Question

Phenotypically wild \(\mathrm{F}_{1}\) female Drosophila, whose mothers had light eyes (It) and fathers had straw (stw) bristles, produced the following offspring when crossed with homozygous light-straw males:$$\begin{array}{lc} \text { Phenotype } & \text { Number } \\ \hline \text { light-straw } & 22 \\ \text { wild } & 18 \\ \text { light } & 990 \\ \text { straw } & \frac{970}{2000} \end{array}$$ Compute the map distance between the light and straw loci.

Short answer

Answer: The approximate map distance between the light and straw loci in Drosophila is 48 centimorgans (cM).

Step-by-step solution

Identify the parental phenotypes

The F1 females are phenotypically wild and have mothers with light eyes (It) and fathers with straw (stw) bristles, so their genotype can be represented as It+ st+ / It stw. The homozygous light-straw males have a genotype of It stw / It stw. The phenotypes of the offspring resulting from this cross will help us calculate the recombination frequency.

Identify recombinant offspring

The recombinant offspring are those that have a phenotype different from the parental phenotypes (it's the result of a crossover event). In this case, light and straw are the recombinant phenotypes with frequencies of 990 and 970/2000 respectively.

Calculate the recombination frequency

The recombination frequency can be calculated as the ratio of the sum of the numbers of recombinant offspring to the total number of offspring. In this case, the recombination frequency is: Recombination frequency = (number of light offspring + number of straw offspring) / total number of offspring Recombination frequency = (990 + (970/2000)) / (22 + 18 + 990 + (970/2000))

Compute the map distance

The map distance in centimorgans (cM) is found by multiplying the recombination frequency by 100. So, Map distance = recombination frequency * 100 First, let's calculate the recombination frequency: Recombination frequency = (990 + (970/2000)) / (22 + 18 + 990 + (970/2000)) Recombination frequency ≈ 0.48 Now we can compute the map distance: Map distance = 0.48 * 100 Map distance ≈ 48 cM The map distance between the light and straw loci in Drosophila is approximately 48 centimorgans (cM).

Key concepts

Recombination Frequency

Recombination frequency is a crucial concept in genetic mapping. It shows us how often genes on the same chromosome are separated by crossing over during meiosis. When two genes are close together on a chromosome, crossover events between them happen less frequently. This means the recombination frequency will be low. However, when genes are far apart, crossovers can happen more easily, giving a higher recombination frequency.

To find the recombination frequency, we add the counts of recombinant offspring. These are the offspring that do not show the parental phenotypes since crossing over has combined different alleles together. The recombination frequency is calculated by dividing this sum by the total number of offspring.

• Recombinant offspring: Phenotype not in the parents (in our case, light and straw).
• Recombination frequency = (Number of recombinant offspring) / (Total offspring).

Understanding recombination frequency helps geneticists to estimate how far apart two genes are, which is expressed in map units or centimorgans.

Map Distance

Map distance is used to measure how far apart genes are on a chromosome, using the recombination frequency to determine this distance. It is expressed in centimorgans (cM), where 1 cM represents a 1% chance that a crossover event will separate the two gene locations during meiosis.

The relationship between map distance and recombination frequency is direct. We can convert the recombination frequency into map distance by simply multiplying it by 100.

• If recombination frequency is 0.48, then:
• Map distance = 0.48 x 100 = 48 cM

Thus, a 48% recombination frequency translates to a map distance of 48 cM. In genetic studies of organisms like Drosophila, map distance gives insight into gene linkage and helps in constructing genetic maps that outline gene location on chromosomes.

Drosophila Genetics

Drosophila melanogaster, commonly known as the fruit fly, is a favorite model organism for geneticists. It's small, easy to breed, and has a short life cycle. Most importantly, its genetic makeup shares many similarities with other species, including humans, making it invaluable for genetic research.

• Used for studying inheritance and genetic mapping.
• Has only four pairs of chromosomes, making genetic linkage analysis simpler.

In studies involving Drosophila, like the one mentioned, researchers often explore genetic linkage and recombination by observing phenotypes in generations of fruit flies, such as eye color or bristle type. These observable characteristics help confirm findings about genetic distance and inheritance patterns. Through controlled breeding experiments, Drosophila helps researchers understand fundamental principles of genetics.

Phenotypic Ratio

In genetics, the phenotypic ratio provides important insights into the genetic variability and dominance of traits in offspring. These ratios describe the observable traits of organisms arising from a particular genetic cross.

• The phenotype is the outward expression like color or shape.
• Phenotypic ratio helps in identifying dominant and recessive alleles.

In our example with Drosophila, observing the phenotypic ratio allows for identification of recombinant phenotypes whose appearance gives clues about the genetic crossover events. This information helps calculate recombination frequency and map distance, confirming hypotheses about gene linkage. For instance, a skewed phenotypic ratio might indicate a strong genetic linkage or other complex interactions occurring at the chromosomal level.