Hardy-Weinberg Principle (Numericals with solutions)

Q 1) In corn, purple kernels are dominant to yellow. A random sample of 100 kernels is taken from a population in Hardy-Weinberg equilibrium. It is found that 9 kernels are yellow and 91 kernels are purple.What is the frequency of the yellow allele in this population?

ANS-0.3

Q 2) a) The frequency of two alleles in a gene pool is 0.19 (A) and 0.81(a). Assume that the population is in Hardy-Weinberg equilibrium. (a)Calculate the percentage of heterozygous individuals in the population.

ANS-According to the Hardy-Weinberg Equilibrium equation, heterozygotes are represented by the 2pq term. Therefore, the number of heterozygous individuals (Aa) is equal to2pq which equals 2 × 0.19 × 0.81 = 0.31 or 31%

b) Calculate the percentage of homozygous recessives in the population.

ANS- The homozygous recessive individuals (aa) are represented by the q2 term in the H-W equilibrium equation which equals 0.81 × 0.81 = 0.66 or 66%

Q 3) An allele W, for white wool, is dominant over allele w, for black wool. In a sample of 900 sheep, 891 are white and 9 are black. Calculate the allelic frequencies within this population, assuming that the population is in H-W equilibrium.

ANS- The allelic frequency of w is represented by the q term and the allelic frequency W is represented by the p term. To calculate the value of q, realize that qq or q2 represents the homozygous recessive individuals or the black sheep in this case. Since there are 9 black sheep, the frequency of black sheep = # individuals /total individuals=9/900=0.01 thus ww = q2 = 0.01 so q=square root of q square= 0.1 Additionally, p + q = 1 thus p = 1 – q or p = 1 – 0.1 or 0.9 ∴ p = W = 0.9 and q = w = 0.1

Q 4)In a population that is in Hardy-Weinberg equilibrium, the frequency of the recessive homozygote genotype of a certain trait is 0.09. Calculate the percentage of individuals homozygous for the dominant allele.

ANS- We know that the frequency of the recessive homozygote genotype is q2and equal to 0.09 and its square root is 0.30 AND we also know that p + q = 1 Thus, p = 1 – q ∴p = 1 – 0.30 = 0.70 ∴The homozygote dominants are represented by p2= (0.70)2 = 0.49 or 49%

Q 5.In a population that is in Hardy-Weinberg equilibrium, 38 % of the individuals are recessive homozygotes for a certain trait. In a population of 14,500, calculate the percentage of homozygous dominant individuals and heterozygous individuals.

ANS- Always start with the homozygous recessive percentage if given, which is equal to q2. ∴ q =Square root of q∴ q =square root of 0.38 ∴q = 0.616 . Solving for p is now straightforward: p + q = 1, thus p = 1 – q = 1 – 0.616 = 0.384 The homozygous dominant individuals are represented by p square=0.384×0.384=0.147 so,14500×0.0147=2,132 homozygous dominant individual.The heterozygotes are represented by the 2pq term and 2×0.616×0.384=0.473 so,14500×0.473=6,859

Q 6)A rare disease which is due to a recessive allele (a) that is lethal when homozygous, occurs within a specific population at a frequency of one in a million. How many individuals in a town with a population of 14,000 can be expected to carry this allele? ANS-Always start with the homozygous recessive percentage if given, which is equal to q2. The frequency of the recessive genotype, q2 is 1/1, 000, 000 = 0.000001 ∴q = square root of q = 0.000001 = 0.001 = the frequency of allele a and p + q = 1 thus, p = 1 – q = 1 – 0.001 = 0.999 Therefore, the frequency of allele A = p = 0.999 and the frequency of allele a = q = 0.001 Carriers are heterozygous and are equal to 2pq. So, 2x 0.999x 0.001 x 14, 000=27.97

Q 7) In a certain African population, 4 % of the population is born with sickle cell anemia (aa). Calculate the percentage of individuals who enjoy the selective advantage of the sickle-cell gene (increased resistance to malaria)?

ANS-The individuals that are heterozygous enjoy the selective advantage of increased resistance to malaria and are carriers of the sickle cell trait. They have a genotype of Aa and are represented by the 2pq term in the H-W equilibrium equations. We are told that 4% of the population is affected with sickle cell anemia, thus q2 = 4% = 0.04 when expressed as a decimal. So, q = square root of q or we can say aquare root of 0.04=0.20 and p + q = 1, thus p = 1 – q = 1 – 0.20 = 0.80 Thus, A allele has a frequency of 0.80 and the a allele has a frequency of 0.20. Therefore, the percentage of the population that is heterozygous (Aa) and are carriers is = 2x 0.20x 0.80=0.32 or 32%.

Q 8) In the United States, approximately one child in 10,000 is born with PKU (phenylketonuria), a syndrome that affects individuals homozygous for the recessive allele (aa). (a)Calculate the frequency of this allele in the population.

ANS-Always start with the homozygous recessive data if given, which is equal to aa and to q2. The frequency of the recessive genotype, aa or q2 is 1/10, 000= 0.0001 Thus, q =sqaure root of q = 0.01

(b)Calculate the frequency of the normal allele.

ANS-A represents the “normal” allele and is also represented by p in the H-W equilibrium equations. p + q = 1 thus, p = 1 – q = 1 – 0.01 = 0.99

(c)Calculate the percentage of carriers of the trait within the population.

ANS-The heterozygotes are the carriers of the trait and are represented by the 2pq term in the H-W equilibrium equations. 2X 0.01 X 0.99=0.0198 which is then multiplied by 10,000 to obtain the number of heterozygous carriers in the population. So, 0.0198 X 10, 000 = 198 individuals in the population that carry the trait.