Hardy Weinberg Problem Set 4. There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. In the highly unlikely event that these traits are genetic rather than environmental, if these traits involve dominant and recessive alleles, and if the four (4%) represent the frequency of the homozygous recessive condition, please calculate the following: aa = 92 A. The frequency of the recessive allele. 0.Z B. The frequency of the dominant allele. 0.8 C. The frequency of each genotype (AA, Aa and aa). 0.64, 0.32 , 0.04 A 50.04 = 0.2 B. 1 - 0.2 = 0.8 C . - AA : 0.82 = 0.64 - Aa : 2 (0 . 8) ( 0 . 2 ) = 0.32 - aa : 0.04 5. Rock pocket mice data: A. If there are 120 rock pocket mice with light colored fur and 48 with dark-colored light - 0.71 fur caught on light colored substrates in a population, what is the frequency of light colored mice in this population? Dark colored? Light colored is recessive. dark- 0.025 B. If there are 3 rock pocket mice with light colored fur and 54 with dark-colored fur caught on dark colored substrates in a population. Calculate the overall frequencies of light colored mice and dark colored mice caught on dark colored substrates. C. Using the Hardy-Weinberg equation and data from letter A, determine the number of mice with the DD and Dd genotypes on the light, rocky, granite substrate. D. Using the Hardy-Weinberg equation and data from letter B, determine the number of mice with the DD and Dd genotypes on the dark, rocky, lava substrate. p2 + 2pq + 92 = 1 A. 92 = 120/168 = 0.71 p + 9 = 1 50 . 71 = 0. 84 = 9 1 - 0.84 = 0.16 = P P2 = 0.0256 0.0256 + 2 (0.84) ( 0.16 ) + 0.71 = 1 B .Hardy Weinberg Problem Set 92 6. Within a population of ladybugs, the color red (B) is dominant over the color orange (b). In a certain population, 46 of 200 ladybugs are orange. Given this information, calculate the following: 2pq A. The percentage of ladybugs in the population that are heterozygous. 0. 4992 x 0.5 B. The frequency of homozygous dominant individuals. 0.2704 x 0.21 C. The frequency of each allele. p=0.52 ; 9=0.48 92 = 46/200= 0.23 2 ( 0. 52 ) (0.48 ) = 0.49 92 = 0.5 93 0.48 P= 0.52 p 2 = 0.2704 7. After winning the lottery, Johnny and 49 of his closest friends charter a plane to go on a round-the-world tour. Unfortunately, they all crash land (safely) on a deserted island. No one finds them and they start a new population totally isolated from the rest of the world. Five of his friends are homozygous recessive for attached earlobes. Assuming that the frequency of this allele does not change as the population grows, what will be the incidence of attached earlobes on this island? 0.46 or 461- p 2 If the population grows to 543, how many will be heterozygous? _238 people 92 = 5/49 = 0.102 9= 0.32 P = 0.68 2 (0.32 ) (0.68 ) = 0.44 or 44% P 2 = 0.4624 543 - 0.44= 238. 92Hardy Weinberg Problem Set 8. 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: 2 White-spotted (AA) =1469 Intermediate (Aa) = 138 Little spotting (aa) =5 1612 = J.tal Calculate the allele frequencies ( p and q ). AA 2130 Give the numbers of each genotype should the population grow to 3000. Aa 9.3 aa 258 1469/ 1612 = 0.91 = P3 5 / 1412 = 0.0031 = 92 138 /1612 = 0.086 = 2 pq P : 0.95 9: 0.06 0- 114 9. The allele for a widow's peak (hairline) is dominant over the allele for a straight P ? hairline. In a population of 500 individuals, 25% show the recessive phenotype. How many individuals would you expect to be homozygous dominant and heterozygous for the trait? What is the frequency of each allele? AA Aa 92= 0.25 AA : 0. 25 . 500= 125 people 9 = 0.5 Aa : 0.5. 500: 250 people P = 0.5 P 2 = 0.25 2 (0.5 ) (0.5 ) = 0.5 To. The allele for a hitchhiker's thumb is recessive compared to straight thumbs, which are dominant. In a population of 1000 individuals, 510 show the dominant phenotype. How many individuals would you expect for each of the three possible genotypes for this trait. p 2 = 510/ 1000= 0.51 P = 0.71 92 = 490/1000= 0.49 9 : 0.7