Question

In the fruit fly, Drosophila melanogaster, three Chicago oriented genes have been found and a female with a Lakers allele at the Uni gene (wildtype Jaguar) female fly is mated to a male that is cubby at the baseball gene (wild type is sox), and Lions at the NFL gene (wildtype Bears). Phenotypically wild-type Fį female progeny were mated to fully homozygous (mutant) males, and the following progeny (1000 total) were observed:

Phenotypes Number Observed
Lakers, Sox, Bears 321
Jaguars, Sox, Bears 38
Lakers, Cubby, Bears 130
Jaguars, Cubby, Bears 18
Jaguars, Cubby, Lions 309
Lakers, Cubby, Lions 32
Jaguars, Sox, Bears 140
Lakers, Sox, Bears 12
(a) Which gene is in the middle?
(b) With respect to the three genes mentioned in the problem, what are the genotypes of the homozygous parents used in making the phenotypically wild-type F1 heterozygote?
(c) What are the map distances for the three genes?

Answer 1

To determine gene order and map distances in a genetics problem involving fruit fly (Drosophila melanogaster), one must analyze progeny phenotypes, identify the least common phenotypic combinations, and calculate the recombination frequencies, which help determine the physical distance between genes on a chromosome.

Step-by-step explanation:

To solve this genetics problem related to Drosophila melanogaster, we need to analyze the results of the crosses and look for the pattern of inheritance that indicates which gene is in the middle and calculate the map distances that show the physical distance between genes on a chromosome.

(a) The gene in the middle can be determined by looking for the least common phenotype combination among the progeny, which would be the result of double crossover events. In this scenario, the least common phenotypes are Jaguars, Cubby, Bears (18) and Lakers, Sox, Bears (12), suggesting that the Sox (baseball) gene is in the middle.

(b) The genotypes of the homozygous parents would be wildtype for all three traits. Hence, the female parent would be Jaguars/Uni, Sox/Baseball, and Bears/NFL, and the male parent would be Lakers/Uni, Cubby/Baseball, and Lions/NFL.

(c) The map distances are determined by calculating the frequency of recombination events between each pair of genes. For example, the recombination frequency between Lakers/Uni and Sox/Baseball can be calculated by adding the number of progeny with the recombinant phenotypes (Jaguars, Sox, Bears + Lakers, Cubby, Bears) and dividing by the total number of progeny, then multiplying by 100 to get a percentage, which can be translated into map units or centiMorgans (cM).