In this question and the next we will work on using chi square with a dihybrid c
ID: 144081 • Letter: I
Question
In this question and the next we will work on using chi square with a dihybrid cross to test the hypothesis of independent assortment. Consider 2 mutant genes in a fruit fly, one causes ebony body color and one causes vestigial wings (atrophied). Wildtype versions of these traits are gray body and normal wings. The mutations are recessive.
You cross two dihybrid mutants to each other. This means each parent is heterozygous for the two mutations. You do not have to make a Punnett square to solve (if you think about ratios that Mendel observed from dihybrid crosses and independent assortment of alleles). However, if you want to practice this, the genotypes of these parents are:
e+e vg+vg and e+e vg+vg (remember, in fruit fly the + denotes the normal version of a gene and the letter without plus is the mutant allele). Because the two mutations are recessive, the phenotype of both parents is gray body and normal wings.
After the cross you observe 84 offspring. The phenotype results you obtain are:
gray body, normal wings: 44
gray body, vestigial wings: 19
ebony body, normal wings: 12
ebony body, vestigial wings: 9
To make sure you are on the right track, first determine the expected number (not percentage or fraction) of gray body, normal wing flies and enter it here. Do not enter units.
Explanation / Answer
Gray Body (E) > Ebony Body (e)
Gray Body genotypes = EE, Ee, eE
Ebony Body genotypes = ee
Normal Wings (V) > Vestigeal Wings (v)
Normal Wings genotypes = VV, Vv, vV
Vestigeal Wings genotypes = v v
Genotypes of the parents:
Heterozygous Gray Body, Normal Wing = EeVv
Heterozygous Gray Body, Normal Wing = EeVv
As per Mendelian Laws of Inheritance (Segregation, Independent Assortment, and Dominance):
As such, the expected number of gray body, normal wing flies would be in 9/16 ratio.
Since there are 84 offsprings, so expected number is (9/16)*84 = 47.25
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