Phenotype for eye colour is determined by a certain locus. E indicates the domin
ID: 2060 • Letter: P
Question
Phenotype for eye colour is determined by a certain locus.E indicates the dominant allele ande indicates the recessive allele. Thecross between a MALE normal fruit fly and a FEMALE white-eyed fruitfly produced this set of offspring:F1 Normal Male: 0 Normal Female: 45 White Eyed Male: 55 White Eyed Female: 0 Brown eyed Female: 1 The normal and white eyed individuals from the F1 generationabove were then crossed to produce the following offspring:
F2 Normal Male: 23 Normal Female: 31 White Eyed Male: 22 White Eyed Female: 24 Brown Eyed Female: 0 Determine the genotypes of the original parents, andexplain your reasoning. Also, what are two types of mutations thatmight have produced the brown eyed female? THANK YOU! This is a fairly difficult problem so I will rateit lifesaver, please help! Phenotype for eye colour is determined by a certain locus.E indicates the dominant allele ande indicates the recessive allele. Thecross between a MALE normal fruit fly and a FEMALE white-eyed fruitfly produced this set of offspring:
F1 Normal Male: 0 Normal Female: 45 White Eyed Male: 55 White Eyed Female: 0 Brown eyed Female: 1 The normal and white eyed individuals from the F1 generationabove were then crossed to produce the following offspring:
F2 Normal Male: 23 Normal Female: 31 White Eyed Male: 22 White Eyed Female: 24 Brown Eyed Female: 0 Determine the genotypes of the original parents, andexplain your reasoning. Also, what are two types of mutations thatmight have produced the brown eyed female? THANK YOU! This is a fairly difficult problem so I will rateit lifesaver, please help! THANK YOU! This is a fairly difficult problem so I will rateit lifesaver, please help!
Explanation / Answer
Before I begin, please try the Punett square crosses yourself tounderstand the ratios I obtained. Given the F1 generation, the total proportion of phenotypes seemsto be 1:1 between normal females and white eyed males (the browneyed mutant is negligible). First we test if the gene isautosomal. We assume that E is the dominant normal type and eis the recessive white eye mutation. The first cross we can is Ee x ee (since one parent is normal andone parent is white eyed). This produces 1Ee:1ee in theF1. But this proportion should be apparent in BOTH sexes ofthe offspring and since this is not the case, we reject this. Similarly, the same happens in the other possible cross, EE xee. This produces 1Ee in the F1. This is inconsistentwith our given ratio, so we reject this as well. Now we can assume that it is sex linked. Pops should benormal and moms should be white eyed, so we can test forXE Y normal male x Xe Xe whiteeyed female. The ratio becomes 1XeY:1XE Xe. This ratio is consistent withour results. Now we cross the F1's, Xe Y white eyed male xXE Xe normal females. We get a 1:1:1:1 ratiobetween normal females, white eyed females, normal males, and whiteeyed males. This is consistent with the given F2 generations'results. Therefore, the genotypes of the original parents are XE Y for pops and Xe Xe formoms. As for the brown eye female mutation, sorry, can't help youthere. Epitasis or pleiotropy (both of which are geneinteractions between our given gene and genes on different loci),maybe?