## Answers

Cross:

Wild type (trihybrid) x Pale, Oval, Short

Ll Rr Tt x ll rr tt

Progeny Phenotype | Genotype (Gamete from Tryhybrid in bold) | No. of Progeny |
---|---|---|

Pale | l R T/l r t | 648 |

Pale, Oval | l r T/l r t | 64 |

Pale, Short | l R t/l r t | 10 |

Pale, Oval, Short | l r t/l r t | 102 |

Oval | L r T/l r t | 6 |

Oval, Short | L r t/l r t | 618 |

Short | L R t/l r t | 84 |

Wild Type | L R T/l r t | 98 |

1630 |

**i) Use the Chi^2 test to determine if l & r assort independently.**

If L and R assort independently, the Trihybrid parent would produce the LR, Lr, lR and lr gametes in equal ratios. Therefore, the Progeny would also have these phenotypic classes in equal ratios. Therefore, the expected number of progeny for each class is 407.5.

Progeny Class | Observed (O) | Expected (E) | O - E | (O - E)2 | (O - E)2/E |
---|---|---|---|---|---|

LR | 84 + 98 = 182 | 407.5 | -225.5 | 50850.25 | 124.786 |

Lr | 6 + 618 = 624 | 407.5 | 216.5 | 46872.25 | 115.024 |

lR | 648 + 10 = 658 | 407.5 | 250.5 | 62750.25 | 153.988 |

lr | 64 + 102 = 166 | 407.5 | -241.5 | 58322.25 | 143.122 |

= 536.92 |

L and R do not assort independently, as the probability of observing these deviations due to chance alone is less than 0.005. For d.f = 3; Critical Value of Chi-square at the 95% confidence level is 7.815.

As 536.92 > 7.815, we can reject the null hypothesis that these loci are assorting independently.

**ii) Use the Chi^2 test to determine if l & t assort independently**.

Similarly to i), since the Chi-square value is greater than the critical value at the 95% confidence, we reject the hypothesis that the L and T alleles assort independently.

**iii) Use the Chi^2 test to determine if r & t assort independently**.

These two loci are also not assorting independently. This is because the Chi-square value for the devation from expected, is much greater than the significaant value at 95% confidence.