A truebreeding black mouse is crossed with an albino mouse p

A true-breeding black mouse is crossed with an albino mouse, producing F1 offspring that are all agouti (grizzled fur). The F1s are undercrossed, producing F2s with the following phenotypes: 48 agouti- 28 albinos, and 18 black. Actual research has revealed that fur color in this population of mice is controlled by two genes. What is the observed phenotypic ratio for the F2 generation? The researchers believe that dominant epistasis is occurring in this cross. Use the chi test to test their hypothesis. Arc they correct? Do you think another type of epistasis is occurring? Test your hypothesis with another chi square. Are you correct?

Solution

1. Observed phenotypic ratio is Agouti : albino : black :: 48 : 28: 18

OR Agouti : albino : black :: 24 : 14: 9

2. Expected frequency of dominant epistasis is 12 : 3 : 1

Observed frequency is 48:28 :18

Divide observed frequency with 5.875 ( 94/16)

Array: Obs Expected

Agouti 8.170213 12

Albino 4.65957 3

Black 3.06383 1

Total 16 16

Chi Table

Agouti (8.170213-12)²/12

Albino (4.65957-3)² /3

Black (3.06383-1)²/1

Chi Table

Agouti 1.222273

Albino 1.039535

Black 4.259393

6.521201

The p value for this chi square at df 2.0 is much smaller than 0.05

Thus, these frequencies does not match.

Conclusion: This is not a case of dominant epistasis

3. Expected frequency of recessive epistasis is 9:4:3

Observed frequency is 48:28 :18

Divide observed frequency with 5.875 ( 94/16)

Array: Obs Expected

Agouti 8.170213 9

Albino 4.65957 4

Black 3.06383 3

Total 16 16

Chi Table

Agouti (8.170213-9)²/9

Albino (4.65957-4)² /4

Black (3.06383-3)²/3

Chi Table

Agouti 0.076505

Albino 0.146673

Black 0.001358

0.224536

The p value for this chi square at df 2.0 is nearly 90%

Thus, observed and expected frequencies match.

Conclusion: This is a case of recessive epistasis