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a) Wild type as XE and white eye as Xe. All females are Xe in the first generation, so they are XeXe. All males are XE in the first generation, so they are XEY, but all of the males in the second generation are Xe while all of the females are XE. Since the gene is X-linked, only males have the white eye. Therefore, the female being homozygous for the white eye gene because the father parent cannot pass the X chromosome to his sons. Since there are no white eye females either, so the male parent has wild-type eyes. Because all females are wild type and all males are white eye, XE is the dominant E and Xe is the recessive e. If Xe was dominant, everyone would have the same phenotype. Parental females are XeXe, and parental males are XEY. F1 females are Ee, and F1 males are eY. F2 result is 1:1:1:1 ratio.

b) There were a total of 100 offspring in F2, so the outcome of the cross is 1:1:1:1 ratio or 25% have each phenotype.
Chi-squared value= ((31-25)^2)/25 + ((23-25)^2)/25+ ((24-25)^2)/25+ ((22-25)^2)/25= 2
There are four terms, so there are 3 degrees of freedom. The critical value is 7.82. The probability of 2 is less than the critical value 7.82 for 0.05 probabilities with 3 degrees of freedom in this cross. This means that the hypothesis of a (1:1:1:1) phenotypic ratio is valid.

c) A mutation is a change in the DNA code. Two types of mutatioa: deletions and insertions Deletions are mutations in which a section of DNA is lost, or deleted. Insertions are mutations in which extra base pairs are inserted into a new place in the DNA. Theses mutations would result in the change in phenotypic expression. For instance, a different gene sequence codes for different RNA, could results in a different protein causing the difference in fly eye color.