Biology Lab

Only available on StudyMode
  • Download(s) : 192
  • Published : October 15, 2012
Open Document
Text Preview
Jong Choi
A.P. Biology
Ms. Lia Krieg
Table 8.1: Phenotypic Proportions of Tasters and Non-tasters and Frequencies of the Determining Alleles | Phenotypes| Allele Frequency Based on the H-W Equation| | Tasters (p2+2pq)| Non-tasters(q2)| P| q|

Class Population| #| %| #| %| 0.57| 0.43|
| 13| 81.25| 3| 18.75| | |
North American Population| 0.55| 0.45| 0.33| 0.67|

Topics for Discussion
1. What is the percentage of heterozygous tasters (2 pq) in your class? ____49.02%____ 2(0.57)(0.43) = 0.4902
2. What percentage of the North American population is heterozygous for the taster trait? ____44.22%____ 2(0.33)(0.67) = 0.4422
Number of A alleles present at the Fifth Generation
Number of offspring with genotype AA __27__ X 2 = ___54___ A alleles Number of offspring with genotype Aa __42__ X 1 = ___42___ A alleles Total = ___96___ A alleles p = Total number of A alleles / Total number of alleles in the population = 96/160 = _0.60_ Number of a alleles present at the Fifth Generation

Number of offspring with genotype aa __11__ X 2 = ___22___ A alleles Number of offspring with genotype Aa __42__ X 1 = ___42___ A alleles Total = ___64___ A alleles q = Total number of A alleles / Total number of alleles in the population = 64/160 = _0.40_

Case I Questions
1. What does the Hardy-Weinberg equation predict for the new p and q? It predicts that the new p and q will be determined by a chance rather than it being picked. 2. Do the results you obtained in this simulation agree? If not, why? No the results do not agree because the population is too small which causes disequilibria. However, it is correct in the sense that the number of A and a are almost the same because there is an equal chance of getting them. 3. What major assumption(s) were not strictly followed...
tracking img