CSS 650 Advanced Plant Breeding
Midterm 2011
Name
KEY
Please show your work.
1) A single recessive gene in meadowfoam causes flower buds to be glabrous (no pubescence).
The species is normally cross-pollinated by honey bees. On a vacation to Salt Spring Island
in British Colombia, a plant breeder discovers an isolated population of meadowfoam and
observes that the frequency of glabrous plants is 0.16.
8 pts
a) If the population is in Hardy-Weinberg Equilibrium, what is the expected frequency of
plants that are homozygous for the dominant allele?
q2 = 0.16
q = 0.4
p= 1-0.4 = 0.6
p2 = 0.36
8 pts
b) She selected 50 pubescent plants and crossed all of them to glabrous plants. How many of
the 50 pubescent plants would you expect to produce progeny that are segregating for the
pubescent/glabrous trait?
Frequency of heterozygotes among pubescent plants = 2pq/(p2+ 2pq)=0.48/0.84=0.57
E(A1A2) = 0.57*50 = 28.57 29
2) In a new synthetic variety of corn, the frequency of an allele ‘A’ is 0.6, and the frequency of
allele ‘B’ at another locus is 0.6. The frequency of ‘AB’ gametes in the current generation is
0.55.
6 pts
a) What is the value of D (the disequilibrium coefficient) in this generation?
D = PAB - pApB = 0.55 – (0.6)(0.6) = 0.19
8 pts
b) Under ideal conditions, the disequilibrium will be reduced by one-half in each subsequent
generation. Name two factors that could slow down the rate of decay of disequilibrium
between these two loci.
Linkage
Selfing or inbreeding
Small population size
New migration or mutations
Selection
1
6 pts
3) You have been given 10 corn kernels from a very large, random-mating population. You
plant the corn in your garden. You are the only person in your neighborhood growing corn.
What is the expected level of inbreeding after one generation of random mating?
F = 1/2N = 1/20 = 0.05
4) Assume that the parents in the mating scheme below are not inbred.
C
A
B
D
E
X
8 pts
Y
a) What is the inbreeding coefficient of Y?
FY=(1/2)3+(1/2)3=1/4
b) What is the coancestry of X and Y?
10 pts
CD and CE are half-sib relationships = 1/8
AC and AD are parent-offspring relationships = 1/4
XY = ¼(CD+CE+AD+AE) = ¼(1/8+1/8+1/4+1/4) = 3/16
Another solution
XD = XE =1/2(AD+CD)= 1/2(1/4+1/8)=3/16
XY = 1/2(XD+XE) = 1/2(3/16+3/16) = 3/16
2
5) A random-mating population of corn is segregating at the ‘A’ locus, which determines days
to flowering. The table below shows the frequency of genotypes and their observed
genotypic values.
a) Fill in the coded values for these genotypes.
Frequency
Value
Coded
A1A1
0.36
58
a=4
A1A2
0.48
55
d=1
A2A2
0.16
50
-a = -4
8 pts
MP = 54
8 pts
b) Calculate the population mean on the original scale.
Average = Σ(fi*Value) = 0.36(58)+0.48(55)+0.16(50) = 55.28
Could also solve from formula: Average = MP + a(p-q)+2pqd
p = 0.36+0.48/2 = 0.6 q = 1-0.6 = 0.4
Average = 54 + 4(0.6-0.4)+2(0.6)(0.4)(1) = 54 + 1.56 = 55.28
8 pts
c) If you cross an A1A1 genotype with an A2A2 genotype, what is the expected breeding
value of the progeny (in reference to the original random-mating population)?
 = a+d(q-p) = 4+1(0.4-0.6) = 4-0.2 = 3.8
All of the progeny will be A1A2
Breeding value = (q-p)α = (0.4-0.6)*3.8 = -0.76
Alternatively, you could take the average breeding value of the two parents:
A1A1 = 2q = 2*0.4*3.8 = 3.04
A2A2 = -2p = -2*0.6*3.8 = -4.56
Average = (3.04-4.56)/2 = -0.76
d) What is the additive genetic variance for this locus?
8 pts
V (X)   f i X i2   μ x 
2
= 0.36(3.042)+0.48(-0.762)+0.16(-4.562)-02 = 6.9312
= 2pq2 = 2*0.6*0.4*3.82 = 6.9312
3
6 pts
7) Consider the graph below showing the regression of genotypic values on number of
favorable Z1 alleles:
genotypic value

0
Z2Z2
1
Z1Z2
2
Z1Z1
The variance among genotypic values = 12 and the variance due to linear regression = 8.
What is the estimate of dominance variance?
G2   A2   D2
8 pts
 D2  G2   A2  12  8  4
8) A nature lover has been visiting a national park every summer for the past ten years. He has
noticed that the frequency of a purple form of his favorite wildflower has increased steadily
over the years, while the pink form has become less prevalent. What possible explanations
could you provide for the changes that he has observed?
 Mutation can change gene frequencies, but is not a likely explanation over a 10 year period
 Random drift due to small population size (although this is not too likely since the change
appears to be consistently in one direction)
 Selection – the park rangers may have planted seed from outside the area, but the purple form
is increasing because it has better local adaptation
 Selection – there may be a change in the environment that is favoring one form over the other
 Selection – Tourists prefer the pink forms and are picking too many of them
 Migration - pollinators are bringing in more purple genes from surrounding areas
 Color is influenced by the environment so climate changes are affecting the appearance of
the plants
4
Extra Credit
6 pts
1) A fellow graduate student sees that you are hard at work using the tabular method to estimate
coancestries for a complex pedigree. He suggests that you should use a popular software
package to solve the problem instead. He runs the kinship program and obtains the following
output:
indiv parent1 parent2
A
B
C
D
E
G
H
X
A
A
B
A
D
G
B
B
C
E
H
A
B
C
D
E
G
H
X
1.0000
.
0.5000
0.5000
.
0.7500
0.2500
0.5000
.
1.0000
0.5000
0.5000
0.5000
0.2500
0.5000
0.3750
0.5000
0.5000
1.0000
0.5000
0.2500
0.7500
0.3750
0.5625
0.5000
0.5000
0.5000
1.0000
0.2500
0.5000
0.6250
0.5625
.
0.5000
0.2500
0.2500
1.0000
0.1250
0.6250
0.3750
0.7500
0.2500
0.7500
0.5000
0.1250
1.2500
0.3125
0.7813
0.2500
0.5000
0.3750
0.6250
0.6250
0.3125
1.1250
0.7188
0.5000
0.3750
0.5625
0.5625
0.3750
0.7813
0.7188
1.1563
Does this appear to be a table of coancestries? What is your evidence?
No. The diagonal values should be the coancestries of each individual with themselves. This
would range from 0.50 for a noninbred individual to 1.0 for a fully inbred individual (we
don’t know if there is prior inbreeding.) Because a coancestry represents the probability that
alleles are identical by descent, a coancestry value cannot exceed 1.0. This table includes
values greater than one, so it cannot be a table of coancestries. The values in this table are
consistent with expectations for covariances with no prior inbreeding.
8 pts
2) Given the formulas for breeding values in the table below, use algebra to show that the
variance of breeding values = 2pqα2 (you can assume that the mean of breeding values is
zero).
Genotype
Frequency
Breeding Value
A1A1
p2
2q
A1A2
2pq
(q - p)
A2A2
q2
-2p
Genotype
Freq.
Breeding Value
(Breeding Value)2
Freq. x Value2
A1A1
p2
2q
4q22
4p2q22
A1A2
2pq
(q - p)
(q - p)22
2pq(q - p)22
A2A2
q2
-2p
4p22
4p2q22
2
σ2A  4p2q2 2  2pq  q  p  2  4p2q2 2
σ A2  2pq 2  2pq  q2  2pq  p2  2pq  2pq 2 p2  2pq  q2   2pq 2
5