PBG 650 Advanced Plant Breeding
First Midterm 2013
Name
KEY
Please show your work.
1)
Homozygous AABB individuals are crossed to homozygous aabb individuals. The F1
generation is selfed to produce the F2 generation. The A and B loci are 20cM apart.
a.
6 pts
What would be the initial value of disequilibrium (D) for these two loci in the
gametes that form the F2?
Parental gametes PAB , Paabb =(0.5)(1-0.2)=0.40
Recombinant gametes PAb , PaB =(0.5)(0.2)=0.10
D = PAB - pApB = 0.40 – (0.5)(0.5) = 0.15
OR
D = PAB Pab - PAb PaB = (0.40) (0.40) – (0.1)(0.1) = 0.15
6 pts
b. Based on the expected rate of decay of disequilibrium, what would the value of D be for
these two loci after one generation if the F2 plants are allowed to mate at random?
Dt+1 = (1-c)Dt = (1-0.2)(0.15) = 0.12
6 pts
c. Consider another locus ‘E’ in this population that is not on the same chromosome as the
A and B loci. Would you expect there to be any disequilibrium between the A and E loci
in the gametes that form the F2 generation? Explain your answer.
No, because there is an opportunity for recombination and independent assortment
between loci A and E in all of the F1 parent plants. All of the F1 plants are double
heterozygotes. In the absence of linkage, there will be no linkage disequilibrium in an F2
population.
Note that if the two parents had been planted as a mixture and allowed to random mate,
there would be initial LD between unlinked loci that would decay by one-half with each
generation of random mating. Thus, the answer that there is no LD because the loci are
not linked and there is random mating is not really a sufficient explanation. The fact that
the population originated from a cross between two inbred lines is critical.
1
2) 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:
A1A1
22
8 pts
A1A2
36
A2A2
42
What is the inbreeding coefficient for this variety?
p = [22 + (1/2)36]/100 = 0.4
q = 0.6
Another way to approach this is to
f(A1A2) = 2pq (1-F)
use the formula F = 1-(Hobs/Hexp)
0.36 = 2(0.4)( 0.6)(1-F)
F = 1-(0.36/0.48) = 0.25
1-F = 0.36/0.48
F = 1 – 0.75 = 0.25
6 pts
3) 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 32
plants have purple stems and 168 plants have green stems.
a. Assuming that the population is in Hardy-Weinberg Equilibrium, what is the initial gene
frequency of the allele for purple stems?
q2 = 32/200 = 0.16 q = 0.4
8 pts
b. If you remove all of the plants with purple stems and allow the remaining plants to
intermate, what is the new frequency of the allele for purple stems that you would expect
in this population?
Initial p = 1-0.4=0.6
New q = pq/(p2+2pq)=0.4*0.6/(0.62+2*0.6*0.4)=0.24/0.84=0.2857
q0
0.4
q1 

 0.2857
Or you could apply the formula
1  q0 1  0.4
2
4) Assume that all parents of individuals C and D in the following pedigree are not inbred.
A
B
C
D
X
Y
Z
8 pts
a) What is the inbreeding coefficient (F) for individual Z?
FZ = (1/2)4(1+FA)+(1/2)4(1+FB) = (1/2)4+(1/2)4 = 1/8
6 pts
(XCAY; XBDY)
b) What does the inbreeding coefficient that you calculated for Z (previous question) tell
you about the coancestry of X and Y?
XY = FZ = 1/8
Proof: XY = ¼(AB+BD+AC+CD) = ¼(0+0.25+0.25+0) = 1/8
8 pts
c) Calculate the coancestry of X and Z. How does your result compare to the coancestry of
parent and offspring for individuals A and C? How would you explain the difference, if
any?
XZ = ½(XX+XY)=1/2(1/2+1/8)=5/16
AC = 1/4
XZ is slightly larger than AC because X and Z share common ancestors (A and B)
through both parents of Z. In other words, the parents of Z are related to each other,
whereas the parents of C are assumed to be unrelated.
Note that Z is the only inbred individual in the pathway.
3
5) You collect a random sample of 30 individuals from a population and determine that there
are 12 AA and 18 Aa genotypes. You perform a Fisher’s Exact test and obtain the following
probability values for each of the possible outcomes for the observed numbers of alleles.
AA
21
20
19
18
12
17
13
16
14
15
10 pts
Aa
0
2
4
6
18
8
16
10
14
12
aa
9
8
7
6
0
5
1
4
2
3
Probability
0.0000
0.0000
0.0003
0.0055
0.0245
0.0427
0.1442
0.1611
0.3091
0.3125
1.0000
Cumulative
0.0000
0.0000
0.0003
0.0058
0.0303
0.0730
Reject
Reject
Reject
Reject
Reject
Accept
Total
a) Would you accept or reject the null hypothesis that the population is in Hardy-Weinberg
equilibrium (using =0.05)? Explain your reasoning.
Possible outcomes (numbers in each genotypic class) are sorted from the least probable to
the most likely based on the allele frequencies in the sample. The cumulative probability
0.0303 for our sample is less than 0.05. Therefore we would reject the null hypothesis
and would conclude that at least one of the assumption for H-W equilibrium is not valid.
Note that the actual probability of the observed sample (0.0245) is not the deciding
factor. It is not a reliable indicator because individual events become less and less
probable as the sample size increases (there are more possible outcomes).
6) What is meant by the “average effect of a gene” in quantitative genetics?
8 pts
The average effect of a gene indicates the average performance of individuals who receive
that gene expressed as a deviation from the population mean.
4
7) Assuming that there is complete dominance for a locus ‘A’, then the average effect of an
allele substitution (replacing A2 with A1) will be greatest when (choose one answer)
6 pts
a)
b)
c)
d)
The coded genotypic value a is large and the frequency of A1 (p) is high.
The coded genotypic values a is large and the frequency of A1 (p) is low.
The coded genotypic value a is small and the frequency of A1 (p) is high.
The coded genotypic value a is small and the frequency of p = q = 0.5.
8) The graph below plots the genotypic values for the Z locus on the Y axis as a function of the
number of favorable Z1 alleles on the X axis.
12
10
Value
8
6
4
2
0
0
1
Z 2Z 2
Z 1Z 2
2
Z 1Z 1
8 pts
a) Without doing any calculations, plot the approximate breeding values for the three
genotypes on the graph above. The exact slope of the line and breeding values will
depend on gene frequencies. The point is that breeding values represent the portion of the
variation among genotypic values that can be explained by linear regression on number
of Z1 alleles.
6 pts
b) Based on the appearance of the graph, circle the answer below that best describes the
observed level of dominance at this locus.
i. No dominance
ii. Partial dominance
iii. Complete dominance
5
iv. Overdominance
6