9 pts
6 pts
6 pts
6 pts
KEY
Please show your work.
1) In an open-pollinated landrace of corn, the purple stem characteristic is determined by a single recessive gene. You grow a sample of this landrace in your garden and observe that 24 plants have purple stems and 126 plants have green stems. Assuming that the population is in
Hardy-Weinberg Equilibrium, how many of the green plants would you expect to be heterozygous for this trait? q
2
= 24/150 = .16 q = 0.4 p = 1-q = 0.6
E(A
1
A
2
) = 2pq*150 = 72
2) A breeder mixes equal quantities of seed of an A
1
A
1
B
1
B
1
genotype and an A
2
A
2
B
2
B
2 genotype to form a new, random-mating breeding population. a) What is the value of D (the disequilibrium coefficient) in the initial generation?
D = P
A1B1
– p
A1 p
B1
= 0.50 – (0.5)(0.5) = 0.25 b) What is the value of r
2
in the initial generation? r 2 = D 2 / p
A1 p
A2
p
B1 p
B2
= 0.25
2 /(0.5*0.5*0.5*0.5)=1 c) What value of D would be expected after one generation of random mating if the loci are
20 cM apart on the same chromosome?
D t=1
= (1-c)D t=0
= 0.8*0.25 = 0.20
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9 pts
8 pts
4 pts
6 pts
3) In an idealized population, the change in inbreeding in a single generation is

F=1/2N.
Consider a breeding program for a cross-pollinating crop. List at least 3 factors that could cause the effective population size N e
to be less than the apparent population size N in a breeding population.
 assortative mating
 different numbers of male and female parents
 different N in successive generations or a bottleneck in an earlier generation
 selection for particular alleles
 partial selfing
4) Assume that the parents in the mating scheme to the right are not inbred.
A B a) Which individuals in the pedigree are inbred? what are their inbreeding coefficients?
F
D
=(1/2)
2
=1/4
F
E
=(1/2) 3 =1/8 b) Use your results from part ‘a’ to determine the coancestry of A and D.

AD
=F
E
=(1/2)
3
=1/8 c) What is the coancestry of C and D?
C
E
D

CD
= ½(

CC
+

BC
) =1/2(1/2+1/4)=3/8
2

6 pts
6 pts
8 pts
6 pts
5) A random-mating population of pearl millet is segregating at the ‘A’ locus, which determines grain protein content. The table below shows the frequency of genotypes and their genotypic values. a) What are the coded values (a, d, -a) for these genotypes?
A
1
A
1
Frequency Value
0.49 14
Coded
Value a = 1
A
1
A
2
A
2
A
2
0.42
0.09
13.5
12 d = 0.5
a = -1
MP = 13 b) Calculate the population mean on the original scale.
Average =
Σ(f i
*Value) = 0.49(14)+0.42(13.5)+0.09(12) = 13.61
Could also solve from formula: Average = MP + a(p-q)+ 2 pqd p = 0.49+0.42/2 = 0.7 q = 1-0.7 = 0.3
Average = 13 + 1*(7-0.3)+2(0.7)(0.3)(0.5) = 13 + 0.61 = 13.61

= a + d ( q p ) = 1+0.5(0.3-0.7) = 1-0.2 = 0.8
Expected breeding value of A
1
A
2
= (q-p)

= (0.3-0.7)(0.8) = -0.32 c) If you crossed an A
1
A
1
genotype with an A
2
A
2
genotype, what would be the expected breeding value of the progeny? d) What is the dominance variance for this locus?

D
2
= 4p
2 q
2 d
2
= 4*0.7
2
*0.3
2
*0.5
2
= 0.0441
3

8 pts
12 pts
6) You harvest seed from a single plant in a large, out-crossing meadowfoam population. What is the expected covariance among the progeny from that plant?
Cov
1
2
HS 4
A
7) You are discussing the concept of heritability with an animal geneticist. He maintains that your definition is incorrect. How would you explain and defend the definition of heritability that is commonly used by plant breeders?
Animal geneticists define heritability on the basis of individuals. Thus, narrow sense heritability is the ratio of the additive genetic variance divided by the total phenotypic variance in a population. Plant breeders generally define heritability on a family mean basis.
Traits are measured on plots that are replicated in blocks and over several environments. The advantage of this definition is that estimates of heritability can be used directly to predict response to selection for various types of germplasm (inbred lines, clones, families in a recurrent selection program). The disadvantage is that the heritability estimates are unique to a particular set of experimental conditions (plot size, number of blocks, number of locations).
Despite this disadvantage, this definition is widely utilized, and information from many trials and traits provide useful insights about the relative importance of genotype and environment in determining phenotype. Many of the traits of greatest interest such as yield have little meaning when expressed on an individual plant basis.
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