11 A cancer researcher has a true breeding mouse line that h

11. 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 F1 mice are all black and do not get tumors. So the breeder sets up a test cross between F1 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 8 pts) Are the genes responsible for white color and tumors segregating independently? Use a chi squared test to support your answer.

Solution

Assume that the gene coding for black colour is B and the gene coding for white is, b; B is dominant over b. Similarly, the gene coding for non-tumour prone is, T and the gene coding for tumours is t; T is dominant over t.

Now, in the parental cross, the non-tumour prone black mouse (BBTT) is crossed to tumour prone white mouse (bb tt), and all the resulting progeny are non-tumour prone black mouse. Means, the parental mouse are true breeders.

BBTT* bb tt --à Bb Tt (100% black, non-tumour prone) --à F1

Test cross of F1: Bb Tt* bb tt ---à Bb Tt (1/4, black, non-tumour prone), Bb tt (1/4, black, tumour prone), bb Tt (1/4, white, non-tumour prone), bb tt (1/4, white, tumour prone)

Means, all the progeny are expected to occur at a ratio of 1:1:1:1

The total progeny are, 60. So, the expected number of each phenotype is, 15.

CHI - SQUARE (X²):

X² = (O - E)² / E

Where O = Observed frequency

E = Expected frequency

Phenotype

O

E

(O-E)

(O-E)^2

(O-E)^2/E

B_T_

22

15

7

49

3.266667

B_ tt

10

15

-5

25

1.666667

bb T_

9

15

1.5

2.25

0.155172

bb tt

19

15

4

16

1.066667

5.088505

The calculated Chi-square value is = 5.0885

The degrees of freedom is = n -1 = 4-1 = 3

The calculated P value is 0.165466, which is not significant at p< 0.05. So, we reject the null hypothesis.

The difference between the expected an observed phenotypic ratios is not significant and the genes are assorting independently.

Phenotype	O	E	(O-E)	(O-E)^2	(O-E)^2/E
B_T_	22	15	7	49	3.266667
B_ tt	10	15	-5	25	1.666667
bb T_	9	15	1.5	2.25	0.155172
bb tt	19	15	4	16	1.066667
					5.088505