Hardy-Weinberg practice
Hardy-Weinberg tutorial (allele frequencies):
If 98 out of 200 individuals in a population express the recessive phenotype, what percent of the population are homozygous dominant?
49%
(wrong)
70%
(wrong)
30%
(wrong)
42%
(wrong)
9%
<==== correct
Solution
:
Recessive phenotype implies homozygous genotype.
q
2
= 98 / 200 = 0.49
q
= 0.7
p
+
q
= 1
p
+ 0.7 = 1
p
= 0.3
p
2
=
0.3
2
= 0.09
If 98 out of 200 individuals in a population express the recessive phenotype, what percent of the population are heterozygotes?
49%
(wrong)
70%
(wrong)
30%
(wrong)
42%
<==== correct
9%
(wrong)
Solution
:
Recessive phenotype implies homozygous genotype.
q
2
= 98 / 200 = 0.49
q
= 0.7
p
+
q
= 1
p
+ 0.7 = 1
p
= 0.3
2
pq
= 2 (0.7 * 0.3) = 0.42
double check:
p
2
+ 2
pq
+
q
2
= 1?
double check: 0.09 + 0.42 + 0.49 = 1.
Brown hair (B) is dominant to blond hair (b). If there are 168 brown hairs in a population of 200:
What is the predicted frequency of heterozygotes?
16%
(wrong)
40%
(wrong)
60%
(wrong)
48%
<==== correct
84%
(wrong)
Solution
:
BB + Bb = 168 / 200 = 0.84 = brown hairs
bb = 1 - 0.84 = 0.16 =
q
2
blond hairs
q
= 0.4
p
= 1 -
q
= 1 - 0.4 = 0.6
2
pq
= 2 (0.4 * 0.6) = 0.48
What is the predicted frequency of homozygous dominant?
16%
(wrong)
40%
(wrong)
60%
(wrong)
48%
(wrong)
36%
<==== correct
Solution
:
p
2
= 0.6
2
= 0.36
What is the predicted frequency of homozygous recessive?
16%
<==== correct
40%
(wrong)
60%
(wrong)
48%
(wrong)
36%
(wrong)
Solution
:
q
2
= 0.4
2
= 0.16
double check:
p
2
+ 2
pq
+
q
2
= 1?
double check: 0.6
2
< + 2 (0.6 * 0.4) + 0.4
2
= 0.36 + 0.48 + 0.16 = 1.
If blonds occur in 36% of the population:
What is the allele frequency for b?
36%
(wrong)
60%
<==== correct
40%
(wrong)
16%
(wrong)
48%
(wrong)
Solution
:
bb =
q
2
= 0.36
q
= b = 0.6
What is the allele frequency for B?
36%
(wrong)
60%
(wrong)
40%
<==== correct
16%
(wrong)
48%
(wrong)
Solution
:
p
+
q
= 1
p
+ 0.6 = 1
q
= 1 - 0.6 = 0.4
What is the predicted frequency of heterozygotes?
36%
(wrong)
60%
(wrong)
40%
(wrong)
16%
(wrong)
48%
<==== correct
Solution
:
2
pq
= 2 (0.6 * 0.4) = 0.48
What is the predicted frequency of homozygous dominant?
36%
(wrong)
60%
(wrong)
40%
(wrong)
16%
<==== correct
48%
(wrong)
Solution
:
p
2
+ 2
pq
+
q
2
= 1
p
2
+ 0.48 + 0.36 = 1
p
2
= 1 - (0.48 + 0.36)
p
2
= 1 - 0.84
p
2
= 0.16