Question

In an experiment using Drosophila melanogaster, white-eyed females were crossed with wild-type (red-eyed) males. The F1 flies were then crossed among themselves. The following data was collected: $224$ white-eyed females, $218$ wild-type females, $243$ white eyed males, $215$ wild type males.
According to the chi-square test, what can be concluded about the validity of these results?

• A. The chi-square value is $2.1$ which is less than the critical value of $7.81$ with $3$ degrees of freedom, $p=.05$. Therefore, there is no significant difference between observed and expected values.
• B. The chi-square value is $7.81$ which is greater than $3$ degrees of freedom, $p=.05$. Therefore, the is a significant difference between observed and expected values.
• C. The chi-square value is $900$ which is greater than the critical value of $7.81,p=.05$. Therefore, there is a significant difference between observed and expected values.
• D. The chi-square value is $2.1$ which is less than the critical value of $9.49$ with $4$ degrees of freedom where $p=.05$. Therefore, there is no significant difference between observed and expected values.

Answer 1

The correct option is A The chi-square value is $2.1$ which is less than the critical value of $7.81$ with $3$ degrees of freedom, $p=.05$. Therefore, there is no significant difference between observed and expected values.

Eye colour in Drosophila is controlled by a sex-linked gene. Red eye colour is dominant than the white one.

Here is the notation for the genotypes of the male, female Drosophila with Red or white eye colour.

X${}^R$Y - wild-type males with red-eye colour

X${}^r$Y - wild-type males with white-eye colour

X${}^R$X${}^R$ or X${}^R$X${}^r$ - wild-type females with red-eye colour

X${}^r$X${}^r$ - wild-type females with white-eye colour

According to the question given above the initial cross was made between white-eyed females and wild-type (red-eyed) males

X${}^R$Y X X${}^r$X${}^r$

	X${}^R$	Y
X${}^r$	X${}^R$X${}^r$	X${}^r$Y
X${}^r$	X${}^R$X${}^r$	X${}^r$Y

So in the F1 generation, there will be 50% females with red eyes (X${}^R$X${}^r$) and 50% males with white eyes (X${}^r$Y )

In the second part of the statement, it says the F1 flies were then crossed among themselves. Ideally, this is what is expected when the F1 cross is done

X${}^R$X${}^r$ X X${}^r$Y

	X${}^r$	Y
X${}^R$	X${}^R$X${}^r$	X${}^R$Y
X${}^r$	X${}^r$X${}^r$	X${}^r$Y

So, the expected outcome is

white-eyed females ( X${}^r$X${}^r$) - 1

wild-type females (X${}^R$X${}^r$) - 1

white eyed males - (X${}^r$Y) - 1

wild-type males - (X${}^R$Y) - 1

So the expected ratio would be 1:1:1:1

In order to do the Chi-square test, we need to calculate the frequencies and the degree of freedom

chi-square = Sum of (Observed frequency - Expected frequency )${}^2$

${}^{---------------------------------}$

Expected Frequency

Expected frequency (E) = Expected ratio X Observed total

Observed total, in this case, is 224+218+243+215 = 900

The expected ratio would be 1/4 since the phenotype ratio is 1:1:1:1

	white-eyed females	wild-type females	white-eyed males	wild-type males
Observed (O)	224	218	243	215
Expected (E)	900 X 1/4 = 225	900 X 1/4 = 225	900 X 1/4 = 225	900 X 1/4 = 225
(O-E)	-1	-7	18	-10
(O-E)${}^2$ ${}^{------}$ E	(-1)${}^2$/225 = 0.004	(-7)${}^2$/225 = 0.217	(18)${}^2$/225 = 1.44	(-10)${}^2$/225 = 0.44

From this table the Chi sqare can be calculated

Chi sqaure = 0.004+0.217+1.44+0.44

Chi square = 2.1

which is already given in the question.

Degrees of freedom is given by n-1 where n is the number of phenotypes. In this case, 4 phenotypes are given, so the degree of freedom = 4-1 = 3

Now with these values of Chi-square and degree of freedom, if we look at the chi-square value table, A value of 2.1 lies between p values of 0.5 and 0.7.

In this case, a value of greater than 6.25 is required for results to be considered statistically significant

If the p-value is greater than 0.05 then the results are considered to be insignificant. So the correct answer is A.