Download Biology 6 Practice Allele Frequency Problems

Biology 6 Practice Allele Frequency Problems
Determine the allele frequencies of each allele in the following pea plant populations in
which P is the dominant allele for purple flower color and p is the recessive allele for white
flower color:
1.
60% PP
30% Pp
10% pp
P = 0.75
2.
¼ PP
½ Pp
P = 0.5
3.
0.15 PP
0.25 Pp
P = 0.275
p = 0.25
¼ pp
p = 0.5
0.60 pp
p = 0.725
Determine the allele frequencies of each allele in the following bird populations in which
feather color is determined by 2 alleles that exhibit incomplete dominance:
black feathers (BB), gray feathers (Bb), whiter feathers (bb)
4.
20% black
40% gray
B = 0.4
5.
3/5 black
1/5 gray
B = 0.7
6.
0.40 black
0.35 gray
B = 0.575
40% white
b = 0.6
1/5 white
b = 0.3
0.25 white
b = 0.425
Using the Hardy-Weinberg equation below, determine the expected frequencies of each
genotype and phenotype in various pea plant populations using the information provided:
p2 + 2pq + q2 = 1
7.
P = 0.2
p = 0.8
PP = 0.04
Pp = 0.32
pp = 0.64
purple = 0.36
white = 0.64
8.
P = 0.7
PP = 0.49
Pp = 0.42
pp = 0.09
purple = 0.91
white = 0.09
9.
36% of the population has white flowers
PP = 0.16
Pp = 0.48
pp = 0.36
purple = 0.64
white = 0.36
Using the Hardy-Weinberg equation, determine the expected frequencies of each
genotype in the hypothetical human populations below and the likelihood of having the
indicated genetic condition or being a carrier:
10.
normal allele (A) = 0.98
albinism allele (a) = 0.02
AA = 0.9604
Aa = 0.0392
aa = 0.0004
carrier = ~1 in 25
albino = 1 in 2500
11.
the frequency of the autosomal recessive Tay-Sachs allele is 0.01
TT = 0.9801
carriers = ~1 in 50
12.
Tt = 0.0198
tt = 0.0001
Tay-Sachs = 1 in 10,000
1 in 400 people have the recessive autosomal disease cystic fibrosis
FF = 0.9025
carrier = ~1 in 10
Ff = 0.095
ff = 0.0025
cystic fibrosis = 1 in 400