Question

In Drosophila, a cross was made between females expressing the three X-linked recessive traits, scute bristles $(sc),$ sable body $(s)$ and vermilion eyes ( $v$ ), and wild-type males. All females were wild type in the ${\mathrm{F}}_1$, while all males expressed all three mutant traits. The cross was carried to the ${\mathrm{F}}_2$ generation and 1000 offspring were counted, with the results shown in the following table. No determination of sex was made in the ${\mathrm{F}}_2$ data. (a) Using proper nomenclature, determine the genotypes of the ${\mathrm{P}}_1$ and $F_1$ parents. (b) Determine the sequence of the three genes and the map distance between them. (c) Are there more or fewer double crossovers than expected? (d) Calculate the coefficient of coincidence; does this represent positive or negative interference?

Short answer

Based on the provided information and step-by-step solution, we can conclude: a) The P1 female parent has the genotype sc/sc, s/s, v/v, while the P1 male parent has the genotype Sc Y, S Y, V Y. The F1 females have the genotype Sc/sc, S/s, V/v, and the F1 males have the genotype sc Y, s Y, v Y. b) To determine the sequence of the three genes and the map distance between them, we would need the F2 data to be provided. With this data, we could identify the middle gene and calculate the map distance using the provided formula. c) By comparing observed and expected double recombinants, we can determine whether there are more or fewer double crossovers than expected. However, without specific F2 data, we cannot provide an answer. d) To compute the coefficient of coincidence and interference, we need specific data on observed and expected double recombinants from the F2 generation, which is not provided in the question.

Step-by-step solution

a) Determine the genotypes of the P1 and F1 parents.

Since we know that the three traits are X-linked recessive, the female parent (P1) must be homozygous recessive for all three traits: $\text{sc/sc},\text{s/s},\text{v/v}$. The wild-type male parent (P1) can only carry dominant alleles since he is wild type, having genotype $\text{Sc Y},\text{S Y},\text{V Y}$, where Y represents the absence of the gene on the Y chromosome. In the F1 generation, since all females are wild-type and all males express all three mutant traits, the F1 females will have the genotype $\text{Sc/sc},\text{S/s},\text{V/v}$, while the F1 males will have the genotype $\text{sc Y},\text{s Y},\text{v Y}$.

b) Determine the sequence of the three genes and the map distance between them.

In order to determine the sequence of the three genes, we need to find which gene is in the middle of the other two. To do this, we need to examine the F2 data and find out which of the double recombinants appears less frequently. Since no sex determination was given, we have to analyze the combination of traits in the 1000 offspring. The least frequent double recombinants will be those where the middle gene "switches" its allele between the outer two. For example, if the sequence is sc - s - v, the least frequent double recombinant classes would be scute+sable wild type ($\text{s}$) and wild type ($\text{Sc}$)+vermilion. We must examine the data and determine which gene behaves as the middle gene. Once we have found the least occurring double recombinant class, we can calculate how many recombinants are attributed to each gene pair and use this to estimate the map distance. Map distance is calculated by the formula: $MapDistance=(No.ofrecombinants/Totalcount)\times 100$ cM.

c) More or fewer double crossovers?

To determine if there are more or fewer double crossovers than expected, we need to compare the number of observed double recombinants with the expected number. Expected double recombinants can be calculated by multiplying the frequency of single recombinants of both gene pairs: $No.ofexpecteddoublerecombinants=F1recombinantfrequency\times F2recombinantfrequency\times Totalcount$ If the observed number of double recombinants is higher than the expected number, then there are more double recombinants in the F2 generation than expected. If the observed number of double recombinants is lower than the expected number, then there are fewer double recombinants than expected.

d) Calculate the coefficient of coincidence and interference.

The coefficient of coincidence (COC) is the ratio of observed double crossovers to the expected double crossovers: $COC=ObservedDoubleCrossovers/ExpectedDoubleCrossovers$ Interference is the difference between the expected frequency of double recombinants and the observed frequency, calculated as follows: $Interference=1-COC$ If there is positive interference, it means that the presence of one crossover event inhibits the occurrence of another nearby crossover, and the value will be positive (greater than 0). If there is negative interference, it means that a crossover event promotes the occurrence of another nearby crossover, and the value will be negative (less than 0).

Key concepts

Genetic Linkage

Genetic linkage refers to the phenomenon where genes that are located close to each other on the same chromosome tend to be inherited together. This occurs because they are less likely to be separated during the process of recombination when germ cells are being formed.
The closer two genes are on a chromosome, the less likely a crossover event will occur between them, allowing these genes to stay linked through generations. In genetic mapping, studying linked genes helps researchers determine the physical proximity of genes on the chromosome.
For instance, in the Drosophila gene mapping exercise, the genes for scute bristles (sc), sable body (s), and vermilion eyes (v) all reside on the X chromosome and are X-linked recessive traits. Because they are on the same chromosome and show linkage, the offspring patterns provide insight into their order and spacing.

Recombination Frequency

Recombination frequency is the measure used to determine the distance between genes on a chromosome. This frequency reflects how often crossover events occur between two genes during meiosis.
It is expressed as a percentage or map units (centiMorgans, cM), where 1% recombination equals 1 cM of distance. The higher the recombination frequency, the further apart the genes are on the chromosome.

• Calculation: To calculate map distance, the formula used is $MapDistance=\left(\frac{\text{Number of recombinants}}{\text{Total count}}\right)\times 100$ cM.
• Example: If 10% of offspring show recombination between two genes, these genes are 10 cM apart.

In the Drosophila exercise, analyzing how often each combination of traits appears allows us to determine distance and order of the genes, as recombination frequencies highlight these linkages.

Coefficient of Coincidence

The coefficient of coincidence (COC) is a measure that describes how the observed occurrence of double crossover events compares to what would statistically be expected if crossovers were independent. Understanding COC helps explain the intricacies of genetic linkage and interactions among crossover events.

• Calculation: $COC=\frac{\text{Observed Double Crossovers}}{\text{Expected Double Crossovers}}$.
• Expected Value: The expected number is found by multiplying the single crossover frequencies for each gene pair involved in the double crossover.

When completing the Drosophila mapping exercise, if COC is less than 1, fewer double crossovers occurred than expected, indicating interference. If it's greater than 1, more double crossovers were observed, suggesting negative interference.

Gene Sequence Determination

Determining the sequence of genes involves identifying their order on a chromosome. It requires analyzing the combinations of genetic traits found in offspring to detect which gene lies in the middle during genetic cross experiments.
One effective approach is to look for the least common recombinant frequency in offspring, which indicates possible double crossover events, as these occurrences are less frequent. This is essential in pinpointing the 'middle' gene that switches alleles during crossover.

• Steps: Examine phenotype combinations to identify the configuration that results from a double recombination affecting the middle gene.
• Example: If scute+, sable wild type, and vermilion appear least frequently, it suggests that the sable gene is in the middle, assuming genes display this switching pattern.

Gene sequence determination thereby establishes the linear sequence of genes, providing an understandable map of genetic relationships.