frac{18}{200} = 0.09 = 9\%\)

Calculate the map distance

Finally, the recombination frequency can be used to determine the map distance between the two genes: Map distance = recombination frequency * 100 = 9 * 100 = 9 centiMorgans (cM) So, the map distance between the gene for the black body and the gene for cinnabar eyes is 9 cM.

Key concepts

Recombination Frequency

When studying genetics, particularly in the context of breeding experiments with organisms like the fruit fly, understanding recombination frequency is crucial. This measure tells us how often new combinations of genes are formed in the offspring that differ from the parent types. Here's the foundational concept: genes that are closer together on a chromosome have a lower chance of being separated during the process of crossover in meiosis. As a result, they are inherited together more frequently. Conversely, genes further apart have a higher chance of recombination.

To put it into perspective with the textbook solution, the recombination frequency between the gene for body color and eye color in Drosophila was calculated by looking at the offspring that exhibited different physical characteristics than their parents. These offspring represent the recombination events. If we divide the number of such recombinant offspring by the total number of offspring, we get the recombination frequency. In this way, it serves as an essential tool for geneticists to understand how genes are linked and how they might behave in future generations.

Gene Mapping

Moving forward from the idea of recombination, gene mapping is an extension of that concept, involving the establishment of the positions of genes on a chromosome. The closer two genes are, the less likely they are to be separated during recombination, implying a low recombination frequency. Gene mapping uses recombination frequencies to quantify the distances between genes, providing a layout of the gene arrangement along chromosomes, often depicted as a genetic map.

Such maps can be invaluable for various purposes, including identifying the location of disease-causing genes, the study of complex genetic traits, and breeding programs in agriculture. In the case of our fruit flies, we use the recombination frequency calculated from the offspring phenotypes to estimate the map distance. The resulting 'map distance' helps provide a visual understanding of gene locations and is usually expressed in centiMorgans, a unit named after geneticist Thomas Hunt Morgan for his pioneering work in Drosophila genetics.

Drosophila Genetics

Drosophila melanogaster, the common fruit fly, is a powerhouse in genetic studies, and for good reason. Drosophila genetics has been a cornerstone of the field, contributing enormously to our understanding of genetic inheritance, gene function, and gene interactions. The fruit fly is a model organism of choice because of its quick life cycle, large number of offspring, easily identifiable mutant phenotypes and, most importantly, its relatively simple chromosome structure.

In the exercise provided, the inheritance of two different traits, body color and eye color, demonstrates a fundamental experiment in Drosophila genetics. From such experiments, geneticists have been able to establish that genes reside on chromosomes and that they are capable of forming new combinations through a process called recombination. The flies' phenotype—a visible trait such as body color or eye color—acts as a guide to deciphering the underlying genotype— the genetic make-up—and how it gets passed on from one generation to the next.