HARDY-WEINBERG PROBLEMS
Easy problems
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For the first three problems calculate the frequency of the three genotypes and then calculate the
frequency of the two phenotypes.
1. A herd of horses with p = 0.25.
2. A school of seahorses with p = 0.65.
3. A school of redhorse suckers (a type of fish) with q = 0.65.
For the next three problems calculate the frequency of the three genotypes. Also calculate the
hypothesized actual number of the genotypes and the phenotypes.
4. A herd of 300 cows (+ a few bulls) with 12 homozygous recessive individuals.
5. A school(?) of 86 seacows includes 24 individuals which have a recessive skin disorder
causing them to be fluorescent yellow.
6. A flock of 125 scarlet macaws that includes 15 individuals with gold colored feathers (a
recessive mutation).
For this problem you are making a prediction about the next generation.
7. A flock of 68 scarlet tanagers, 14 of which have a white cap (a recessive mutation). They
breed and successfully raise 94 chicks. How many of these chicks do you expect to have a
white cap?
More difficult problems
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For problems 8 & 9 you are not sure which phenotype is recessive so calculate each problem both
ways.
8. A flock of bluebirds, 125 with orange bills and 63 with brown bills.
9. A population of bluejays, 3,428 with black bars on their tail feathers and 1,853 with plain
blue tail feathers.
Determine the allele frequency of the current group of individuals, calculate the expected
genotypic frequencies of their offspring and the frequency of each phenotype.
10. A flock of 342 scarlet ibis that includes 102 individuals with green colored legs (a
dominant mutation).
11. A flock of 1,200 cowbirds with 867 homozygous dominant individuals.
12. A colony of adult lab rats is variable for the length of their tail. They have been carefully
bred so that the rats are all homozygous, some dominant, some recessive. Long tails are
dominant to short tails. 25 have long tails and 75 have short tails.
A. Calculate p and q for this population.
B. Someone decided that they should be "liberated" from their cages and let all of the
rats out. They bred randomly. If 400 rat pups are born as a result, predict how many
of the offspring had long tails and how many had short tails.
13. What are the allele frequencies in an isolated field of 382 pink, 355 white and 103 red
snapdragons.
Hardy-Weinberg Answers
Easy answers
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For the first three problems calculate the frequency of the three genotypes and then calculate
the frequency of the two phenotypes.
1. A herd of horses with p = 0.25.
p = 0.25, q = 0.75 (0.25 + 0.75 = 1.0)
p2 = 0.0625
2pq = 0.375
q2 = 0.5625 (0.0625 + 0.375 + 0.5625 = 1.0)
A phenotype = 0.0625 + 0.375 = 0.4375
a phenotype = 0.5625 (0.4375 + 0.5625 = 1.0)
2. A school of seahorses with p = 0.65.
p = 0.65, q = 0.35 (0.65 + 0.35 = 1.0)
p2 = 0.4225
2pq = 0.455
q2 = 0.1225 (0.4225 + 0.455 + 0.1225 = 1.0)
A phenotype = 0.4225 + 0.455 = 0.8775
a phenotype = 0.1225 (0.8775 + 0.1225 = 1.0)
3. A school of redhorse suckers (a type of fish) with q = 0.65.
p=0.35, q= 0.65 (0.35 + 0.65 = 1.0)
p2 = 0.1225
2pq = 0.455
q2 = 0.4225 (0.1225 + 0.455 + 0.4225 = 1.0)
A phenotype = 0.1225 + 0.455 = 0.5775
a phenotype = 0.4225 (0.5775 + 0.4225 = 1.0)
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For the next three problems calculate the frequency of the three genotypes. Also calculate the
hypothesized actual number of the genotypes and the phenotypes.
4. A herd of 300 cows (+ a few bulls) with 12 homozygous recessive individuals.
p=0.8, q=0.2 (0.8 + 0.2 = 1.0)
p2 = 0.64
2pq = 0.32
q2 = 0.04 (0.64 + 0.32 + 0.04 = 1.0)
AA = 300 x 0.64 = 192
Aa = 300 x 0.32 = 96
aa = 300 x 0.04 = 12
"A" phenotype = 192 + 96 = 288
"a" phenotype = 12 (288 + 12 = 300)
5. A school(?) of 86 seacows includes 24 individuals which have a recessive skin
disorder causing them to be fluorescent yellow.
p=0.472, q=0.528 (0.472 + 0.528 = 1.0)
p2 = 0.223
2pq = 0.498
q2 = 0.279 (0.223 + 0.498 + 0.279 = 1.0)
AA = 86 x 0.223 = 19.2
Aa = 86 x 0.498 = 42.8
aa = 86 x 0.279 = 24
"A" phenotype = 19.2 + 42.8 = 62
"a" phenotype = 24 (62 + 24 = 86)
6. A flock of 125 scarlet macaws that includes 15 individuals with gold colored
feathers (a recessive mutation).
p=0.654, q=0.346 (0.654 + 0.346 = 1.0)
p2 = 0.427
2pq = 0.453
q2 = 0.12 (0.427 + 0.453 + 0.12 = 1.0)
AA = 125 x 0.427 = 53.4
Aa = 125 x 0.453 = 56.6
aa = 125 x 0.12 = 15
"A" phenotype = 53.4 + 56.6 = 110
"a" phenotype = 15 (110 + 15 = 125)
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For this problem you are making a prediction about the next generation.
7. A flock of 68 scarlet tanagers, 14 of which have a white cap(a recessive
mutation). They breed and successfully raise 94 chicks. How many of these
chicks do you expect to have a white cap?
q2 = 0.206
aa = 94 x 0.206 = 19.4
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More difficult answers
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For problems 8 & 9 you are not sure which phenotype is recessive so calculate each problem
both ways.
8. A flock of bluebirds, 125 with orange bills and 63 with brown bills.
orange recessive 125/188 = 0.665 = q2
q = 0.815, p = 0.185
p2 = 0.034
2pq = 0.302
q2 = 0.665 (0.034 + 0.302 + 0.665 = 1.0)
AA = 188 x 0.034 = 6.4
Aa = 188 x 0.302 = 56.8
aa = 188 x 0.665 = 125
"A" phenotype = 6.4 + 56.8 = 63.2
"a" phenotype = 125 (63 + 125 = 188)
brown recessive 63/188 = 0.335 = q2
q = 0.579, p = 0.421
p2 = 0.1772
2pq = 0.488
q2 = 0.335 (0.177 + 0.488 + 0.335 = 1.0)
AA = 188 x 0.177 = 33.3
Aa = 188 x 0.488 = 91.7
aa = 188 x 0.335 = 63.0
"A" phenotype = 33.3 + 91.7 = 125
"a" phenotype = 63 (125 + 63 = 188)
9. A population of bluejays, 3,428 with black bars on their tail feathers and 1,853
with plain blue tail feathers.
black bars recessive 3,428/5,281 = 0.649 = q2
q = 0.806, p = 0.194
p2 = 0.038
2pq = 0.313
q2 = 0.649 (0.038 + 0.313 + 0.649 = 1.0)
AA = 5,281 x 0.038 = 200.7
Aa = 5,281 x 0.313 = 1653
aa = 5,281 x 0.649 = 3,427.4
"A" phenotype = 200.7 + 1653 = 1853.7
"a" phenotype = 3,427.4 (1853.7 + 3,427.4 = 5281.1)
plain blue recessive 1,853/5,281 = 0.351 = q2
q = 0.592, p = 0.408
p2 = 0.1665
2pq = 0.483
q2 = 0.351 (0.167 + 0.483 + 0.351 = 1.0)
AA = 5,281 x 0.167 = 881.9
Aa = 5,281 x 0.483 = 2,550.7
aa = 5,281 x 0.351 = 1,853.6
"A" phenotype = 881.9 + 2,550.7 = 3432.6
"a" phenotype = 1,853.6 (3432.6 + 1,853.6 = 5,286.2)
(I probably rounded off to much.)
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Determine the allele frequencies of the current group of individuals, calculate the expected
genotypic frequencies of their offspring and the frequency of each phenotype.
10. A flock of 342 scarlet ibis that includes 102 individuals with green colored legs (a
dominant mutation).
342 - 102 = 240 recessive phenotypes
240/342 = 0.702 = q2
q = 0.838, p = 0.162 (0.838 + 0.162 = 1.0)
p2 = 0.026
2pq = 0.272
q2 = 0.702 (0.026 + 0.272 + 0.702 = 1.0)
"A" phenotype = 0.026 + 0.272 = 0.298
"a" phenotype = 0.702 (0.298 + 0.702 = 125)
11. A flock of 1,200 cowbirds with 867 homozygous dominant individuals.
867/1200 = 0.723 = p2
p = 0.85, q = 0.15
2pq = 0.255
q2 = 0.022
"A" phenotype = 0.723 + 0.255 = 0.978
"a" phenotype = 0.022
12. A colony of adult lab rats is variable for the length of their tail. They have been
carefully bred so that the rats are all homozygous, some dominant, some
recessive. Long tails are dominant to short tails. 25 have long tails and 75 have
short tails.
A. Calculate p and q for this population.
50/(50 + 150) = 0.25 = p
150/(50 + 150) = 0.75 = q (0.25 + 0.75 = 1.0)
B. Someone decided that they should be "liberated" from their cages and let
all of the rats out. They bred randomly. If 400 rat pups are born as a
result, predict how many of the offspring had long tails and how many
had short tails.
p2 = 0.0625
2pq = 0.375
q2 = 0.5625 (0.0625 + 0.375 + 0.5625 = 1.0)
"A" phenotype = 0.0625 + 0.375 = 0.4375 x 400 = 175 long-tailed
pups
"a" phenotype = 0.5625 x 400 = 225 short-tailed pups
What are the allele frequencies in an isolated field of 382 pink, 355 white and 103 red
snapdragons.
(103 x 2) + 382 = 588 red alleles
(355 x 2) + 382 = 1,092 white alleles
588/(588 + 1,092) = 0.35 red alleles
1,092/(588 + 1,092) = 0.65 white alleles