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Quantitative Genetics Practice Problems (page 4)

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

Solution 12

Practice 13

Fifty gilts (female pigs) born each year in a given herd can be used for proving sires. Average litter size at birth is 10 with 10%mortality to maturity. Only the 5 boars (males) with the highest sire index will be saved for further use in the herd. If each test requires 18 mature progeny, how much culling can be practiced among the progeny-tested boars; i.e., what proportion of those tested will not be saved?

Solution 13

Each gilt will produce an average of 10 – (0:1) (10) = 9 progeny raised to maturity. If 18 mature progeny are required to prove a sire, then each boar should be mated to 2 gilts. (50 gilts) / (2 gilts per boar) = 25 boars can be proved. 20/25 = 4/5 = 80% of these boars will be culled.

Practice 14

Given the following pedigree with butterfat records on the cows and equal-parent indices on the bulls, estimate the index for young bull X   (a) using information from A and B,   (b) when the record made by B is only one lactation, and that made in another herd.

Solution 14

  1. The midparent index (estimate of transmitting ability) for X is (750 + 604)/2 = 677.
  2. Since we cannot rely on B's record, we should use information from C and D, recalling that X is separated by two Mendelian segregations from the grandparents. Then X = 750/2 + 820/ 4 + 492/4 = 703.

Practice 15

Calculate the inbreeding coefficient for A in the following pedigree.

Solution 15

First we must convert the pedigree to an arrow diagram.

There is only one pathway from B to C and that goes through ancestor E. However, ancestor E is himself inbred. Note that the parents of E are full sibs; i.e., G and H are 50% related (see Example 8.17). By formula (8.23),

FE = (1/2)RGH = 1/2 (0:5) = 0:25

The inbreeding coefficient of A is given by equation (8.26),

FA = (1/2)n (1 + Fancestor) = (1/2)3 (1 + 0:25) = 0:156

where n is the number of arrows connecting the individual (A) through one parent (B) back to the common ancestor (E) and back again to the other parent (C).

Practice 16

The average plant heights of two inbred tobacco varieties and their hybrids have been measured with the following results: inbred parent (P1 ) = 47:8 in, inbred parent (P2) = 28:7 in, F1 hybrid (P1 × P2) = 43:2 in. ´ (a) Calculate the amount of heterosis exhibited by the F1.   (b) Predict the average height of the F2.

Solution 16

  1. The amount of heterosis (formula (8.27)) is expressed by the excess of the F1 average over the midpoint between the two parental means.
  2. As a general rule, the F2 shows only about half the heterosis of the F1: 1/2 (4:95) = 2:48. Hence, the expected height of F2 plants = 38:25 + 2:48 = 40:73 in.
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