HARDY WEINBERG PRACTICE 1. If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene? Ss or 2pq = ? ss or q2 = 9% or 0.09 q = 0.09 = 0.3 p =1–q = 1 – 0.3 = 0.7 2pq = 2 (0.7) (0.3) = 0.42 = 42% 2. This is a classic data set on wing coloration in the scarlet tiger moth (Panaxia dominula). Coloration in this species had been previously shown to behave as a single-locus, two-allele system with incomplete dominance. Data for 1612 individuals are given below: White-spotted (AA) =1469 Intermediate (Aa) = 138 Little spotting (aa) =5 Calculate the allele and genotypic frequencies: p =? q =? AA = p2 = ? Aa= 2pq = ? aa = q2 = ? aa = q2 = 5/1612 = 0.003 q = 0.003 = 0.05 p =1-q = 1 – 0.05 = 0.95 AA = p2 = (0.95)2 = 0.90 Aa = 2pq = 2 (0.95) (0.05) = 0.095 3.Seventy percent of a population has the dominant trait while 30% are recessive. What percentage of the population are carriers? A_ = 70% aa = q2 = 30% Aa = 2pq = ? aa = q2 = 0.30 q = 0.30 = 0.548 p=1–q = 1- 0.548 = 0.452 Aa = 2pq = 2 (0.452) (0.548) = 0.495 4. It is determined in a certain population that 1% of the people do not have freckles. What percentage of the population are carriers for this trait? Note: Freckles are caused by a dominant allele! ff = q2 = 1% = 0.01 Ff = 2pq = ? q2 = 0.01 q = 0.01 = 0.1 p =1–q = 1 – 0.1 = 0.9 Ff = 2pq = 2 (0.9) (0.1) = 0.18 = 18% 5. For foxes there exists a single gene that controls coat thickness. Allele C confers a thick coat while allele c a thin coat. In a certain population of 540 foxes, 49 have a thin coat. What are the allelic and individual frequencies? C – thick coat c – thin coat Population size = 540 foxes cc - 49 foxes Find p, q, p2, 2pq, and q2. cc = q2 = 49/540 = 0.09 q = 0.09 = 0.3 p=1–q = 1 – 0.3 = 0.7 CC = p2 = (0.7)2 = 0.49 Cc = 2pq = 2 (0.7) (0.3) = 0.42 6. In a certain population of flowers, when red and white flowers are crossed they produce pink flowers. A certain field has 1190 red flowers, 1070 pink flowers and 240 white flowers. What is the frequency of individuals and alleles? CRCR = 1190 flowers CRCW = 1070 flowers CWCW = 240 flowers Total flowers = 2500 Find p, q, p2, 2pq, and q2. CWCW = q2 q = 0.096 = 0.31 p =1–q = 1 – 0.31 = 0.69 = 240/2500 = 0.096 CRCR = p2 = (0.69)2 = 0.48 CRCW = 2pq = 2 (0.69) (0.31) = 0.43 7. Pea seeds can be round (R: dominant) or wrinkled (r: recessive). In a field of wild peas ninety-six percent (96%) of the peas are round. What are the allelic and genotypic frequencies? R – round r – wrinkled R_ = 96% Find p, q, p2, 2pq, and q2. If R_ = 96%, then rr = q2 = 4% = 0.04 q = 0.04 = 0.2 p =1–q = 1- 0.2 = 0.8 RR = p2 = (0.8)2 = 0.64 Rr = 2pq = 2 (0.8) (0.2) = 0.32