Question

In Drosophila, a cross was made between females-all expressing the three X-linked recessive traits scute bristles (sc), sable body (s), and vermilion eyes (v)-and wild-type males. In the \(\mathrm{F}_{1}\), all females were wild type, 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. $$\begin{array}{ll} \text { Phenotype } & \text { Offspring } \\ \text { se } s \quad v & 314 \\ +++ & 280 \\ +s v & 150 \\ s c++ & 156 \\ s c+v & 46 \\ +s+ & 30 \\ s c s+ & 10 \\ ++v & 14 \end{array}$$ No determination of sex was made in the data. (a) Using proper nomenclature, determine the genotypes of the \(P_{1}\) and \(F_{1}\) parents. (b) Determine the sequence of the three genes and the map distances between them. (c) Are there more or fewer double crossovers than expected? (d) Calculate the coefficient of coincidence. Does it represent positive or negative interference?

Short answer

a) The genotypes of the P1 and F1 parents are: - P1 female: X^{sc s v} X^{sc s v} - P1 male: X⁺ Y - F1 female: X^{sc s v} X⁺ - F1 male: X^{sc s v} Y b) The gene sequence is sc - s - v, with map distances of 34.6 cM between sc and s, and 36.6 cM between s and v. c) There are more double crossovers than expected, as the observed (44) is higher than the expected (12.67). d) The coefficient of coincidence is 3.47, which represents negative interference. This means that the presence of one crossover event increases the chances of another crossover event occurring nearby.

Step-by-step solution

a) Determining the genotypes of the P1 and F1 parents

The given cross was between females expressing the three X-linked recessive traits (sc, s, v) and wild-type males. Since the offspring's traits are X-linked, the phenotypes of the parents can be represented as: - P1 female: X^{sc s v} X^{sc s v} - P1 male: X⁺ Y In the F1 generation, all females are wild-type, while all males express all three mutant traits. Hence, the genotypes of the F1 parents are: - F1 female: X^{sc s v} X⁺ - F1 male: X^{sc s v} Y

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

To determine the sequence of the three genes and the map distances between them, we'll calculate the number of recombinants between each gene pair and divide that by the total number of offspring to get the recombination frequencies. - Number of recombinants between sc and s: (150 + 156 + 30 + 10) = 346 - Recombination frequency between sc and s: 346 / 1000 = 0.346 - Number of recombinants between s and v: (156 + 150 + 14 + 46) = 366 - Recombination frequency between s and v: 366 / 1000 = 0.366 The map distance is calculated as 100 times of the recombination frequency: - Map distance between sc and s: 0.346 × 100 = 34.6 cM - Map distance between s and v: 0.366 × 100 = 36.6 cM Therefore, the gene sequence is sc - s - v, with map distances of 34.6 cM and 36.6 cM.

c) Are there more or fewer double crossovers than expected?

To determine if there are more or fewer double crossovers than expected, we need to compare the observed double crossovers with the expected double crossovers. Observed double crossovers (scv and +sc): - scv: 14 offspring - +sc: 30 offspring - Total: 44 offspring Expected double crossovers can be calculated by multiplying the recombination frequencies between each gene pair: - Expected double crossovers: 0.346 × 0.366 × 100 = 12.67 offspring Since the observed double crossovers (44) are higher than the expected double crossovers (12.67), there are more double crossovers than expected.

d) Calculate the coefficient of coincidence. Does it represent positive or negative interference?

The coefficient of coincidence (c.o.c) can be calculated by dividing the observed double crossovers by the expected double crossovers: c.o.c = (Observed double crossovers) / (Expected double crossovers) c.o.c = 44 / 12.67 = 3.47 Interference can then be calculated as: Interference = 1 - c.o.c Interference = 1 - 3.47 Interference = -2.47 Because the interference is negative (-2.47), it represents negative interference, meaning that the presence of one crossover event increases the chances of another crossover event occurring nearby.

Key concepts

X-linked recessive traits

X-linked recessive traits are genetic characteristics that are associated with genes located on the X chromosome, one of the two sex chromosomes in mammals. In the context of genetics, 'recessive' indicates that two copies of the mutant gene are necessary for the trait to be expressed in females (who have two X chromosomes), while only one copy is needed for the trait to manifest in males (who have only one X chromosome).

In the exercise provided, we observe X-linked recessive traits in Drosophila, specifically scute bristles (sc), sable body (s), and vermilion eyes (v). Because males have only one X chromosome, any male with the mutant allele will display the recessive traits. Meanwhile, females will only show the traits if both of their X chromosomes carry the mutant alleles. This creates distinct patterns in inheritance that can be used to track the genes through generations.

Recombination frequency

Recombination frequency is a measure of the likelihood that two alleles will be separated during the crossing over of chromosomes in meiosis. It's an essential concept in understanding genetic linkage and gene mapping. This value is calculated by dividing the number of recombinant offspring by the total number of offspring.

For instance, in the textbook problem, the recombinant frequencies between the scute bristles (sc) and sable body (s) genes, and between sable body (s) and vermilion eyes (v) genes were used to infer gene positions. The recombination frequency is then used to calculate map distances, showing the physical distance between genes on a chromosome. The lower the recombination frequency, the closer the genes are to each other, which usually implies they are genetically linked.

Gene mapping

Gene mapping involves determining the relative positions of genes on a chromosome and the distance between them. It is a foundational tool in genetics that facilitates the study of inheritance patterns for specific traits.

The gene mapping exercise example demonstrates how recombination frequencies are converted into map distances, measured in units called centiMorgans (cM). These map distances represent the average number of crossovers between two loci, hence serving as a genetic map that can predict how likely it is for genes to be inherited together. Having accurate gene maps allows geneticists to diagnose and study various genetic conditions.

Coefficient of coincidence

The coefficient of coincidence (c.o.c) quantifies the degree to which the actual frequency of double crossovers in a particular area of the chromosome deviates from the expected frequency. It is calculated by dividing the observed number of double crossover individuals by the expected number based on the individual recombination frequencies of gene pairs involved.

In the exercise, the c.o.c is greater than 1, indicating that more double crossovers were observed than expected. This unusual result requires careful interpretation, as it may signal an anomaly in data or point towards other complex genetic phenomena affecting crossover rates.

Interference

Interference is the phenomenon where the occurrence of one crossover event in meiosis influences the probability of another crossover happening nearby. Typically, crossover interference results in a lower number of double crossovers than expected, which reflects a 'positive interference.' More simply, one crossover can inhibit the formation of another in its vicinity.

However, in our exercise, the interference value was found to be negative, suggesting an increase in the probability of a second crossover in the presence of the first, known as 'negative interference'. This phenomenon is less commonly observed but can occur under specific genetic circumstances.