## Answers

a)

Null hypothesis: The pea plant flower color is a simple Mendelian trait.

Alternative hypothesis: The pea plant flower color is not a simple Mendelian trait.

b)

df = n - 1. where n = number of expected phenotype.

So, df = 2 - 1 = 1

c)

Total number of observed flowers,

Purple = 80 , White = 25,

So, total number of flowers = 80 + 25 = 105.

The expected Mendelian ratio for monohybrid cross = 3 : 1.

So, the expected number of purple flowers = 105 * (3/4) = 78.75.

and the expected number of white flowers = 105 * (1/4) = 26.25.

Phenotype | Expected Frequency | Observed (O) | Expected (E) | O-E | (O-E)^{2} /E |

Purple | 3/4 | 80 | 78.75 | 1.25 | 0.02 |

White | 1/4 | 25 | 26.25 | -1.25 | 0.06 |

Total | 1 | 105 | 105 | 0 | 0.08 |

d) So, the optained p value = 0.08

e) For df = 1 and significance level 0.05 the maximum p value we can count is 3.841.

Here, p value = 0.08.

So, we can not reject the null hypothesis. { We can only reject the null hypothesis if the p value comes beyond 3.841}

So, the flower color trait is a Mendelian trait.

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