Genetics and Genetic Prediction in Plant Breeding

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Genetics and Genetic Prediction in Plant Breeding
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Class Test 2, March, 1998
Eight Questions worth 100 points total Bonus
Point worth 10 points Show all calculations 50
minutes
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Question 1a
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 1b
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 1c
From a properly designed field trial that
included P1, P2 and F1 families, the following
yield estimates were obtained.  P1 1928 Kg
P2 1294 Kg F1 1767 Kg  From these family
means, estimate the expected value of F2, B1, B2
and Fa, based on the additive/dominance model of
inheritance 3 points.  
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Question 1c
P1 1928 Kg P2 1294 Kg F1 1767 Kg
a P1 P2/2 1928-1294/2 317
m P1 a 1928 317 1611
d F1 m 1767 1611 156
F2 m ½d 1611 78 1689 B1 m½a½d
1611158.578 1397.5 B2 m½a½d
1611-158.578 1530.5
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Question 2
A spring barley breeding program has major
emphasis in developing cultivars which are short
in stature and with yellow stripe resistance. It
is known that the inheritance of short plants is
controlled by a single completely recessive gene
(tt) over tall plants (TT), and that yellow
stripe rust resistance is controlled by a single
completely dominant gene (YY), over a recessive
susceptible gene (yy). The tall gene locus and
yellow rust gene locus are located on different
chromosomes. Given that a tall resistant plant
(TTYY) is crossed to a short susceptible plant
(ttyy), both parents being homozygous, what would
be the expected proportion of genotypes and
phenotypes in the F1 and F2 families 12
points.  
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Tall, Res. (TTYY) x Short, Susc. (ttyy)
F1 TtYy Tall/Resistant
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Tall, Res. (TTYY) x Short, Susc. (ttyy)
Gametes from female parent Gametes from male parent Gametes from male parent Gametes from male parent Gametes from male parent
Gametes from female parent TY Ty tY ty
TY TTYY TTYy TtYY TtYy
Ty TTYy TTyy TtYy Ttyy
tY TtYY TtYy ttYY ttYy
ty TtYy Ttyy ttYy ttyy
1 TTYY2 TTYy1 TTyy2 TtYY4 TsYy2 Ttyy1
ttYY2 ttYy1 ttyy
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Tall, Res. (TTYY) x Short, Susc. (ttyy)
Gametes from female parent Gametes from male parent Gametes from male parent Gametes from male parent Gametes from male parent
Gametes from female parent TY Ty tY ty
TY TTYY TTYy TtYY TtYy
Ty TTYy TTyy TtYy Ttyy
tY TtYY TtYy ttYY ttYy
ty TtYy Ttyy ttYy ttyy
9 T_Y_ 3 T_yy 3 ttY_ 1 ttyy
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Question 2
How many F3 plants would need to be assessed to
ensure, with 99 certainty, that at least one
plant would exist that was short and homozygous
yellow rust resistant (i.e. ttYY) 8 points.
ttYY 9/64 0.1406
Ln1-0.99 Ln(0.01)
Ln1-0.1406 Ln(0.8594)
30.39, need 31 F3 plants
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Question 3a
Two genetically different homozygous lines of
canola (Brassica napus L.) were crossed to
produce F1 seed. Seed from the F1 family was
self pollinated to produce F2 seed. A properly
designed experiment was carried out involving
both parents (P1 and P2, 10 plants each), the F1
(10 plants) and the F2 families (64 plants) was
grown in the field and plant height of individual
plants (inches) recorded. The following are
family means, variances and number of plants
observed for each family Family Mean Variance
Plants P1 52 1.97 10 P2 41
2.69 10 F1 49 3.14 10 F2 43 10.69 34  Co
mplete a statistical test to determine whether an
additive/dominance model of inheritance is
appropriate to adequately explain the inheritance
of plant height in canola 7 points.
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Question 3a
Family Mean Variance Plants
P1 52 1.97 10
P2 41 2.69 10
F1 49 3.14 10
F2 43 10.69 64
C-scaling test 4F2 2F1 P1 P2
C 172-98-52-41 19
V(C) 16V(F2)4V(F1)V(P1)V(P2)
V(C) 171.0412.561.972.69 188.26
se(C) ?188.26 13.72
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Question 3a
Family Mean Variance Plants
P1 52 1.97 10
P2 41 2.69 10
F1 49 3.14 10
F2 43 10.69 64
C 172-98-52-41 19
se(C) ?188.26 13.72
t90df 19/13.72 1.38 ns
Therefore, additive/dominance model is adaquate
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Question 3a
Family Mean Variance Plants
P1 52 1.97 10
P2 41 2.69 10
F1 49 3.14 10
F2 43 10.69 64
P1 52
P2 41
m 46.5
F2 43
F1 49
?
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Question 3b
If the additive/dominance model is inadequate,
list three factors which could cause the lack of
fit of the model 3 points.

• Abnormal chromosomal behavior where the
  heterozygote does not contributes equal
  proportions of its various gametes to the gene
  pool.
• Cytoplasmic inheritance where the character is
  determined by non-nuclear genes.
• Epistasis where alleles at different loci are
  interacting.

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Question 4a
F1, F2, B1, and B2 families were evaluated for
plant yield (kg/plot) from a cross between two
homozygous spring wheat parents. The following
variances from each family were found s2F1
123.7 s2F2 496.2 s2B1 357.2 s2B2
324.7 Calculate the broad-sense (h2b) and
narrow-sense (h2n) heritability for plant yield
10 points.
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Question 4a
s2F1 123.7 s2F2 496.2 s2B1 357.2 s2B2 324.7
h2b Genetic variance Total variance
E V(F1) 123.7
h2b 496.2 123.7 496.2
h2b 0.75
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Question 4a
s2F1 123.7 s2F2 496.2 s2B1 357.2 s2B2 324.7
E V(F1) 123.7
D 4V(B1)V(B2)-V(F2)-E 4357.2324.7-496.2-123
.7 248
A 2V(F2)-¼D-E 2496.2-62-123.7 621
h2n ½A/V(F2) 310.5/496.2 0.63
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Question 4b
Given the heritability estimates you have
obtained, would you recommend selection for yield
at the F3 in a wheat breeding program, and why?
2 points.
A narrow-sense heritability greater than 0.6
would indicate a high proportion of the total
variance was additive in nature. Selection at F3
is often advesly related to dominant genetic
variation (caused by heterozygosity), here D is
small compared to A. Additive genetic variance is
constant over selfing generations and so
selection would result in a good response.
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Question 5a
Four types of diallel crossing designs have been
described by Griffing. Briefly outline the
features of each Method 1, 2, 3, and 4 4
points.
1. Complete diallel with selfs, Method 1. 2. Half
diallel with selfs, Method 2. 3. Complete
diallel, without selfs, Method 3. 4. Half
diallel, without selfs, Method 4.
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Question 5a
Why would you choose Method 3 over Method 1? 1
point. In instances where it was not possible to
produce selfed progeny (i.e. in cases of strong
self-incompatibility in apple and rapeseed). Why
would you choose Method 2 over Method 1? 1
point. In cases where there is no recipricol or
maternal effects.
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Question 5b
A full diallel, including selfs is carried
involving five chick-pea parents (assumed to be
chosen as fixed parents), and all families
resulting are evaluated at the F1 stage for seed
yield. The following analysis of variance for
general combining ability (GCA), specific
combining ability (SCA) and reciprocal effects
(Griffing analysis) is obtained
Source df MSq
GCA 5 30,769
SCA 10 10,934
Recipricol 10 9,638
Error 49 5,136
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Question 5b
Source df MSq F
GCA 5 30,769 5.99
SCA 10 10,934 2.12 ns
Recipricol 10 9,638 1.87 ns
Error 49 5,136
GCA is significant at the 99 level, while SCA
and reciprical differences were not significant.
This indicates that a high proportion of
phenotypic variation between progeny is additive
rather than dominant or error.
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Question 5b
Source df MSq F
GCA 5 30,769 2.81 ns
SCA 10 10,934 2.12 ns
Recipricol 10 9,638 1.87 ns
Error 49 5,136
If a random model is chosen then the GCA term is
tested using the SCA mean square. In this case
the GCA term is not formally significant, and the
indication overall would be that there is no
significant variation between progeny in the
diallel.
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Question 5b
Plant height was also recorded on the same
diallel families and an additive/dominance model
found to be adequate to explain the genetic
variation in plant height. Array variances Vi's
and non-recurrent parent covariances (Wi's) were
calculated and are shown along-side the general
combining ability (GCA) of each of the five
parents, below
Vi Wi GCA
Parent 1 491.4 436.8 -0.76
Parent 2 610.3 664.2 12.92
Parent 3 302.4 234.8 -14.32
Parent 4 310.2 226.9 -15.77
Parent 5 832.7 769.4 17.93
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Question 5b
Vi Wi GCA
Parent 1 491.4 436.8 -0.76
Parent 2 610.3 664.2 12.92
Parent 3 302.4 234.8 -14.32
Parent 4 310.2 226.9 -15.77
Parent 5 832.7 769.4 17.93
Visual inspection of Vi and Wi values would
indicate a linear relationship with slope
approximatly equal to 1, which would indicate a
additive/dominance model of inheritance. Parents
with lowest Vi and Wi values (those with greatest
frequency of dominant alleles) have negative GCA
values indicating short stature in chick pea is
dominant over tall stature.
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Question 6a
Two homozygous barley parents were crossed to
produce an F1 family. One parent was tall with
awns and the other was short and awnless. Tall
plants are controlled by a single dominant gene
and awned plants are also controlled by a single
dominant gene. The F1 family was crossed to a
plant which was short and awnless and the
following number of phenotypes observed
Phenotype observed
Tall, awned 954
Short, awned 259
Tall, awnless 221
Short, awned 966
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Question 6a
Phenotype observed
Tall, awned 954
Short, awned 259
Tall, awnless 221
Short, awned 966
Recombination 259221/2400 0.20
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Gametes from female parent Gametes from male parent Gametes from male parent Gametes from male parent Gametes from male parent
Gametes from female parent TA-0.4 Ta-0.1 tA-0.1 ta-0.4
TA 0.4 TTAA 0.16 TTAa 0.04 TtAA 0.04 TtAa 0.16
Ta 0.1 TTAa 0.04 TTrr 0.01 TtAa 0.01 Ttaa 0.04
tA 0.1 TtAA 0.04 TtAa 0.01 ttAA 0.01 ttAa 0.04
ta 0.4 TtRr 0.16 Ttaa 0.04 ttAa 0.04 ttaa 0.16
16 TTAA8 TTAa1 TTaa8 TtAA34 TtAa8 Ttaa1
ttAA8 ttAa16 ttaa
66 T_A_ 9 T_aa 9 ttA_ 16 ttaa
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Question 6a
What is the difference between linkage and
pleiotropy? 2 points. Linkage is when alleles
at two loci do not segregate independantly and
hence there is linkage disequilibrium in
segregating populations. The cause is that the
two loci are located on the same chromosome.
Pleiotropy is where two characters are controlled
by alleles at a single locus.
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Question 7a
Two homozygous squash plants were hybridized and
an F1 family produced. One parent was long and
green fruit (LLGG) and the other was round and
yellow fruit (llgg). 1600 F2 progeny were
examined from selfing the F1's and the following
number of phenotypes observed
L_G_ L_gg llG_ llgg
891 312 0 397
Explain what may have caused this departuure from
a 9331 expected frequency of phenotypes 4
points.
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Question 7a
L_G_ L_gg llG_ llgg
891 312 0 397
Explain what may have caused this departuure from
a 9331 expected frequency of phenotypes 4
points.
This departure from a 9331 ratio could be
caused by recessive epistasis, where ll is
epistatic to G, so llG_ and llgg have the same
phenotype.
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Question 7a
A appropriate test to use would be a chi-square
test.
L_G_ L_gg llG_ llgg
Observed 891 312 0 397
Expected 900 300 - 400
Difference 9 12 - 3
D2/exp 0.09 0.48 - 0.02
?2 2df 0.59 ns ?2 2df 0.59 ns ?2 2df 0.59 ns ?2 2df 0.59 ns ?2 2df 0.59 ns
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Bonus Question
A 4x4 half diallel (with selfs) was carried out
in cherry and the following fruit yield of each
possible F1 family observed.
Sm. Red
Sm. Reds 12 Big Yld
Big Yld 27 36 Jims D
Jims D. 21 35 27 Fellman
Fellman 18 27 26 21
Calculate narrow-sense heritability.
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Bonus Question
Mid-Parent Off-spring ?x2 ?xy
24 27 576 648
19.5 21 380 410
16.5 18 272 297
31.5 35 992 1102
28.5 27 812 770
24 26 576 624
144 155 3,609 3,850
SS(x) ?x2 (?x)2/n 153.0 SP(x,y) ?xy (?x
?y)/n 130.5 b 130.5/153.0 0.8529 h2n
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The End
Thank you all Good Luck on Friday