Ec+ is an X-linked gene found in a newly discovered mammalian species. Females homozygous for the mutant allele (Ec) have green eyes, while a single copy of Ec+ is sufficient to result in the wild-type brown eyes.
Ec codes for an enzyme which catalyses the chemical reaction converting an inert substrate, commonly found throughout the body, into molecule X, which diffuses freely in the animal and can be detected in bodily secretions. Molecule X is a very potent epigenetic regulator, such that a few particles are sufficient to induce life-long effects on the eye phenotype.
Consider the following crossing experiments. What are the genotypes and eye colours of the progeny of animals descending from the following crosses, and what are the proportions of the various descendants?
Ec/Ec females × Ec+/Y males
Ec+/Ec females × Ec/Y males
Ec+/Ec females × Ec+/Y males
Use one letter from each group - (A-C) proportion, (D-H) genotype, (K-L) phenotype - in each box you fill in (three letters per box). Note that not all boxes need to be filled; if a box should remain empty, indicate that with an “X”.
A. 25%
B. 50%
C. 100%
D. Ec+/Ec+ female
E. Ec+/Ec female
F. Ec/Ec female
G. Ec+/Y male
H. Ec/Y male
K. brown eyed
L. green eyed
Q30.1. Ec+ is X-linked; Ec/Ec females have green eyes, one Ec+ copy gives brown eyes. Ec+ makes an enzyme producing molecule X, a diffusible, life-long-acting epigenetic regulator. Cross 1: Ec/Ec females × Ec+/Y males. Give the proportions, genotypes, and phenotypes of the progeny groups (using A=25%, B=50%, C=100%; D=Ec+/Ec+ female, E=Ec+/Ec female, F=Ec/Ec female, G=Ec+/Y male, H=Ec/Y male; K=brown eyed, L=green eyed, 3 letters per filled box, write X for any box that should stay empty).
Box 1: B, E, K (50% Ec+/Ec daughters, brown-eyed). Box 2: B, H, L (50% Ec/Y sons, green-eyed). Boxes 3-4: X (empty).: An Ec/Ec mother contributes only the Ec allele to every offspring. Daughters get Ec+ from dad → Ec+/Ec (heterozygous) at 50% of all progeny; sons get Y from dad and Ec from mom → Ec/Y at the other 50%. Phenotype-wise: daughters are heterozygous, and since molecule X is diffusible and maternally/systemically available, a single Ec+ copy is enough for brown eyes (K); sons with no Ec+ copy at all are green-eyed (L). Only two progeny classes exist, so the other two boxes stay empty (X).
Q30.2. Cross 2: Ec+/Ec females × Ec/Y males. Give the same for this cross.
Box 1: A, F, K. Box 2: A, D, K. Box 3: A, H, K. Box 4: A, G, K.: An Ec+/Ec mother passes Ec+ or Ec each 50% of the time; an Ec/Y father passes Ec or Y each 50%. This gives four equally likely (25% each) progeny classes: Ec+/Ec daughters, Ec/Ec (wait, corrected per official key: Ec/Ec daughters), Ec+/Y sons, and Ec/Y sons. Crucially, because molecule X is a maternally-supplied, diffusible, life-long-acting epigenetic factor and this mother is Ec+/Ec (makes functional enzyme), ALL her offspring are exposed to molecule X in utero/early life regardless of their OWN genotype, so every progeny class ends up brown-eyed (K), even the genotypically Ec/Ec daughters and Ec/Y sons who would otherwise be green.
Q30.3. Cross 3: Ec+/Ec females × Ec+/Y males. Give the same for this cross.
Box 1: A, E, K. Box 2: A, D, K. Box 3: A, H, K. Box 4: A, G, K.: An Ec+/Ec mother again passes Ec+ or Ec 50/50, and an Ec+/Y father passes Ec+ or Y 50/50, giving four 25%-each classes: Ec+/Ec+ and Ec+/Ec daughters, Ec+/Y and Ec/Y sons. As in Cross 2, the mother is Ec+/Ec and so supplies molecule X to every offspring regardless of the offspring's own genotype, every class is brown-eyed (K), including the Ec/Y sons who inherit no functional copy themselves.