A Drosophila melanogaster female with cinnabar eyes and normal wings is crossed
ID: 163912 • Letter: A
Question
A Drosophila melanogaster female with cinnabar eyes and normal wings is crossed to a D. melanogaster male with red eyes and vestigial (small) wings. The F1 are all red-eyed with normal wings.
The F1 are allowed to cross with each other, and produce the following offspring.
57 red eyed normal wings
14 red eyes vestigial wings
13 cinnabar eyes normal wings
6 cinnabar eyes vestigial wings
One of the cinnabar-eyed females with vestigial wings is crossed with one of the male F1. The results are as follows.
27 red eyed normal wings
24 red eyes vestigial wings
23 cinnabar eyes normal wings
26 cinnabar eyes vestigial wings
a) What genetic system do the results of this cross indicate? How many loci, how many alleles?
b) Diagram both crosses and explain the results.
Explanation / Answer
As the most observed phenotype produced in all crosses is red-eyed/normal wings, those would be the dominant traits, therefore:
Parental (P): female cinnabar-eyed/normal wings (rrWW), and male red-eyed/small wings (RRww)
gametes: rW, and Rw
F1 cross: rW x Rw
F1 offspring: RrWw
F2 cross: RrWw x RrWw
Ratios: 9/16 red-eyed/normal wings, 3/16 red-eyed/small wings, 3/16 cinnabar-eyed/normal wings, and 3/16 cinnabar-eyed/small wings.
Observed: 57 red-eyed/normal wings, 14 red-eyed/small wings, 13 cinnabar-eyed/normal wings, and 6 cinnabar-eyed/small wings. Total n° of individuals = 57 + 14 + 13 + 6 = 90
Expected: 9/16 x 90 = 51 red-eyed/normal wings, 3/16 x 90 = 17 red-eyed/small wings, 3/16 x 90 = 17 cinnabar-eyed/normal wings, and 1/16 x 90 = 6 cinnabar-eyed/small wings.
F3 cross: female cinnabar-eyed/small wings (rrww) x F1 male (RrWw)
gametes: rw x RW, Rw, rW, rw
Ratios: 1/4 red-eyed/normal wings, 1/4 red-eyed/small wings, 1/4 cinnabar-eyed/normal wings, and 1/4 cinnabar-eyed/small wings.
Observed: 27 red-eyed/normal wings, 24 red-eyed/small wings, 23 cinnabar-eyed/normal wings, and 26 cinnabar-eyed/small wings. Total n° of individuals = 27 + 24 + 23 + 26 = 100
Expected: 1/4 x 100 = 25 red-eyed/normal wings, 1/4 x 100 = 25 red-eyed/small wings, 1/4 x 100 = 25 cinnabar-eyed/normal wings, and 1/4 x 100 = 25 cinnabar-eyed/small wings.
a) What genetic system do the results of this cross indicate? Taking into account that the observed phenotypes for all the crosses are very similar to the expected, the type of inheritance for both traits is autosomal recessive and they assort independently in all cases. F1 is a hybridization, a cross of 2 pure breeding individuals (rrWW x RRww). F2 is a dihybrid cross, a cross between two individuals heterozygous for 2 traits (expected ratios = 9:3:3:1), and F3 is a dihybrid testcross, a cross between two individuals one heterozygous and one homozygous recessive for 2 traits (expected ratios = 1:1:1:1).
How many loci, how many alleles? 2 alleles (1 dominant, 1 recessive), and 2 loci as the color eye gene and wing gene are located on a different chromosome or very far from each other on the same chromosome for independent assortment.
b) Diagram both crosses and explain the results.
Punnett square diagrams are shown above
RW Rw rW rw RW RRWW RRWw RrWW RrWw Rw RRWw RRww RrWw Rrww rW RrWW RrWw rrWW rrWw rw RrWw Rrww rrWw rrwwRelated Questions
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