Plant Science 446/546
Class Test #2 March 23, 1998
Ag.Sci. Room 339
9.30pm to 10.30pm
Name :
Answer all 7 questions
A total of 100 points are available
A Bonus question is available for an extra 10 points
Points available from each part of each question
are shown in bold square parenthesis
Try to be as brief and concise as possible
Please write in a legible form
Show any working/calculations
Make sure that any additional paper used
is attached to the questionnaire
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
P2
F1
1b.
From these expectations, what would be the expected values for F2, B1 and B2 based
on m, [a] and [d]. [3 points].
F2
B1
B2
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].
F2 =
B1 =
B2 =
1
2a.
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].
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].
2
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
P1
P2
F1
F2
52
41
49
43
Variance
# Plants
1.97
10
2.69
10
3.14
10
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].
If the additive/dominance model is inadequate, list three factors which could cause
the lack of fit of the model [3 points].
1.
2.
3.
3
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
σ2F2
σ2B1
σ2B2
= 123.7
= 496.2
= 357.2
= 324.7
Calculate the broad-sense (h2b) and narrow-sense (h2n) heritability for plant yield
[10 points].
h2b =
h2n =
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].
4
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.
2.
3.
4.
Why would you choose a Method 3 over a Method 1? [1 point]
Why would you choose a Method 2 over a Method 1? [1 point]
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 of variation
df
M.Sq
GCA
SCA
Reciprocal
Replicate error
5
10
10
30,769
10,934
9,638
49
5,136
Complete the analysis of variance and explain your conclusions from the analysis
[10 points].
5
5c.
Given that the parents were chosen at random, how would this change the results
and your conclusions [4 points].
5d.
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:
Parent 1
Parent 2
Parent 3
Parent 4
Parent 5
Vi
491.4
610.3
302.4
310.2
832.7
Wi
436.8
664.2
234.8
226.9
769.4
GCA
-0.76
+12.92
-14.32
-15.77
+17.93
Without further calculations, what can be deduced about the inheritance of plant
height in chick-pea? [6 points].
6
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
tall, awned
short, awned
tall, awnless
short, awnless
Number observed
954
259
221
966
Given the above information, determine the genotypic and phenotypic frequency of
plant types that would be expected at F2, given that the F1 family was self pollinated
[10 points].
6b.
Briefly explain the difference between linkage and pleiotropy [2 points].
7
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:
LLGG L-gg
891
llG312
llgg
0
397
Explain what may have caused this departure from a 9:3:3:1 expected frequency of
phenotypes [4 points].
7b.
Use an appropriate statistical test to prove your hypothesis [6 points].
8
Bonus question
Worth + 10 points
A 4x4 half diallel design (with selfs) was carried out in cherry and the following fruit yields
of each possible F1 family was observed:
Small reds
Big yields
Jim's delight
Fellman's best
12
27
21
28
36
35
27
27
26
Small
reds
Big
yields
Jim's
delight
21
Fell's
best
From the above data, determine the narrow-sense heritability for yield in cherry [possible
10 points].
9