Chapter 2, Problem 32P

Summary Introduction

To analyze:

In a sample of $\text{120}$ families having $\text{4}$ children each, the parents are carriers of an autosomal recessive mutation i.e. cystic fibrosis (CF). This results in the following distribution of children having CF and lacking CF.

$\begin{array}{cccccc}\text{No}\text{.}\text{of}\text{families}& \text{16}& \text{52}& \text{32}& \text{18}& \text{2}\\ \text{Children}\text{with}\text{CF}& 0& 1& 2& 3& 4\\ \text{Children}\text{without}\text{CF}& 4& 3& 2& 1& 0\end{array}$

a. By calculating the chi-square analysis, check whether the number of children with CF in these families are reliable to the expected ratio?

b. Under expectations of binomial probability, what is the expected distribution of CF children in families with $0$ through $4$?

c. Under expectations of binomial probability, is the above distribution of CF children in families with $0$ through $4$ reliable to the expected ratios of binomial probability?

Introduction:

In genetic crosses, to determine whether the data is significant or whether it fits in any of the Mendelian ratios, the Chi-square test ${\text{(χ}}^{\text{2}}\text{)}$ is performed. When a genetic cross is performed, the genotypes are given and the phenotypic ratio is predicted by a hypothesis. Chi-square test ${\text{(χ}}^{\text{2}}\text{)}$ allows determining the accuracy of the prediction or hypothesis on the observed and expected values or ratios. For Mendelian genetic crosses, Chi-square ${\text{(χ}}^{\text{2}}\text{)}$ Goodness of Fit (one sample) test is performed.

The test is performed in a series of steps- 1. Hypotheses (Null and Alternative), 2.Calculating chi-square test statistic, 3. Determine the Degree of Freedom df and p-value, 4. The decision about the null hypothesis, 5.Conclusion.

The formula for test statistics ${\text{χ}}^{\text{2}}\text{=}{\displaystyle \sum \frac{\text{(O-Ε)2}}{\text{Ε}}}$.