By Susan Elrod, Ph.D. | William Stansfield, Ph.D. — McGraw-Hill Professional

Updated on Aug 23, 2011

**Selection Methods Questions**

- The length of an individual beetle is 10.3mm or 0.5 when expressed in "standardized" form. The average measurement for this trait in the beetle population is 10.0 mm. What is the variance of this trait?
- Given the swine selection index I = 0.14W – 0.27S, where W is the pig's own 180-day weight and S is its market score. (
*a*) Rank the following three animals according to index merit: - A beef cattle index (I) for selecting replacement heifers takes the form I = 6 + 2WW' + WG', where WW' is weaning weight in standardized form and WG' is weaning grade in standardized form. The average weaning weight of the herd = 505 lb with a standard deviation of ±34:5 lb. The average weaning grade (a numerical score) is 88.6 with a standard deviation of ±2:1. Which of the following animals has the best overall merit?
- Suppose 360 ewes (female sheep) are available for proving sires. All ewes lamb; 50% of ewes lambing have twins. The 10 rams with the highest progeny test scores will be kept as flock sires. How much selection can be practiced among the progeny-tested individuals; i.e., what proportion of those tested can be saved if a test requires (
*a*) 18 progeny, (*b*) 12 progeny, (*c*) 6 progeny? - During the same year, three dairy bulls were each mated to a random group of cows. The number of pounds of butterfat produced by the dams and their daughters (corrected to a 305-day lactation at maturity with twice daily milking) was recorded as shown below.

(*b*) If differences in index score are 20% heritable, and parents score 3.55 points higher than the average of the population, how much increase in the average score of the progeny is expected?

(*a*) Calculate the sire index for each of the three sires. (*b*) Which sire would you save for extensive use in your herd?

**Mating Methods Questions**

- A is linebred to B in the following pedigree. Calculate the inbreeding coefficient of A.
- Calculate the inbreeding of A in the following. (
*Hint*: There are nine pathways between B and C) - The yield of seed (in bushels per acre) and plant height (in centimeters) was measured on several generations of corn. Calculate by formula (8.27), (
*a*) the amount of heterosis in the F_{1 }resulting from crossing the parental varieties with the inbreds, (*b*) the yield and height expectations of the F_{2}.

**Answers**

**Vocabulary**

- continuous variation
- normal (or Gaussian) distribution
- variance
- overdominance
- heritability (broad definition)
- regression coefficient
- correlation coefficient
- progeny test
- inbreeding
- heterosis

**Multiple-Choice**

- c
- e (environmental)
- e (1/4)
- a
- b
- c
- c
- b
- a
- c

**Polygenic Traits**

- Eight pairs of factors
- Three or four loci (pairs of alleles)

**The Normal Distribution**

- (
*a*) 216 lb (*b*) ±6.58 lb (*c*) 229.16lb - ±1000 lb/acre
- 4.16 or approximately four gene pairs

**Type of Gene Action**

- (
*a*) Both parents, F_{1}= 36 units (*b*) Both parents = 36 units, F_{1}= 48 units

**Heritability**

- 0.64
- (
*a*) Heritability estimates measure the proportion of the total phenotypic variation that is due to genetic variation for a trait among individuals of a population. There is no genetic variation in a pure line (heritability = 0), but blood groups (for example) would still be 100% determined by genes. (*b*) Suppose that a group of identical twins were divided, one member of each pairto the two populations A and B. Each population would then have the same genetic constitution. If population A is not given equal social, educational, and vocational opportunities with population B, then A might be expected to show lower average IQ. In other words, the average IQs of these populations would be reflective solely of nongenetic (environmental) differences regardless of the heritability estimates made in each population. (*c*) The answer to part (*b*) demonstrates that the difference between phenotypic averages of two populations does not necessarily imply that one population is genetically superior to the other. Important environmental differences may be largely responsible for such deviations. We could imagine that a pure line (heritability = 0) would be very well suited to a particular environment, whereas a highly genetically heterogeneous population with a high heritability for the same trait might be relatively poorly adapted to that same environment. In other words, the high heritabilities for IQ within populations A and B reveal nothing about the causes of the average phenotypic differences between them. - (
*a*) V_{E}= 42.0, V_{A}= 77.0, V_{D}= 11.5 (*b*) 0.55 (*c*) 0.59 - (
*a*)*x*= 4 (*b*)*x*= 4,*y*= 2 (*c*)*x*= 2 - (
*a*)*b*= 0.22 (*b*) h^{2}= 2b = 0.49 - 0.35
- (
*a*) 7.17–12.83 lb (*b*) 0.3 - (
*a*) 40 lb^{2}(*b*) 100 lb^{2} - 0.88 in
- (
*a*) 0.22 (*b*) 880, 000 lb^{2}(*c*) 16, 000–20, 000 lb - (
*a*) = 25 cg,*s*= ±3.94 cg (*b*) 15.53 cg^{2}, note that this is the square of the phenotypic standard deviation in part (*a*). In pure lines, all of the variance is environmentally induced. (*c*) h^{2}= 0, since a pure line is homozygous; there is no genetic variability. (*d*) = 25 cg; no genetic gain can be made by selecting in the absence of genetic variability.

**Selection Methods**

- 0.36 mm
^{2} - (
*a*) Y = 23.34, Z = 19.90, X = 17.84 (*b*) 0.71 point - I
_{A}= 6.526, I_{B}= 6.041, A excels in overall merit. - (
*a*) 1/3 (*b*) 22.2% (*c*) 1/9 - (
*a*) A = 630.0, B = 615.0, C = 597.5 (*b*) Sire A

**Mating Methods**

- 0.25
- 0.4297
- (
*a*) Heterosis for seed yield = 22.2 bushels/acre, for plant height = 28 cm (*b*) 60.3 bushels/acre, 243 cm

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From Schaum's Outlines of Genetics. Copyright © 2010 by The McGraw-Hill Companies. All Rights Reserved.

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