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Fly Lab

Essay by   •  June 4, 2011  •  Lab Report  •  388 Words (2 Pages)  •  943 Views

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Fly Lab

The objective of this was to study patterns of inheritance through fruit flies. A gene is a part of DNA. A single gene establishes one thing about your body. For example there is a gene for hair color, eye color, and skin color. A specific type of a single gene for one trait is called an allele. The set of alleles present for a specific trait is called the genotype. Some alleles are dominant over others. When an allele is dominant, it will be exhibited, covering up the presence of any other allele present in the genotype. Since a human or a fly has two alleles for each trait, they pass on one of their alleles to their offspring, while the mate passes on half of theirs to complete

the genome (complete set of genes) of the offspring. Chromosomes, or the condensed form of the genes, are passed on (with their genes) independently of each other, and this method is

called "Independent Assortment". When talking about fruit flies, a "wild type" fly is a normal fly - this fly has the most commonly seen traits, the alleles of which are dominant. These traits are red eyes, tan bodies, and normal wings. Any other trait noticed is a mutation. Mutations can cause white eyes, ebony bodies,vestigial wings (shrivelled), and/or complete lack of wings. In my case I had ebony bodies and vestigial wings. The P-Generation (parent generation) is the first male and female that mate. In my case I guessed that they were both Heterozygous. The offspring of these specific parents, referred to as the F1- Generation (fillial-1 generation). The third generation, the F2-Generation, are the offspring of any 2 individuals from the F1-Generation.

Hypothesis: My hypothesis was that the actual ratios would turn out somewhere near the expected ones(from Mendel's experiment).


Date Counted W/N W/eb vg/W vg/eb

5/23 17 10 7 0

5/25 60 28 23 4

5/31 30 6 5 1

Total 107 44 35 5

Actual Ratio 8.9 3.6 2.9 .4

Expected Ratio 9 3 3 1

A=normal a=ebony B=normal b=vestigial

P1:Possible genotypes: AaBb


B b


b Bb bb

A a


a Aa aa

F2: Has every possible genotype.

Data: Chi Square:



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