Name: Chi-Square Test
Uploaded: 2019-07-19
Duration: 11 min 53 s
Description: Learn how to calculate the chi-squared value to test the null hypothesis, explain the importance of the critical value and define the degrees of freedom.

Question

What ratio of phenotypes would you expect from the cross of a male Drosophila from the F₁ generation discussed with a female with ebony body and aristapedia antennae if the genes for body colour and antennae shape were not linked- that is, they were on different chromosomes? Explain using a genetic diagram.

Accepted Answer

When two or more gene loci are on the same chromosome, they do not assort independently in meiosis as they would if they were on different chromosomes. In the given cross, the gene responsible for the nature of the body and antennae are not linked or they were in different chromosomes. In Drosophila, the striped body (E) is dominant over ebony (black) and normal antennae (A) is dominant over aristapedia antennae.

The genetic diagram of the cross between F₁ male (striped body with normal antennae) and female of ebony body and aristapedia antennae is given below:

The two genes (E and A) are not linked, hence, crossing over takes place and formed four types of gametes from a heterozygous male. Therefore, the phenotypic ratio is 1 : 1 : 1 : 1. Both parental and non-parental offspring are also found in F₂generation with almost equal numbers.

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