For each problem in this homework your assignment is to dete
For each problem in this homework, your assignment is to determine modes of inheritance for a set of phenotypes each of which is determined by genotypes at two independently assorting loci. The data come from an alien species of fly whose genetics is identical to that of earth flies. As part of your assignment, you must confirm your hypothesized modes of inheritance by applying a chi-square analysis. NOTE ABOUT DEGREES OF FREEDOM: In situations where the observed counts and the expected counts for a category are both 0, that category does not figure into the chi-square calculation, so that the degrees of freedom is reduced by one in such situations. The possible modes of inheritance are: (i) Autosomal recessive; (ii) Homozygous lethal (dominant); (iii) Sex-linked recessive. The Sex-linked recessive mode can only occur in females parents. Counts of phenotype pairs for the F1 generation give you clues as to the mode of inheritance of each phenotype. Good luck! ONE PHENOTYPE PER PARENT, TWO PHENOTYPES TOTAL For the Tratagorian Kumbus Fly, that lives on the planet Zeon, parental matings occur only among flies in which each parent has a disease phenotype at one locus and is wild-type at the other locus.
Problem 1. A female with Flonasean eye color mates with a male with Rheingoldenbach eye color, producing the following counts: F1 Generation Males WT-WT Disease-WT WT-Disease Disease-Disease 489 0 0 0 Females WT-WT Disease-WT WT-Disease Disease-Disease 513 0 0 0
Solution
F1 Generation:
Males
WT-WT Dis-WT WT-Dis Dis-Dis
489 0 0 0
Females
WT-WT Dis-WT WT-Dis Dis-Dis
513 0 0 0
The disease is autosomal recessive at one locus and also showing recessive epistasis of another locus.
DdWw x DdWw
DW Dw dW dw
DW DDWW DDWw DdWW DdWw
Dw DDWw DDww DdWw Ddww
dW DdWW DdWw ddWW ddWw
dw DdWw Ddww ddWw ddww
Phenotype
Exp Obs
D_W_ 9 1
D_ww 3 0
ddD_ 3 0
ddww 1 0
D_ww: died because of epistasis
ddW_: Died because of disease
ddww: Died because of epistasis or disease
Degree of freedom is zero. Therefore Chi-square can\'t be applied.
