Evolution & Ecology
Hardy-Weinberg Answers

Easy answers

  • 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)

  • For the next three problems calculate the frequency of the three genotypes. Also calculate the hypothesized actual number of the genotypes and the phenotypes.

    1. 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)

    2. 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)

    3. 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)

  • For this problem you are making a prediction about the next generation.

    1. 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


More difficult answers

  • For problems 8 & 9 you are not sure which phenotype is recessive so calculate each problem both ways.

    1. 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)

    2. 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.)


  • Determine the allele frequencies of the current group of individuals, calculate the expected genotypic frequencies of their offspring and the frequency of each phenotype.

    1. 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)

    2. 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

    3. 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.

      1. 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)

      2. 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

    4. 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

The hardest problems

  • Read carefully and be thoughtful.

    1. I had a room full of 666 cockroaches which included 24 albinos. These albinos were so scary looking that I squashed them and let the remaining cockroaches breed. The next generation of cute little baby cockroaches totaled 13,485. Assuming that the allele for albinism is recessive how many of these youngsters were albino?

        Before selection, 24/666 = 0.036 = q2
        q = 0.19, p = 0.81

        p2 = 0.656
        2pq = 0.308
        q2 = 0.036 (0.656 + 0.308 + 0.036 = 1.0)

        AA = 666 x 0.656 = 436.9
        Aa = 666 x 0.308 = 205.1, 437 + 205 = 642 = 666-24

        After selection, 205/1,284 alleles are "a", q = 0.16
        q2 = 0.0256, 0.0256 x 13,485 = 345.2 albino babies

    2. Texas Wesleyan University used to have a colony of gerbils that were used for genetics research. I was particularly interested in two coat color genes. One gene causes gerbils to be brown (B-) or black (bb). The other is called dilution and it causes colors to look lighter. For example, genetically black gerbils that are also dd look grey, genetically brown but dd gerbils are called "cinnamon". In 1997, I had the following gerbils:
      • 35 grey
      • 24 black
      • 22 cinnamon
      • 12 brown

      A high school teacher gave me two litters of her gerbils. They were

      • 5 brown
      • 3 cinnamon
      • 5 black
      • 3 grey

      1. What was the frequency of the B and b alleles of my colony after the new gerbils were added?

          (35 grey + 24 black) + (3 grey + 5 black) = 67 bb
          67/(35+24+22+12)+(5+3+5+3) = 67/109=0.6147 = q2

          q = 0.784, p = 0.216

      2. What was the frequency of the D and d alleles of my colony after the new gerbils were added?

          (35 grey + 22 cinnamon) + (3 grey + 3 cinnamon) = 63 dd
          63/(35+24+22+12)+(5+3+5+3) = 63/109=0.5780 = q2

          q = 0.76, p = 0.24

    3. A human trait involving finger length in humans is recessive in males but dominant in females (This is true!). A survey of 938 college students found that 385 of 438 males and 104 of 500 females had dominant phenotypes. What are the allele frequencies?

    4. The gene for color-blindness is on the X chromosome. Men have one X chromosome (and a Y). Women have two X chromosomes. I have personally tested a large number of students for color-blindness and have found that, on average, one male will be color-blind in a class of 30. Assuming that such a class is 1/2 male and 1/2 female, how many women would I have to test before I found one who was color-blind?

        15 males, 1/15 = 0.0667 = q
        q2 = 0.0044

        44/10,000 women are color-blind or 1 in 227.3

    5. Many cattle breeders obtain bulls of one breed to sire calves with cows of another breed. One of the most common crosses is black angus bulls with hereford cows. Purebred herefords all have a white face that is caused by a dominant allele. As a result of being inbred you can assume that they are all homozygous. Purebred black angus never have a white face. What percentage of the offspring in such a herd will have a white face?

        AA x aa = 100% Aa


kmnelson@uta.edu /