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Drosophila Work Essay

1147 words - 5 pages

Vial1 contains the first filial (F1) generation of the offspring of the parents which were crossed, wild type (+/+) x vestigial (v/v).

Alleles |
v |
v |

+ |
+v |
+v |
+ |
+v |
+v |

Fig1. Shows a monohybrid cross examining a single trait for the F1 generation

Using this monohybrid cross, we are able to predict as best we can, the phenotype of the offspring. Here we see that the wild type phenotype will be present 100% of the time (fig1).

From the results gathered on the day, observations showed that there was only one phenotype present on ...view middle of the document...

Alleles |
+ |
c |

+ |
++ |
+c |

+ |
++ |
+c |
The ratio which I gather from vial two was 12 wild type to 5 mutant; which account to approximately 71 %wild type: 29% mutant phenotype. The class results showed a ratio to of (3:1). The results which I gathered do not exactly account for a 3:1 ratio, however this could have been due to some experimental errors. If this experiment was to be repeated, only one person should do the counting instead of two in every vial.

Comparing my result and the class results, to the expected pattern from the monohybrid crossed, with two F1 heterozygous alleles (+/v) x (+/v) to produce the F2 off springs, I think that the proportion of wild type to the mutant flies in vial 2 is what I expected. That the ratio of wild type to mutant was 3:1. For every 4 flies, one would be mutant and three would be wild type, this meant that there was a 25% mutant phenotype in vial 2 according to the monohybrid inheritance pattern.

In this vial we see three genotypic ratios, 1(+/+): 2(+/v): 1(v/v). The dominant allele in this cross is + (wild type). Therefore we have three of the offspring which will have a phenotype (wild type) as they carry one of the (+) alleles which is masking the recessive (v) allele (Brooker et al., 2014). However due to this cross being done with two heterozygous parents, we see one of the offspring having a recessive trait (v/v), vestigial phenotype (Omics international, 2014).

Wild type (+/+) x curly (+/c) parents were crossed and in vial 3 contained the F1 generation of offspring.

Fig 3. Shows a monohybrid cross examining a single trait (dominant cross).

The ratio which I gather from vial 3 was approximately; 1 wild type phenotype: 1 curly phenotype (mutant). The class results also showed an approximate ratio to my individual result (1:1).

Comparing my result and the class results, to the expected pattern from the monohybrid, I think that

the proportion of wild type to the mutant flies in vial 3 is what I expected. That the ratio of wild type to mutant was 1:1. Therefore, for every 4 flies, two would be mutant and two would be wild type, this meant that there was a 50% mutant phenotype in vial 3. This is due to the parent with the dominant phenotype being a Heterozygous (+/c), this will produce half of the offspring mutant (+/c) and the other wild type (+/+). If the mutant was a homozygote (c/c) as it’s the dominant phenotype, this would have produced all mutants, whilst crossed with the wild type (+/+) (Brooker et al., 2014).

The proportion of mutant flies in the F2 generation of the...

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