Chapter 24.3, Problem 35AQP

Summary Introduction

To calculate:

The regression coefficient of mean offspring length against mean parental length and also to estimate the narrow-sense heritability for the trait.

Introduction:

The regression line is mathematically expressed as follows:

$y=a+bx$

Here, y is the mean offspring length of the middle joint of right index finger.

$x$  is the mean parental length of the middle joint of the right index finger.

$b$  is the slope of the regression line (regression coefficient).

$a$  is the intercept of the line.

Explanation of Solution

Regression coefficient $\left(b\right)$ can be calculated as follows:

$b=\frac{{\mathrm{cov}}_{xy}}{s_x^2}$

Here, ${\mathrm{cov}}_{xy}$ is the covariance of the x and y.

$s_x^2$ is the variance of x.

xi	yi	$\left(x_i-\overline{x}\right)$	$\left(y_i-\overline{y}\right)$	${\left(x_i-\overline{x}\right)}^2$	${\left(y_i-\overline{y}\right)}^2$	$\left(x_i-\overline{x}\right)\times \left(y_i-\overline{y}\right)$
30	31	-1.7	-1.4	2.89	1.96	2.38
35	36	3.3	3.6	10.89	12.96	11.88
28	31	-3.7	-1.4	13.69	1.96	5.18
33	35	1.3	2.6	1.69	6.76	3.38
26	27	-5.7	-5.4	32.49	29.16	30.78
32	30	0.3	-2.4	0.09	5.76	-0.72
31	34	-0.7	1.6	0.49	2.56	-1.12
29	28	-2.7	-4.4	7.29	19.36	11.88
40	38	8.3	5.6	68.89	31.36	46.48
33	34	1.3	1.6	1.69	2.56	2.08
Total =317	324			${\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}$=140.1	${\displaystyle \sum {\left(y_i-\overline{y}\right)}^2}$= 114.4	${\displaystyle \sum \left(x_i-\overline{x}\right)\left(y_i-\overline{y}\right)}$= 112.2

Mean of x_i and y_i are as follows:

$\begin{array}{l}\text{Mean }\left(\overline{x}\right)=\frac{{\displaystyle \sum x_i}}{n}\\ \text{Mean}\left(\overline{x}\right)=\frac{317}{10}\\ \text{Mean}\left(\overline{x}\right)=31.7\\ \text{Mean }\left(\overline{y}\right)=\frac{{\displaystyle \sum y_i}}{n}\\ \text{Mean}\left(\overline{y}\right)=\frac{324}{10}\\ \text{Mean}\left(\overline{y}\right)=32.4\end{array}$

Covariance is calculated as follows:

$\begin{array}{l}{\mathrm{cov}}_{xy}=\frac{{\displaystyle \sum \left(x_i-\overline{x}\right)\left(y_i-\overline{y}\right)}}{n-1}\\ {\mathrm{cov}}_{xy}=\frac{112.2}{10-1}\\ {\mathrm{cov}}_{xy}=\frac{112.2}{9}\\ {\mathrm{cov}}_{xy}=12.47\end{array}$

Now, the variance of x can be calculated as follows:

$\begin{array}{l}{s}_x{}^2=\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}\\ s_x{}^2=\frac{140.1}{10-1}\\ s_x{}^2=\frac{140.1}{9}\\ s_x{}^2=15.57\end{array}$

Now, putting the values of variance and covariance in the formula of the regression coefficient as follows:

$\begin{array}{l}b=\frac{{\mathrm{cov}}_{xy}}{s_x^2}\\ b=\frac{12.47}{15.57}\\ b=0.80\end{array}$

The value of the regression coefficient is 0.80 and the narrow-sense heritability is equal to the regression coefficient. So, narrow-sense heritability is also 0.80.

Conclusion

The value of the regression coefficient and narrow-sense heritability is 0.80.