PBG 650 Advanced Plant Breeding
First Midterm 2015
Name
KEY
Please show your work.
For multiple choice questions, circle the one best answer
6
6 pts
pts
1) You cross an AA genotype to an aa genotype. The F1 hybrid is backcrossed to the AA parent
to produce the first backcross generation (BC1). If the BC1 seed is planted and the
population is random-mated, what will be the expected genotype frequencies in the next
generation?
Harder Method (not recommended)
Consider all possible matings in BC1,
which is ½ AA, ½ Aa
½ AA x ½ AA  (1/4)AA
½ AA x ½ Aa  (1/8)AA, (1/8)Aa
½ Aa x ½ AA  (1/8)AA, (1/8)Aa
½ Aa x ½ Aa  (1/16)AA, (1/8)Aa, (1/16)aa
Collect terms:
(9/16)AA, (3/8)Aa, (1/16)aa
Gene frequencies in BC1
f(A) = p = 0.75
f(a) = q = 0.25
Random mating  HWE
f(AA) = p2 = (0.75)(0.75) = 9/16 = 0.5625
f(Aa) = 2pq= 2(0.75)(0.25) = 6/16 = 0.375
f(aa) = q2 = (0.25)(0.25) = 1/16 = 0.0625
2) A breeder crosses an AAbb genotype to an aaBB genotype. The F1 hybrid is random-mated
to produce the F2. When the F2 generation is planted, 1/100 plants show the double
recessive phenotype (aabb).
5 pts
a) Assuming that the A and B loci have no effect on fitness, is there evidence for
disequilibrium in this population? Explain your answer.
If there is independent assortment (no linkage), then we would expect the frequency of ‘ab’
alleles to be 0.25. We would then expect that 1/16 of the F2 population would be double
recessive aabb. It appears that there is a deficiency of recombinant ‘ab’ gametes, because
they occur at a frequency of 0.10. The disequilibrium is 0.10 - 0.25 = -0.15 (or +0.15 if you
consider the ‘a’ and ‘b’ alleles to be in repulsion phase linkage).
b) Given that crosses were made by the breeder, what would be the most likely cause of
disequilibrium in this population? (you can answer this as a hypothetical question
regardless of your answer to question 2a)
5 pts
With a total of 20% recombinant gametes, these loci appear to be located about 20 cM
apart on the same chromosome. Any disequilibrium that is not due to linkage should have
been eliminated through recombination when gametes were produced by the AaBb F1
hybrid.
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3) When there is no linkage, the expected rate of decay of disequilibrium with each generation
of random-mating is
5 pts
a) 25 %
b) 50 %
c) 100 %
d) It depends on gene frequencies
4) Circle one statement about random genetic drift that is NOT true.
5 pts
a) The variance in gene frequency among sub-populations increases over time
b) There is no change in the average gene frequency among sub-populations over time
c) The direction of the change in gene frequency in a sub-population can be predicted
d) There is a decrease in heterozygosity within sub-populations over time
5) Which of the following would NOT have a coancestry = 1.0?
5 pts
a) Plants from the same pureline variety of barley
b) Plants from the same clone of Douglas Fir (the state tree of Oregon)
c) Plants from the same inbred line of corn
d) All of the above (because none of them have a coancestry = 1.0)
7 pts
6) Explain the difference between stabilizing selection and directional selection. Use the
example of flowering time (e.g., days from planting until first bloom) in an outcrossing
species to illustrate why a breeder might want to use these types of selection to achieve
specific breeding goals.
With stabilizing selection, selection is for intermediate phenotypes and against the
extremes. This might be used to achieve greater uniformity in flowering time, which could
be important in a crop that is harvested all at one time, or for an ornamental plant where
flowering is expected at a particular time. Directional selection could be used to make the
crop earlier or later in maturity. This could be desirable because it might allow the crop to
be used more effectively in a particular environment or crop rotation system.
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7) An heirloom variety of carrots has been maintained by several members of a gardener’s
cooperative. Every year they each plant a small patch in their gardens, and allow it to crosspollinate, taking care that there is no other variety of carrots pollinating nearby. You plant
small quantities of seed from each gardener and score them for an SSR marker in the
laboratory. You tally the frequency of genotypes across all of the samples:
A 1A 1
60
6 pts
A 1A 2
60
A 2A 2
30
What is the inbreeding coefficient for this variety?
p = [60 + (1/2)60]/150 = 0.6 q = 0.4
Another way to approach this is to
f(A1A2) = 2pq (1-F)
use the formula F = 1-(Hobs/Hexp)
60/150 = 0.4 = 2(0.6)( 0.4)(1-F)
F = 1-(0.40/0.48) = 1-(5/6) = 1/6 = 0.16667
1-F = 0.40/0.48 = 5/6 = 0.833333
F = 1 – (5/6) = 1/6 = 0.16667
8) In an open-pollinated landrace of corn, the purple stem characteristic is determined by a
single recessive gene. The frequency of the recessive allele is 0.3.
6 pts
a) You grow a population of this landrace and remove all of the plants with purple stems.
What proportion of the plants that are remaining in your field will be heterozygous for
the purple trait?
2pq
0.42

 0.46
p  2pq 0.49  0.42
46% are heterozygous
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b) If you allow the remaining plants to intermate, what is the new frequency of the allele
for purple stems that you would expect in this population?
6 pts
Initial p = 1-0.3=0.7
New q = pq/(p2+2pq)=0.7*0.3/(0.72+2*0.7*0.3)=0.21/0.91=0.231
Or you could apply the formula
q1 
3
q0
0.3

 0.231
1  q0 1  0.3
9) Assume that there is no prior inbreeding for any individuals shown in the pedigree below.
A
C
B
D
X
Z
a) Which individuals in the pathway are inbred?
5 pts
D, X, and Z
b) What is the coancestry of B and D?
6 pts
5 pts
BD = (1/2)(BB+BC) = (1/2) ((1/2) + (1/8)) = 5/16 = 0.3125
c) The coancestry of B and D is equal to the inbreeding coefficient for ___X___.
d) What is the inbreeding coefficient for Z?
6 pts
(XBACD; XBD; XD)
FZ = (1/2)5(1+FA)+(1/2)3(1+FB) )+(1/2)2(1+FD) = (1/2)5+(1/2)3 +(1/2)2(1+1/8)
= (1/32) + (1/8) + (1/4)(9/8) = (1/32) + (4/32) + (9/32) = 14/32 = 7/16 = 0.4375
Another way FZ = DX = (1/2)(BD+DD)
BD = 5/16 (from question 9b)
DD = ½(1 + FD) = ½(1 + (1/8)) = 9/16
FZ = DX = (1/2)(BD+DD) = (1/2)((5/16)+(9/16)) = 7/16 = 0.4375
Another way FZ = DX = (1/4)(BB + BC + BD + CD) = (1/4)((1/2) + (1/8) + (5/16) + (5/16))
(1/4)((8/16) + (2/16) + (5/16) + (5/16)) = (1/4)(5/4) does not give correct answer
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10) A meadowfoam breeding population is grown in a field that is 10 miles from the nearest
commercial meadowfoam crop. Plants are spaced 3 feet apart and pollinated with
honeybees (i.e., they are randomly intermated). Seeds are harvested separately from each
plant and are evaluated in replicated progeny trials in the subsequent season. Offspring
from parent plant X are found to produce an average of 50 kg/ha more seed in comparison
to the mean for all progeny from the population.
a) What type of progeny were evaluated?
5 pts
i.
Half-sib families
ii.
Full-sib families
iii.
S1 families
iv.
We can’t tell
b) The breeding value of parent plant X is
5 pts
i.
0
ii.
25 kg/ha
iii.
50 kg/ha
iv.
100 kg/ha
11) 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:
Frequency Value
A 1A 1
0.36
60
A 1A 2
0.48
58
A 2A 2
0.16
50
a)
What are the coded genotypic values for each genotype (a, d, and –a)?
Pbar = (60 + 50)/2 = 55
6 pts
a = +5
b)
6 pts
d = +3
-a = -5
What is the population mean on the original scale?
Average = Σ(fi*Value) = 0.36(60)+0.48(58)+0.16(50) = 57.44
Could also solve from formula: Average = Pbar + a(p-q)+2pqd
p = 0.36+0.48/2 = 0.6 q = 1-0.6 = 0.4
Average = 55 + 5(0.6-0.4)+2(0.6)(0.4)(3) = 55 + 2.04 = 57.44
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