1. 39 red eyed males, 83 red eyed females, and 42 white eyed males. Assuming tha
ID: 180162 • Letter: 1
Question
1. 39 red eyed males, 83 red eyed females, and 42 white eyed males. Assuming that all genotypes from this cross should have equal survival rates, what are the genotypes of the parent flies?
(a. XwY and Xw+Xw+
b.XwY and Xw+Xw
c. Xw+YandXw+Xw
d. Xw+Y and XwXw
e. Xw+Y w+ and X w+Xw
1. 39 red eyed males, 83 red eyed females, and 42 white eyed males. Assuming that all genotypes from this cross should have equal survival rates, what are the genotypes of the parent flies?
(a. XwY and Xw+Xw+
b.XwY and Xw+Xw
c. Xw+YandXw+Xw
d. Xw+Y and XwXw
e. Xw+Y w+ and X w+Xw
Explanation / Answer
Red- and white-eyed males have been produced in almost equal frequencies, whereas only red-eyed females have been produced at double the frequency of red-eyed males.
Males inherit their X chromosome from the female parent only; therefore, the female parent must have produced gametes with the red-eyed and white-eyed alleles in equal frequencies. This suggests that the female parent was Xw+Xw.
All female offspring have red eyes, and we have already deyermined that the female parent is Xw+Xw. Therefore, the male parent must be Xw+Y (because if male parent is XwY, some of the female progeny would have been XwXw, i.e. white-eyed).
Hence, the parents are Xw+Y and Xw+Xw, i.e. option c is the most appropriate choice.
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