Quantitative Genetics Practice Problems (page 3)

based on 4 ratings
By — McGraw-Hill Professional
Updated on Aug 23, 2011

Solution 9

  1. Since all of the phenotypic variance within pure lines and their genetically uniform F1 progeny is environmental, the mean of these variances produces the best estimate of the environmental variance (VE), assuming that there has been no change in the environment from one generation to the next.
  2. The phenotypic variance of the F2 has a genetic and an environmental component.
  3. The total genetic variance has both an additive and a dominance component. From page 270 we have
  4. where A = sum of all squared additive deviations from the mean, 1/2A = VA and 1/4D = VD. Multiplying formula (1) by a factor of 2 and subtracting formula (2), we have

  5. The narrow heritability estimate (h2) is
  6. h2 = VA/VF2 = 29:06/40:35 = 0:72

    Notice that h2 is smaller than H2, calculated in part (b). This indicates that most of the genetic variance is additive and relatively less is due to dominance variance.

  7. The dominance variance is calculated thus:
  8. In step (c) the following formula was given
  9. Thus, if (1/2)A = VA and (1/2)D = VD, then A = 2VA and D = 4VD, and the degree of dominance (formula (8.10)) is

Practice 10

The pounds of grease fleece weight was measured in a sample from a sheep population. The data listed below is for the average of both parents (X, midparent) and their offspring (Y).

(a) Calculate the regression coefficient of offspring on midparent and estimate the heritability of grease fleece weight in this population.   (b) Plot the data and draw the regression line.   (c) Calculate the correlation coefficient and from that estimate the heritability.

Solution 10

  1. The regression of offspring on midparent is an estimate of heritability:

    h2 = 0:67

  2. Data plot and regression line:
  3. The regression line goes through the intersection of the two means (, ); = 99:8/10 9:98; = 67:3/10 = 6:73. The regression line intersects the Y axis at the Y intercept (a; formula 8.12).

    a = b = 6:73 – 0:673 (9:98) = 0

    Now let us choose a value of X that is distant from (, ) but easily plotted on the graph (e.g., X = 8:0). The corresponding value of Y is estimated to be

    = a + bX = 0:01 + 0:673 (8:0) = 5:39

    These two points [(, ) and ] establish the regression line with slope b = 0:67. For every 1 lb increase in midparent values, offspring tend to produce 0.67 lb.

  4. The correlation coefficient (r) is calculated using formula (8.18):
  5. Therefore, the X and Y values are very highly positively correlated. Note that two variables can be highly correlated without also being nearly equal. Two variables are perfectly correlated if for one unit change in one variable there is a constant change (either plus or minus) in the other. Negative correlations for heritability estimates are biologically meaningless. Different traits, however, may have negative genetic correlations (e.g., total milk production vs. butterfat percentage in dairy cattle); many of the same genes that contribute positively to milk yield also contribute negatively to butterfat content.

Practice 11

The total genetic variance of 180-day body weight in a population of swine is 250 lb2. The variance due to dominance effects is 50 lb2. The variance due to epistatic effects is 20 lb2. The environmental variance is 350 lb2. What is the heritability estimate (narrow sense) of this trait?

Solution 11

Practice 12

The heritability of feedlot rate of gain in beef cattle is 0.6. The average rate of gain in the population is 1:7 lb/day. The average rate of gain of the individuals selected from this population to be the parents of the next generation is 2:8 lb/day. What is the expected average daily gain of the progeny in the next generation?

View Full Article
Add your own comment

Ask a Question

Have questions about this article or topic? Ask
150 Characters allowed