1. The Chi Square (c2) test:
Expected Counts (Ei)
Observed Counts (Oi)
(di) = (Ei-Oi)
di2 = (Ei-OI)2
a. c2 = S (di2/Ei) = 0.009
b. Results indicate that the null hypothesis supported.
2. Human Phenotypes:
a. The data:
b. Do the ratios of the traits show any particular pattern? Should they? i. No, the ratios should not show any particular pattern. Although it might be expected that dominant phenotypes will dominate among the individuals, it is not the case all the time. Consequently, there is not particular pattern being displayed here. As one can observe, the 1st, 4th and 5th traits show domination of the dominant phenotype. On the other hand, the 2nd and 3rd traits the recessive phenotype occurs at higher rates. c. For each trait, is the dominant phenotype always more abundant among your class members? i. No
d. Assuming my class is representative of the general population, what can you conclude about the dominant and recessive phenotypes and their frequencies in the population? i. The dominant phenotypes do not always occur more commonly than recessive phenotypes do; that is, the dominancy of a phenotype in the population is independent of the dominance of the corresponding allele. In other words, the phenotype that has the higher allele frequency will be more abundant in the population. e. Newlyweds Bill and Sue are non-freckled. Since each had one parent who had freckles, they wonder what the possibility is of their children having freckles. What would you tell them? i. The two newlyweds have the recessive phenotypes (ff); that is, there are no dominant alleles in their genotypic constitution. Consequently, the possibility of their children having freckles is 0%.
f. Mary has freckles but her husband Dick does not. Mary’s father has freckles but her mother does not. What is the probability that Mary and Dick’s child will have freckles? i. Dick has recessive phenotype (ff). Mary has dominant phenotype. However, she is heterozygous (Ff) because her mother was non-freckled. The probability is 50%.