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Genetic Disorders: Annotated Bibliography

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The P1 cross was between four wmf females and nine wild-type males. The F1 progeny consisted of 12 wild-type females, and four triple-mutant males. The P2 cross resulted in 13 females, and 3 males, all with the wild-type phenotype (Table 1). The two parental crosses identify that the mutations are X-link recessive. The triple-mutant females of the P1 cross produce mutant male offspring, but wild-type females. The F1 females would be heterozygous for the mutations, but don’t express the mutations because they still have a wild-type X chromosome. However, the F1 males only have one X chromosome that comes from a mutant mother. The offspring for P1 were crossed again to make and F1 cross. The F1 cross would be X+/Y and X+/X. The F1 cross resulted in 100 F2 progenies over the course of 7 days. …show more content…

There were eight different phenotypes among the progeny. The highest phenotypic frequency was the w+m+f+ at 40% of the progeny. The lowest was the w+mf+ with only 2 % of the progeny (Table 3). The sum of the recombinant frequencies between genes, table 4, was used to determine the gene distance. The recombinant frequency was determined by counting the number of individuals whose genes differed from that of the parental type. For example, how many individuals white eye gene, and miniature wing gene, differed from both wild-type or both mutants. Recombination occurred between the white and miniature gene 33 times. Recombination occurred between the miniature and the forked genes 31 times. Recombination occurred between the white and forked genes 44 time. Double recombination occurred 10 times. Therefore, genes w and f are 64 m.u. apart, m and w are 33 m.u. apart, and m and f are 31 m.u. apart (Figure

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