Hardy-Weinberg worksheet

1. Recessive allele frequency (q): 0.3

1. Turn the percentage into a decimal. This is the share of aa individuals, written q²

   $\frac{9}{100}$$=0.09$

2. If the population is in balance, the aa share equals q × q. So take the square root to get q, the share of a alleles

   $q$$=\sqrt{0.09}$$=0.3$

3. Every allele is either A or a, so p and q add up to 1

   $p$$=1-0.3$$=0.7$

4. Expected share of AA (two dominant copies): the chance of getting A twice, p × p

   $p^2$$={0.7}^2$$=0.49$

5. Expected share of Aa (one copy of each): A then a, or a then A, so 2 × p × q

   $2\mathrm{pq}$$=2\times 0.7\times 0.3$$=0.42$

6. Expected share of aa (two recessive copies): the chance of getting a twice, q × q

   $q^2$$={0.3}^2$$=0.09$

2. Dominant allele frequency (p): 0.5842

1. Add up the individuals to get N, the sample size

   $325+425+170$$=920$

2. Find p, the share of A alleles. Each AA carries two A alleles and each Aa carries one, out of 2N alleles in total

   $p$$=\frac{2\times 325+425}{2\times 920}$$=0.584239$

3. Do the same for q, the share of a alleles. p and q always add up to 1

   $q$$=\frac{2\times 170+425}{2\times 920}$$=0.415761$

4. Expected share of AA (two dominant copies): the chance of getting A twice, p × p

   $p^2$$={0.584239}^2$$=0.341335$

5. Expected share of Aa (one copy of each): A then a, or a then A, so 2 × p × q

   $2\mathrm{pq}$$=2\times 0.584239\times 0.415761$$=0.485808$

6. Expected share of aa (two recessive copies): the chance of getting a twice, q × q

   $q^2$$={0.415761}^2$$=0.172857$

7. Turn each expected share into a head count by multiplying by N

   $\text{AA 314.03, Aa 446.94, aa 159.03}$

8. Measure how far the real counts are from the expected ones. For each genotype, square the gap and divide by the expected count, then add

   $\frac{{\left(325-314.0285\right)}^2}{314.0285}+\frac{{\left(425-446.9429\right)}^2}{446.9429}+\frac{{\left(170-159.0285\right)}^2}{159.0285}$$=2.217548$

9. Find the p-value: the chance of a gap this big by luck alone. It uses 1 degree of freedom (3 groups − 1 − 1 for estimating p)

   $P\text{(χ² ≥}2.2175\text{)}$$=0.13644959$

3. Dominant allele frequency (p): 0.5366

1. Add up the individuals to get N, the sample size

   $275+550+200$$=\mathrm{1,025}$

2. Find p, the share of A alleles. Each AA carries two A alleles and each Aa carries one, out of 2N alleles in total

   $p$$=\frac{2\times 275+550}{2\times \mathrm{1,025}}$$=0.536585$

3. Do the same for q, the share of a alleles. p and q always add up to 1

   $q$$=\frac{2\times 200+550}{2\times \mathrm{1,025}}$$=0.463415$

4. Expected share of AA (two dominant copies): the chance of getting A twice, p × p

   $p^2$$={0.536585}^2$$=0.287924$

5. Expected share of Aa (one copy of each): A then a, or a then A, so 2 × p × q

   $2\mathrm{pq}$$=2\times 0.536585\times 0.463415$$=0.497323$

6. Expected share of aa (two recessive copies): the chance of getting a twice, q × q

   $q^2$$={0.463415}^2$$=0.214753$

7. Turn each expected share into a head count by multiplying by N

   $\text{AA 295.12, Aa 509.76, aa 220.12}$

8. Measure how far the real counts are from the expected ones. For each genotype, square the gap and divide by the expected count, then add

   $\frac{{\left(275-295.122\right)}^2}{295.122}+\frac{{\left(550-509.7561\right)}^2}{509.7561}+\frac{{\left(200-220.122\right)}^2}{220.122}$$=6.388504$

9. Find the p-value: the chance of a gap this big by luck alone. It uses 1 degree of freedom (3 groups − 1 − 1 for estimating p)

   $P\text{(χ² ≥}6.3885\text{)}$$=0.01148618$

4. Dominant allele frequency (p): 0.6016

1. Add up the individuals to get N, the sample size

   $350+425+160$$=935$

2. Find p, the share of A alleles. Each AA carries two A alleles and each Aa carries one, out of 2N alleles in total

   $p$$=\frac{2\times 350+425}{2\times 935}$$=0.601604$

3. Do the same for q, the share of a alleles. p and q always add up to 1

   $q$$=\frac{2\times 160+425}{2\times 935}$$=0.398396$

4. Expected share of AA (two dominant copies): the chance of getting A twice, p × p

   $p^2$$={0.601604}^2$$=0.361928$

5. Expected share of Aa (one copy of each): A then a, or a then A, so 2 × p × q

   $2\mathrm{pq}$$=2\times 0.601604\times 0.398396$$=0.479353$

6. Expected share of aa (two recessive copies): the chance of getting a twice, q × q

   $q^2$$={0.398396}^2$$=0.158719$

7. Turn each expected share into a head count by multiplying by N

   $\text{AA 338.4, Aa 448.2, aa 148.4}$

8. Measure how far the real counts are from the expected ones. For each genotype, square the gap and divide by the expected count, then add

   $\frac{{\left(350-338.4024\right)}^2}{338.4024}+\frac{{\left(425-448.1952\right)}^2}{448.1952}+\frac{{\left(160-148.4024\right)}^2}{148.4024}$$=2.504223$

9. Find the p-value: the chance of a gap this big by luck alone. It uses 1 degree of freedom (3 groups − 1 − 1 for estimating p)

   $P\text{(χ² ≥}2.5042\text{)}$$=0.11354147$

5. Recessive allele frequency (q): 0.2449

1. Turn the percentage into a decimal. This is the share of aa individuals, written q²

   $\frac{6}{100}$$=0.06$

2. If the population is in balance, the aa share equals q × q. So take the square root to get q, the share of a alleles

   $q$$=\sqrt{0.06}$$=0.244949$

3. Every allele is either A or a, so p and q add up to 1

   $p$$=1-0.244949$$=0.755051$

4. Expected share of AA (two dominant copies): the chance of getting A twice, p × p

   $p^2$$={0.755051}^2$$=0.570102$

5. Expected share of Aa (one copy of each): A then a, or a then A, so 2 × p × q

   $2\mathrm{pq}$$=2\times 0.755051\times 0.244949$$=0.369898$

6. Expected share of aa (two recessive copies): the chance of getting a twice, q × q

   $q^2$$={0.244949}^2$$=0.06$

6. Recessive allele frequency (q): 0.3317

1. Turn the percentage into a decimal. This is the share of aa individuals, written q²

   $\frac{11}{100}$$=0.11$

2. If the population is in balance, the aa share equals q × q. So take the square root to get q, the share of a alleles

   $q$$=\sqrt{0.11}$$=0.331662$

3. Every allele is either A or a, so p and q add up to 1

   $p$$=1-0.331662$$=0.668338$

4. Expected share of AA (two dominant copies): the chance of getting A twice, p × p

   $p^2$$={0.668338}^2$$=0.446675$

5. Expected share of Aa (one copy of each): A then a, or a then A, so 2 × p × q

   $2\mathrm{pq}$$=2\times 0.668338\times 0.331662$$=0.443325$

6. Expected share of aa (two recessive copies): the chance of getting a twice, q × q

   $q^2$$={0.331662}^2$$=0.11$

7. Dominant allele frequency (p): 0.5758

1. Add up the individuals to get N, the sample size

   $400+720+200$$=\mathrm{1,320}$

2. Find p, the share of A alleles. Each AA carries two A alleles and each Aa carries one, out of 2N alleles in total

   $p$$=\frac{2\times 400+720}{2\times \mathrm{1,320}}$$=0.575758$

3. Do the same for q, the share of a alleles. p and q always add up to 1

   $q$$=\frac{2\times 200+720}{2\times \mathrm{1,320}}$$=0.424242$

4. Expected share of AA (two dominant copies): the chance of getting A twice, p × p

   $p^2$$={0.575758}^2$$=0.331497$

5. Expected share of Aa (one copy of each): A then a, or a then A, so 2 × p × q

   $2\mathrm{pq}$$=2\times 0.575758\times 0.424242$$=0.488522$

6. Expected share of aa (two recessive copies): the chance of getting a twice, q × q

   $q^2$$={0.424242}^2$$=0.179982$

7. Turn each expected share into a head count by multiplying by N

   $\text{AA 437.58, Aa 644.85, aa 237.58}$

8. Measure how far the real counts are from the expected ones. For each genotype, square the gap and divide by the expected count, then add

   $\frac{{\left(400-437.5758\right)}^2}{437.5758}+\frac{{\left(720-644.8485\right)}^2}{644.8485}+\frac{{\left(200-237.5758\right)}^2}{237.5758}$$=17.928091$

9. Find the p-value: the chance of a gap this big by luck alone. It uses 1 degree of freedom (3 groups − 1 − 1 for estimating p)

   $P\text{(χ² ≥}17.9281\text{)}$$=0.00002294$

8. Recessive allele frequency (q): 0.3162

1. Turn the percentage into a decimal. This is the share of aa individuals, written q²

   $\frac{10}{100}$$=0.1$

2. If the population is in balance, the aa share equals q × q. So take the square root to get q, the share of a alleles

   $q$$=\sqrt{0.1}$$=0.316228$

3. Every allele is either A or a, so p and q add up to 1

   $p$$=1-0.316228$$=0.683772$

4. Expected share of AA (two dominant copies): the chance of getting A twice, p × p

   $p^2$$={0.683772}^2$$=0.467544$

5. Expected share of Aa (one copy of each): A then a, or a then A, so 2 × p × q

   $2\mathrm{pq}$$=2\times 0.683772\times 0.316228$$=0.432456$

6. Expected share of aa (two recessive copies): the chance of getting a twice, q × q

   $q^2$$={0.316228}^2$$=0.1$

9. Recessive allele frequency (q): 0.2828

1. Turn the percentage into a decimal. This is the share of aa individuals, written q²

   $\frac{8}{100}$$=0.08$

2. If the population is in balance, the aa share equals q × q. So take the square root to get q, the share of a alleles

   $q$$=\sqrt{0.08}$$=0.282843$

3. Every allele is either A or a, so p and q add up to 1

   $p$$=1-0.282843$$=0.717157$

4. Expected share of AA (two dominant copies): the chance of getting A twice, p × p

   $p^2$$={0.717157}^2$$=0.514315$

5. Expected share of Aa (one copy of each): A then a, or a then A, so 2 × p × q

   $2\mathrm{pq}$$=2\times 0.717157\times 0.282843$$=0.405685$

6. Expected share of aa (two recessive copies): the chance of getting a twice, q × q

   $q^2$$={0.282843}^2$$=0.08$

10. Dominant allele frequency (p): 0.5638

1. Add up the individuals to get N, the sample size

   $280+500+160$$=940$

2. Find p, the share of A alleles. Each AA carries two A alleles and each Aa carries one, out of 2N alleles in total

   $p$$=\frac{2\times 280+500}{2\times 940}$$=0.56383$

3. Do the same for q, the share of a alleles. p and q always add up to 1

   $q$$=\frac{2\times 160+500}{2\times 940}$$=0.43617$

4. Expected share of AA (two dominant copies): the chance of getting A twice, p × p

   $p^2$$={0.56383}^2$$=0.317904$

5. Expected share of Aa (one copy of each): A then a, or a then A, so 2 × p × q

   $2\mathrm{pq}$$=2\times 0.56383\times 0.43617$$=0.491852$

6. Expected share of aa (two recessive copies): the chance of getting a twice, q × q

   $q^2$$={0.43617}^2$$=0.190244$

7. Turn each expected share into a head count by multiplying by N

   $\text{AA 298.83, Aa 462.34, aa 178.83}$

8. Measure how far the real counts are from the expected ones. For each genotype, square the gap and divide by the expected count, then add

   $\frac{{\left(280-298.8298\right)}^2}{298.8298}+\frac{{\left(500-462.3404\right)}^2}{462.3404}+\frac{{\left(160-178.8298\right)}^2}{178.8298}$$=6.236701$

9. Find the p-value: the chance of a gap this big by luck alone. It uses 1 degree of freedom (3 groups − 1 − 1 for estimating p)

   $P\text{(χ² ≥}6.2367\text{)}$$=0.01251293$