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(Solved) I recently visited a local orchard where I collected 100 individual flies, and took them back to the lab.


 I recently visited a local orchard where I collected 100 individual flies, and took them back to the lab. There, I sequenced the DNA of each fly to determine their specific genotype at the autosomal gene locus that codes for alcohol dehydrogenase (ADH; an enzyme that is necessary for flies to metabolize and detoxify ethanol).

The results of this study revealed that there were 65 individuals that were homozygous for the dominant allele E, 20 individuals that were heterozygous, and 15 individuals that were homozygous for the recessive allele e (and are thus incapable of metabolizing and detoxifying ethanol).

Is this population at Hardy-Weinberg equilibrium, and if not how are they different than would be expected?


Question 5 options:

The population is not at Hardy-Weinberg equilibrium. There are more heterozygotic and homozygotic recessive individuals than I would expect to see.

The population is not at Hardy-Weinberg equilibrium. There are more homozygotic individuals (of both types) than I would expect to see.

The population is not at Hardy-Weinberg equilibrium. There are more heterozygotic individuals (of both types) than I would expect to see.

The population is at Hardy-Weinberg equilibrium.

The population is not at Hardy-Weinberg equilibrium. There are more heterozygotic and homozygotic dominant individuals than I would expect to see.

 


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