Please help with #2 1) A newly identified fruit fly mutant, cyclops eye (large a

Question

Please help with #2

1) A newly identified fruit fly mutant, cyclops eye (large and single in the middle of the head), is hypothesized to be autosomal dominant. The experimenter started with homozygous wild type females (yes, virgins) and homozygous cyclops males. The data from the F2 generation was 44 wild type males, 60 wild type females, 110 cyclops males and 150 cyclops females. Do these data suggest that you should reject your model? Use chi square to prove your position.

2.) Another fictitious mutant, bloodshot eyes, is hypothesized to be autosomal recessive. Again the experimenter used homozygous wild type virgin females but this time the males had homozygous blood shot eyes. The F2 data was 75 wild type males, 60 wild type females, 31 bloodshot males and 45 bloodshot females. Do these data suggest that you should reject your model? Use chi-square to prove your position.1

Explanation / Answer

#2

Bloodshot eye is autosomal we can lump the wild type males and wild type females together (likewise for the bloodshot flies).

Therefore, we observed:

75 + 60 = 135 wild type males

35 + 45 = 76 bloodshot eyes females

The expecteds for this problem are determined from the total number of flies observed

75 + 60 + 31 + 45 = 211 Total flies

If bloodshot eyes are in fact autosomal recessive 1/4 of the F2 generation should have bloodshot eyes and 3/4 of the F2 generation should have wild type eyes.

Therefore:

1/4 x 211 = 158 wild type flies expected

3/4 x 211 = 53 bloodshot eyes flies.

Filling out the X2 equation then:

X2 = [(expected wild - observed wild)2] / expected wild + [(expected bloodshot - observed bloodshot)2] / expected bloodshot

X2 = (158-135)2 /158 + (53-76)2 / 53

X2 = (23)2 / 158 + (-23)2/ 53

X2 =529 / 158 + 529 / 53

X2 =3.3 + 10.0

X2 = 13.3

Since there are two groups of data (and two terms) there are 2-1 = 1 degree of freedom.

Entering the chi square table above in the first row (1 degree of freedom) we move to the right until we find that 13.3 (our chi square value) lies beyond 10.8 (the upper right end of the table.

Moving directly up we find that the probability of this value occuring by chance alone is less than 0.01 (0.1%). Since the standard must frequently used to divide significant from insignificant is 0.05 (5%) we can say that this deviation (difference) is highly significant.

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