Transcript Test 2-1998-A

Genetics and
Genetic Prediction
in Plant Breeding
Class Test #2, March, 1998
Eight Questions worth 100 points total
Bonus Point worth 10 points
Show all calculations
50 minutes
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
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
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 Fα, based on the
additive/dominance model of inheritance
[3 points].
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 = m+½a+½d = 1611+158.5+78 = 1397.5
B2 = m–½a+½d = 1611-158.5+78 = 1530.5
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].
Tall, Res. (TTYY) x Short, Susc. (ttyy)
F1
TtYy
Tall/Resistant
Tall, Res. (TTYY) x Short, Susc. (ttyy)
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 TTYY:2 TTYy:1 TTyy:2 TtYY:4 TsYy:2 Ttyy:1 ttYY:2 ttYy:1 ttyy
Tall, Res. (TTYY) x Short, Susc. (ttyy)
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
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
#=
Ln[1-0.99] =
Ln[1-0.1406]
Ln(0.01)
Ln(0.8594)
= 30.39, need 31 F3 plants
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
Complete 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].
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) = 171.04+12.56+1.97+2.69 = 188.26
se(C) = 188.26 = 13.72
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
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
P2
41
P1
52
m
46.5
F2
43
F1
49
?
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.
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:
σ2F1= 123.7; σ2F2= 496.2; σ2B1= 357.2; σ2B2= 324.7
Calculate the broad-sense (h2b) and narrow-sense
(h2n) heritability for plant yield [10 points].
Question 4a
σ2F1= 123.7; σ2F2= 496.2; σ2B1= 357.2; σ2B2= 324.7
h2b = Genetic variance
Total variance
E = V(F1) = 123.7
h2b = 496.2 – 123.7
496.2
h2b = 0.75
Question 4a
σ2F1= 123.7; σ2F2= 496.2; σ2B1= 357.2; σ2B2= 324.7
E = V(F1) = 123.7
D = 4[V(B1)+V(B2)-V(F2)-E]
4[357.2+324.7-496.2-123.7] = 248
A = 2[V(F2)-¼D-E] = 2[496.2-62-123.7] = 621
h2n = ½A/V(F2) = 310.5/496.2 = 0.63
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.
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.
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.
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
GCA
df
5
MSq
30,769
SCA
Recipricol
Error
10
10
49
10,934
9,638
5,136
Question 5b
Source
GCA
SCA
Recipricol
Error
df
5
10
10
49
MSq
30,769
10,934
9,638
5,136
F
5.99 **
2.12 ns
1.87 ns
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.
Question 5b
Source
GCA
SCA
Recipricol
Error
df
5
10
10
49
MSq
30,769
10,934
9,638
5,136
F
2.81 ns
2.12 ns
1.87 ns
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.
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 nonrecurrent 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
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.
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
Question 6a
Phenotype
# observed
Tall, awned
954
Short, awned
259
Tall, awnless
221
Short, awned
966
% Recombination
[259+221]/2400 = 0.20
Gametes from
female parent
TA – 0.4
Ta – 0.1
tA – 0.1
ta – 0.4
Gametes from male parent
TA-0.4
Ta-0.1
tA-0.1
ta-0.4
TTAA
0.16
TTAa
0.04
TtAA
0.04
TtRr
0.16
TTAa
0.04
TTrr
0.01
TtAa
0.01
Ttaa
0.04
TtAA
0.04
TtAa
0.01
ttAA
0.01
ttAa
0.04
TtAa
0.16
Ttaa
0.04
ttAa
0.04
ttaa
0.16
16 TTAA:8 TTAa:1 TTaa:8 TtAA:34 TtAa:8 Ttaa:1 ttAA:8 ttAa:16 ttaa
66 T_A_ : 9 T_aa : 9 ttA_ : 16 ttaa
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.
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_
891
L_gg
312
llG_
0
llgg
397
Explain what may have caused this departuure from a
9:3:3:1 expected frequency of phenotypes [4 points].
Question 7a
L_G_
891
L_gg
312
llG_
0
llgg
397
Explain what may have caused this departuure from a
9:3:3:1 expected frequency of phenotypes [4 points].
This departure from a 9:3:3:1 ratio could be caused by
recessive epistasis, where ll is epistatic to G, so llG_
and llgg have the same phenotype.
Question 7a
A appropriate test to use would be a chi-square test.
Observed
Expected
Difference
D2/exp
L_G_
891
900
9
0.09
L_gg
312
300
12
0.48
2 2df = 0.59 ns
llG_
0
-
llgg
397
400
3
0.02
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
Jim’s D
Jim’s D.
21
35
27
Fellman
Fellman
18
27
26
21
Calculate narrow-sense heritability.
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
The End
Thank you all
Good Luck on Friday