M = Mutant allele, W = Wild-type allele, we may determine the phenotypes and cou
ID: 96849 • Letter: M
Question
M = Mutant allele,
W = Wild-type allele,
we may determine the phenotypes and counts of the F1 offspring and F2 offspring produced by an F1-intercross mating (using the Fly Lab software).
What values would you use for the expected counts of the Red-eyed flies and the Purple-eyed flies in a Chi-square goodness of fit test, where the observed data are the F2 phenotype counts (all counts are rounded to nearest integer)?
A.
Male flies with red eyes: 384;
Female flies with red eyes: 384;
Male flies with purple eyes: 128;
Female flies with purple eyes: 128.
B.
Male flies with red eyes: 375;
Female flies with red eyes: 375;
Male flies with purple eyes: 125;
Female flies with purple eyes: 125.
C.
Male flies with red eyes: 250;
Female flies with red eyes: 250;
Male flies with purple eyes: 250;
Female flies with purple eyes: 250.
D. THIS ISNT THE ANSWER
Male flies with red eyes: 128;
Female flies with red eyes: 128;
Male flies with purple eyes: 384;
Female flies with purple eyes: 384.
E.
Male flies with red eyes: 125;
Female flies with red eyes: 125;
Male flies with purple eyes: 375;
Female flies with purple eyes: 375.
Explanation / Answer
According to the number of progeny in F2 generation, we can say that progenies rely on 1:1:1:1 ratio.
The total number of F2 progeny is – [381 + 362 + 149 + 131] = 1023
So, the expected number of male flies with red eyes: [1023*1/4] = 255.75
The expected number of female flies with red eyes: [1023*1/4] = 255.75
The expected number of male flies with purple eyes: [1023*1/4] = 255.75
The expected number of female flies with purple eyes: [1023*1/4] = 255.75
Thus, the correct option is – (C).
Male flies with red eyes: 250;
Female flies with red eyes: 250;
Male flies with purple eyes: 250;
Female flies with purple eyes: 250.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.