Solutions For An Introduction to Genetic Analysis Chapter 19 Problem 21P

The overall variance = 1/n? _i (X _i­ X) ²

The overall variance is the mean of the squared deviations.

The deviation of each individual is calculated by subtraction the overall mean from the individual readings.

The overall variance = (228-192.5)² + (186-192.5) ² + (204-192.5) ²+(142-192.5)²+(226-192.5)²+(217-192.5)²+(207-192.5)²+(185-192.5)²+(179-192.5)²+(170-192.5)²+(222-192.5)²+(152-192.5⁾²+(220-192.5)²+(185-192.5)²+(210-192.5)²+(190-192)²+(226-192.5)²+(213-192.5)²+(159-192.5)²+(129-192.5)² /20.

Total variance = 849.8.

Covariance between twins = E (X'X")

Covariance is the expected value of (X' × X").

Mean value of X' set of twins = 228+286+204+142+226+217+207+185+179+170/10

= 194.4

Mean value of X"set of twins = sum of the readings/number of readings = 190.6

Now for every pair of twins, the mean value of that particular set is subtracted from every individual's serum cholesterol. The two values for each set will be multiplied.

For the first pair of twins the calculation is as follows:

The serum cholesterol of the first individual in the first pair of twins =228

The mean value of the set X' =194.5

The serum cholesterol of the second individual in the first pair of twins=222

The mean value of the set X" = 190.6

= (229-194.5)(222-190.6)=1055

Like this the values are calculated for all the set of twins.

X' X" calculated values

1. 228 222 1055.0

2. 186 152 324.2

3. 204 220 282.2

4. 142 185 293.4

5. 226 210 613.0

6. 217 190 -13.6

7. 207 226 446.0

8. 185 213 -210.6

9. 179 159 486.6

10. 170 129 1503.0

Mean total of X' =194.4

Mean total of X"= 190.6

Mean total of the calculated values= 4779.2. This value is divided by total number of individuals= 239.0. This is called covariance.

The broad-sense heritability:

H₂ = COV_{X'X" /} V_X

COVx'x"= covariance between the twins =239.0

Vx= overall variance =849.8

The broad-sense heritability = 239.0/849.8 = 0.28.