AP Biology Practice Exam 2 ❮ 303PART B: GRID-IN QUESTIONS- .0025—For this specific gene in this specific
population, there are a total of 40 alleles, two of
which are the recessive cf allele (2/40 =0.05=q).
Sinceyou need to be homozygous recessive to
have cystic fibrosis, (q)×(q)=q^2 =(0.05)^2 =
.0025. In other words, 25 out of 10,000 people
(0.25 percent)will have cystic fibrosis.# OBSERVED (O) # EXPECTED (e) (o -e) (o -e)^2 (o- e)^2 /e
wild-type flies 135 158 –23 529 3.35
recessive flies 76 53 23 529 9.98− / / /1 2 3 4 5 6 7 8 90 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
0(^0025)
1 2 3 4 5 6 7 8 9
- 13.33—If both parents were heterozygous and if
this trait is indeed recessive, you would expect the
next generation to show 75 percent normal-looking
flies and 25 percent of the flies with the recessive
trait. Based on a total of 211 flies, that would mean
you would expect 158 normal flies and 53 recessive
flies. Your observed numbers were, instead, 135
normal flies and 76 recessive flies.Since your chi-squared value (13.33) is higher
than the critical value of 6.64 (based on 1 degree
of freedom), you have to reject your hypothesis.Something other than an autosomal recessive
trait is going on.− / / /1 2 3 4 5 6 7 8 90 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
0(^3133)
3 4 5 6 7 8 9
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