Question

A student working in a genetics lab was given the task of studying a new Drosophila mutation called eyeless. In the time allotted, they were only able to perform the crosses shown below. NOTE: EY refers to the eyeless phenotype and + refers to the wild type phenotype. Results of Cross #1 Parents (Female: EY) x (Male: +) Offspring Phenotype Number Proportion Ratio Female: EY 526 0.5112 1.046 Male: EY 503 0.4888 1.000 Total 1029 -------------------------------------------------------------------------- Results of Cross #2 Parents (F1 Female: EY) x (F1 Male: EY) Offspring Phenotype Number Proportion Ratio Female: EY 353 0.3451 2.758 Male: EY 396 0.3871 3.094 Female: + 146 0.1427 1.141 Male: + 128 0.1251 1.000 Total 1023 -------------------------------------------------------------------------- 1. What is the genetic basis for the eyeless mutation? 2. Using Drosophila nomenclature, present symbols to represent the alleles for this gene and note the phenotype each is responsible for.

Answer 1

Drosophila melanogaster flies are ideal for genetic study. This is because of the ability to observe the mutation of the DNA through phenotypic manifestations. It was hypothesized that a mutation within the eyeless gene of D. melanogaster produces a phenotypic change in the organisms, and through in-depth analysis of the eyeless gene sequenced, dissimilarities in the mutant's DNA can be utilized to propose mutational mechanisms. Through DNA sequencing of the mutant eyeless sequence, the gene was localized to occupy a portion of exon 2, extending through intron 2 and a portion of exon 3 of the D. melanogaster wild type eyeless gene. This confinement allowed for proposed placement of the mutation, promoting hypothetical mutational mechanisms.

Pax-6 genes, known to be essential for eye development, encode an evolutionarily conserved transcription factor with two DNA-binding domain.

The advantage of Drosophila nomenclature method is that the wild type allele can be immediately identified. A cross between two pure-breeding parents, where one is (+/+) and other is (EY/EY) would yield on F1 generation. If these F1 are then crossed, we will get 4 phenotypic categories in equal numbers: eyeless, ebony; eyeless, wild; wild, ebony; wild, wild. (symbols- EY, + or +, EY).

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