A cancer researcher has a true breeding mouse line that has

A cancer researcher has a true breeding mouse line that has a tendency to get tumors. The tumor-prone mouse is also white. The breeder has another black mouse that tends to not get tumors. The breeder crosses the two mice and the FI mice are all black and do not get tumors. So the breeder sets up a test cross between FI mice and the white parent. The resulting progeny have the following phenotypes: Black with tumors: 10 Black without tumors: 22 White with tumors: 19 White without tumors: 9 Are the genes responsible for white color and tumors segregating independently? Use a chi squared test to support your answer.

Solution

White and Tumor prone X black and not prone to tumor

F1 – All are black with without tumors

Black and without T X White and with T

10 = black with T

22 = black without T

19 = white with T

9 = white without T

The segregation of white and tumor traits are independent according to the usual ratio is 1:1:1:1. It comes under single gene segregation.

A test cross must have 1:1:1:1 ratio (expected)

But, this cross has 10:22:19:9 ratio (observed)

The observed ratio can approximated as 1:2:2:1

Chi-square value = sum of all (observed-expected)² /expected

Degrees of freedom = 4-1=3

Chi-square value = (1-1)²/1 + (2-1)²/1 + (2-1)²/1 + (1-1)²/1

= 0/1 + 1/1 + 1/1 + 0/1 = 2

If we look at the chi-square table for the value of 2 against 3 as degrees of freedom, we can find that the probability of the correctness lies between 0.5 and 0.75.

In the statistical convention, the chi-square value lying at the probability of 0.05 is considered as critical value for a particular degree of freedom. If the value is less than the critical value, the hypothesis is accepted. Otherwise it is rejected.

The critical value at 3 as degree of freedom in the table is 7.81. The calculated chi-square value is 2 which is less than the critical value. So, the hypothesis is accepted which means that the observed ratio fits into the expected ratio of 1:1:1:1.