Genetics and Genetic Prediction in Plant Breeding

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Genetics and Genetic Prediction in Plant Breeding
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(No Transcript)
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Class Test 2, March 10, 2000
Eight Questions worth 100 points total Bonus
Point worth 10 points Show all calculations 50
minutes
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Question 1a.
A cross is made between two homozygous barley
plants. One parent is tall and with short leaf
margin hairs (TTss), and the other is short with
long leaf margin hairs (ttSS). Single genes
control both plant height and leaf margin hair
length. Tall plants (T_) being completely
dominant to short (tt), and long leaf margin
hairs (S_) dominant to short (ss). If a sample
of F1 plants from this cross were self
pollinated, a large population of F2 plants
grown, and their phenotype for height and margin
hairs noted, what would be the expected ratio of
phenotypes based on the following genetic
situations. 8 points.  
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Question 1a.
Genetic situation Plant Phenotype Plant Phenotype Plant Phenotype Plant Phenotype
Genetic situation T_S_ T_ss ttS_ ttss
Complete dominance T dominant to t, S dominant to s.
Duplicate recessive epistasis tt epistasis to S and ss ss epistatic to T and tt.
Recessive epistasis tt epistatic to S and ss.
Duplicate dominant epistasis T epistasis to S and ss S epistasis to T and tt.
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Question 1a.
Genetic situation Plant Phenotype Plant Phenotype Plant Phenotype Plant Phenotype
Genetic situation T_S_ T_ss ttS_ ttss
Complete dominance T dominant to t, S dominant to s. 9 3 3 1
Duplicate recessive epistasis tt epistasis to S and ss ss epistatic to T and tt. 9 0 0 7
Recessive epistasis tt epistatic to S and ss. 9 3 0 4
Duplicate dominant epistasis T epistasis to S and ss S epistasis to T and tt. 15 0 0 1
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Question 1b
In the above experiment, a sample of F2 plants
were self-pollinated and 900 F3 plants evaluated
for plant height and leaf margin hair length.
After evaluating the phenotypes it was found that
there were 335 tall plants with long leaf margin
hairs 220 tall plants with short margin hairs
and 345 short plants with short margin
hairs. Explain what could have caused the
observed frequency of phenotypes and test your
theory using a suitable statistical test. 6
points.
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Question 1b.
T_S_ T_ss ttS_ ttss
335 220 0 345
Explain what may have caused this departure from
a 2515159 expected frequency of phenotypes
6 points.
This departure from a 2515159 ratio could be
caused by recessive epistasis, where tt is
epistatic to S, so ttS_ and ttss have the same
phenotype.
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Question 1b.
A appropriate test to use would be a chi-square
test.
L_G_ L_gg llG_ llgg
Observed 335 220 0 345
Expected 352 211 - 337
Difference -17 9 - 8
D2/exp 0.821 0.384 - 0.190
?2 2df 1.395 ns ?2 2df 1.395 ns ?2 2df 1.395 ns ?2 2df 1.395 ns ?2 2df 1.395 ns
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Question 2.
A spring wheat breeding program aims to develop
cultivars that are resistant to foot-rot,
controlled by a single completely dominant gene
(FF) which confers resistance, and that are
resistant to yellow strip rust, which is
controlled by a single dominant (YY) gene
conferring resistance. A cross is made between
two parents where one parent is resistant to foot
rot and susceptible to yellow strip rust (FFyy)
while the other is resistant to yellow strip rust
but susceptible to foot rot (ffYY). A sample of
F1 plants was self-pollinated, without selection,
and a large sample of F2 plants were grown and
allowed to self-pollinate. At harvest, only
plants that were phenotypically resistant to both
diseases (foot-rot and yellow strip rust) are
retained and used to plant a large population of
F3s.
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Question 2.
What proportion of these F3 plants would be
expected to be Resistant to foot rot and yellow
strip rust Resistant to foot rot but
susceptible to yellow strip rust Susceptible
to foot rot but resistant to yellow strip
rust Susceptible to both foot rot and yellow
strip rust 10 points.
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(FFyy) x (ffYY)
F3 F2 F2 F2 F2 F2 F2 F2 F2 F2 F2
F3 FFYY FFYy FFyy FfYY FfYy Ffyy ffYY ffYy ffyy Total
F3 1/16 2/16 1/16 2/16 4/16 2/16 1/16 2/16 1/16 Total
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(FFyy) x (ffYY)
F3 F2 F2 F2 F2 F2 F2 F2 F2 F2 F2
F3 FFYY FFYy FFyy FfYY FfYy Ffyy ffYY ffYy ffyy Total
F3 4/64 8/64 4/64 8/64 16/64 8/64 4/64 8/64 4/64 Total
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(FFyy) x (ffYY)
F3 F2 F2 F2 F2 F2 F2 F2 F2 F2 F2
F3 FFYY FFYy FFyy FfYY FfYy Ffyy ffYY ffYy ffyy Total
F3 4/64 8/64 4/64 8/64 16/64 8/64 4/64 8/64 4/64 Total
FFyy 4 2 - 2 1 - - - - 9
FFYy - 4 - - 2 - - - - 6
FFyy - 2 4 - 1 2 - - - 9
FfYY - - - 4 2 - - - - 6
FfYy - - - - 4 - - - - 4
Ffyy - - - - 2 4 - - - 6
ffYY - - - 2 1 - 4 2 - 9
ffYy - - - - 2 - - 4 - 6
ffyy - - - - 1 2 - 2 4 9
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(FFyy) x (ffYY)
F3 F2 F2 F2 F2 F2 F2 F2 F2 F2 F2
F3 FFYY FFYy FfYY FfYY Total
F3 4/64 8/64 8/64 16/64 Total
FFYY 4 2 2 1 9
FFYy - 4 - 2 6
Ffyy - 2 - 1 3
FfYY - - 4 2 6
FfYy - - - 4 4
Ffyy - - - 2 2
ffYY - - 2 1 3
ffYy - - - 2 2
Ffyy - - - 1 1
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Question 2.
What proportion of these F3 plants would be
expected to be Resistant to foot rot and yellow
strip rust 25  Resistant to foot rot but
susceptible to yellow strip rust
5  Susceptible to foot rot but resistant to
yellow strip rust 5  Susceptible to both foot
rot and yellow strip rust 1  10 points.
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Question 3.
Assume the situation in question 3 (above) where
a foot rot wheat parent (FFyy) is crossed to a
yellow rust resistant parent (ffYY). The
heterozygous F1 is crossed to a double
susceptible homozygous line with genotype ffyy
and 2,000 BC1 progeny evaluated for disease
resistance. The following numbers of phenotypes
were observed Resistant to both foot rot and
yellow strip rust 90 Resistant to foot rot
but susceptible to yellow strip rust 893
Susceptible to foot rot but resistant to yellow
strip rust 907 Susceptible to both foot rot and
yellow strip rust 110 Determine the
percentage recombination between the foot rot and
yellow strip rust loci. 4 points.
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Question 3.
Recombination (R) Number of Recombinants Number
of observations
90110 0.10 10 2000
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Question 3.
Gametes from female parent Gametes from male parent Gametes from male parent Gametes from male parent Gametes from male parent
Gametes from female parent FY-0.05 Fy-0.45 fY-0.45 fy-0.05
FY 0.05 FYFY 0.0025 FFYy 0.0225 FfYY 0.0225 FfYf 0.0025
Fy 0.45 FFYy 0.0225 FFyy 0.2025 FfYy 0.2025 Ffyy 0.0225
fY 0.45 FfYY 0.0225 FfYy 0.2025 ffYY 0.2025 ffYy 0.0225
fy 0.05 FfYy 0.0025 Ffyy 0.0225 ffYy 0.0225 ffyy 0.0025
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Question 3.
25 FFYY450 FFYy2025 FFyy 450 FfYY4100
FfYy450 Ffyy 2025 ffYY450 ffYy25 ffyy
F_Y_ 5,025 F_yy 2,475 ffY_ 2,475 ffyy
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Question 3.
Estimate the number of F2 plants that would need
to be evaluated to be 99 sure of identifying one
F2 plant that is resistant to both foot rot and
yellow strip rust. 3 points.
n Ln(1-p)/Ln(1-x)
n Ln(1-0.99)/Ln(1- 0.5025) n 6.60 need 7 F2
plants to be 99 sure of one.
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Question 4a.
Potato cyst nematode resistance is controlled by
a single completely dominant allele (R). A cross
is made between two auto-tetraploid potato
cultivars where one parent is resistant to potato
cyst nematode and the other is completely
susceptible. Progeny from the cross are examined
and it is found that 83.3 of the progeny are
resistant to potato cyst nematode. What can be
determined about the resistant parent in the
cross? 4 points.
One parent must be nulliplex (rrrr), and one
resistant (Rrrr, RRrr, RRRr, or RRRR). If RRRr or
RRRR all progeny are resistant. If Rrrr 50 are
resistant (11). If RRrr then 83 (51) are
resistant. Answer parent is duplex (i.e.. RRrr x
rrrr).
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Question 4b.

A breeding program aims to produce potato
parental lines that are either quadriplex or
triplex for the potato cyst nematode resistant
gene R. A cross is made between two parents,
which are know to be duplex for the resistant
gene R. What would be the expected genotype and
phenotype of progeny from this diploid x diploid
cross? 4 points.
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Question 4b.

1 RR 4 Rr 1 rr
1 RR 1 RRRR 4 RRRr 1 RRrr
4 Rr 4 RRRr 16 RRrr 4 Rrrr
1 rr 1 RRrr 4 Rrrr 1 rrrr
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Question 4b.

1 RR 4 Rr 1 rr
1 RR 1 RRRR 4 RRRr 1 RRrr
4 Rr 4 RRRr 16 RRrr 4 Rrrr
1 rr 1 RRrr 4 Rrrr 1 rrrr
1 RRRR 8 RRRr 18 RRrr 8 Rrrr 1 rrrr
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Question 5b.

Why are plant breeders interested in conducting
scaling tests for quantitatively inherited
characters of importance in the breeding scheme?
3 points
Many of the important statistics that breeders
are concerned with (i.e. response to selection)
are based on the additive/dominance model.
Scaling tests are used to determine if this model
is adequate, and hence the predictions are
accurate.
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Question 5b.

A properly designed experiment was carried out in
canola where the high yielding parent (P1), is
grown alongside F1 plants and B1 plants. The
average yield of plants from each of the three
families, the variance of each family and the
number of plants evaluated from each family were
Family Mean Yield Variance of yield Number of plants
P1 1923 74 16
F1 1629 69 16
B1 1985 132 31
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Question 5b.
Family Mean Yield Variance of yield Number of plants
P1 1923 74 16
F1 1629 69 16
B1 1985 132 31

A 2B1 F1 P1 3970-1629-1923 418
V(A) 4VB1 VF1 VP1 5286974 671 se(A)
?V(A) 29.90
t 151530 df A/se(A) 1.40 ns
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Question 6a.

The following is a model for the analysis of
diallels (Griffing analysis). Yijk ? gi gj
sij eijk, Explain what gi and sij represent
in the model 4 points.
gi is the general combining ability of the ith
parent, sij is the specific combing ability (not
explained by GCA) between the ith and the jth
parent.
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Question 6a.

Briefly explain the difference between parents
chosen at random (random parent effects) and
parents specifically chosen (fixed parent
effects). 4 points.
If parents are chosen at random it is assumed
that inference from the analyses are to govern
the situation of all possible parental cross
combinations. When parental are fixed than it is
assumed that the analysis is unique only to the
parents in the design.
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Question 6a.
A 5 x 5 full diallel (including selfs) design was
conducted and yield of the parent selfs and the
F1 progeny were obtained from a properly
randomized and replicated experiment. A Griffing
analysis of variance was carried out and sum of
squares and degrees of freedom are shown below.
Complete the analysis assuming that the parents
are fixed and briefly outline your conclusions
from the analysis. 4 points.

Source df S.Sq
GCA 4 7,900
SCA 10 6,010
Reciprocal 10 5,100
Error 50 17,500
Total 74 35,710
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Question 6a.
Source df S.Sq M.Sq F
GCA 4 7,900 1957 5.64
SCA 10 6,010 601 1.18 ns
Reciprocal 10 5,100 510 1.46 ns
Error 50 17,500 350
Total 74 35,710

Reciprocal effects were not significantly
different from error so there were no maternal or
cytoplasmic effects for yield. SCA was also
non-significant while GCA was significant at the
99 level indicating a high proportion of
additive genetic variance. From the analysis
there would be good opportunity to determine
progeny worth from GCA values of parents.
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Question 6a.

What difference would you make if the parents in
this analysis were chosen at random 2 points.
Source df S.Sq M.Sq F
GCA 4 7,900 1957 3.25 ns
SCA 10 6,010 601 1.18 ns
Reciprocal 10 5,100 510 1.46 ns
Error 50 17,500 350
Total 74 35,710
Use the SCA M.Sq to test the GCA term. In this
case the GCA is not quite formally significant at
the 5 level.
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Question 6d.

A Hayman Jinks analysis was conducted and Vi
and Wi values estimated for each parent. The sum
of products between Vi and Wi (?Vi x Wi) was
found to be 165 the sum of Vis was 26 sum of
Wis was 22 sum of squares of Vi (? Vi2) was
205 and the sum of squares of Wi (? Wi2) was
118. Complete a regression analysis of Vi on to
Wi. 4 points.
SP(Vi,Wi) 165 26 x 22/5 50.6 SS(Vi)
205 262/5 69.8 SS(Wi) 118 222/5
21.2
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Question 6d.

SP(Vi,Wi) 165 26 x 22/5 50.6 SS(Vi) 205
262/5 69.8 SS(Wi) 118 222/5 21.2
b1 SP(Vi,Wi)/SS(Vi) 50.6/69.8 0.72
se(b1) SS(Wi) b1SP(Vi,Wi)/(n-2)SS(Vi)
118 36/209.4 0.39 b0 4.4
0.72 x 5.2 0.67
t3df 1-0.72/0.39 0.71 ns
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Question 6d.

What can be determined from this analysis
regarding the adequacy of the additive/dominance
model and the importance of additive genetic
variance (A) compared to dominant genetic
variance (D). 4 points.
The regression slope (b1) is not significantly
different from one, therefore, the
additive/dominance model is adequate to explain
the variation observed for yield in the diallel.
The intercept (b0) is greater than zero so A is
greater than D. However, b0 is almost equal to
zero so A D.
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Question 7.
A crossing design involving two homozygous pea
cultivars is carried out and both parents are
grown in a properly designed field experiment
with the F2, B1 and B2 families. Given the
following standard deviations for both parents
(P1 and P2), the F2, and both backcross progeny
(B1 and B2), determine the broad-sense
heritability and narrow-sense heritability for
seed size in dry pea 10 points.

Family Standard Deviation
P1 3.521
P2 3.317
F2 6.008
B1 5.450
B2 5.157
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Question 7.

Family Standard Deviation
P1 3.521
P2 3.317
F2 6.008
B1 5.450
B2 5.157
VP112.4 VP211.0 VF236.1 VB129.7 VB226.6
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Question 7.
VP112.4 VP211.0 VF236.1 VB129.7 VB226.6
h2b Genetic variance Total variance
E VP1VP2/2 11.7
h2b 36.1 11.7 36.1
h2b 0.67
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Question 7.
VP112.4 VP211.0 VF236.1 VB129.7 VB226.6
E VP1VP2/2 11.7
D 4V(B1)V(B2)-V(F2)-E 429.726.6-36.1-11.7
8.5
A 2V(F2)-¼D-E 236.1-2.1-11.7 22.3
h2n ½A/V(F2) 11.15/36.1 0.31
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Question 8a.
Assuming an additive/dominance mode of
inheritance for a polygenic trait, list expected
values for P1, P2, and F1 in terms of m, a and
d. 3 points
P1 m a P2 m a F1 m d
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Question 8b.
From these expectations, what would be the
expected values for F2, B1 and B2 based on m, a
and d. 3 points
F2 m ½d B1 m ½a ½d B2 m ½a ½d
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Question 8c.
From a properly designed field trial that
included P1, P2 and F1 families, the following
yield estimates were obtained.  B1 42.0 lb/a
B2 26.0 lb/a F1 38.5lb/a  From these
family means, estimate the expected value of P1,
P2 and F2, based on the additive/dominance model
of inheritance 6 points.  
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Question 8c.
P1 m a P2 m a F1 m d F2 m ½
d B1 m ½ a ½ d B2 m ½ a ½ d
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Question 8c
 B1 42.0 lb/a B2 26.0 lb/a F1 38.5lb/a
B1 B2 m ½a ½d m (-½a) ½d a B1 B2
F1 2m d m - d m
F1 m d
B1 B2 42.0 26.0 36.0 16 a B1 B2
F1 29.5 m F1 m 9.0 d
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Question 8c.
B1 B2 42.0 26.0 36.0 16 a B1 B2
F1 29.5 m F1 m 9.0 d
P1 m a 29.5 16.0 45.5 P2 m a
29.5 16.0 13.5 F2 m ½d 29.5 4.5
25.0
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Quantitative Genetics Models
P2 m
38.5 P1
26.0
42.0
13.5 29.5
38.5 45.5
26.0 34.0
42.0
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Bonus Question
It is important in quantative genetics to know
whether the additive/dominance model based on m,
a, and d is appropriate to explain the
variation observed in this study. Given that you
have available progeny means and variances from
both parents (P1 and P2), the B1, B2, and F3
families. Devise a suitable scaling test (hint
as yet we have not talked about this one)
involving these five families. 10 bonus points.
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Bonus Question
P1 m a P2 m a F3 m ¼ d
B1 m ½ a ½ d B2 m ½ a ½ d
4F3 4md
B1B2P1P2 4m d
4F3 B1 B2 P1 P2 ? 0
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City of the Dead Moscow March 10, 2000