Plant Science 446/546
Class Test #2 Friday March 10, 2000
Ag.Sci. Room 339
9:30pm to 10:20pm
Name:
Answer all 8 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
1.
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].
Genetic situation
T_S_
Plant phenotype
T_ss
ttS_
ttss
Complete dominance: T is dominant to t, and
S is dominant to s.
Duplicate recessive epistasis: tt epistatic to S,
and ss; and ss epistatic to T and tt.
Recessive epistasis: tt epistatic to S and ss.
Duplicate dominant epistasis: T epistatic to S
and ss; S epistatic to T and tt.
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].
1
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 F3’s. 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].
2
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
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
= 90
= 893
= 907
= 110
Determine the percentage recombination between the foot rot and yellow strip rust loci.
[4 points].
What would be the frequency of phenotypes and genotypes with the above percentage
recombination? [7 points].
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].
3
4a.
4b.
5a.
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].
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].
Why are plant breeders interested in conducting scaling tests for quantitatively
inherited characters of importance in the breeding scheme? [3 points].
4
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
P1
F1
B1
1923
1629
1785
Variance of
yield
74
69
132
Number of
plants
16
16
31
Use an appropriate test to determine whether the additive/dominance model of inheritance
is adequate to explain the variation in yield in canola. [7 points].
6a.
The following is a model for the analysis of diallels (Griffing analysis).
Yijk =  + gi + gi + sij + eijk
Explain what gi and sij represent in the above model. [4 points].
5
6b.
Briefly explain the difference between parents chosen at random (random parent
effects) and parents specifically chosen (fixed parent effects). [4 points].
6c.
A 5 x 5 full diallel (including selfs) design was conducted and yield of the parent self’s
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.
Source
GCA
SCA
Reciprocal
Error
Total
df
4
10
10
50
74
Sum of Squares
7,900
6,010
5,100
17,500
35,710
Complete the analysis assuming that the parents are fixed and briefly outline your
conclusions from the analysis. [4 points].
What difference would you make if the parents in this analysis were chosen at random.
[2 points].
6
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 Vi’s was 26; sum of Wi’s 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].
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].
7
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.
Family
P1
P2
F2
B1
B2
Standard deviation
3.521
3.317
6.008
5.450
5.157
[10 points].
8
8a.
8b.
8c.
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
=
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
=
From a properly designed field trial that included B1, B2 and F1 families, the following
yield estimates were obtained.
B1 = 42.0 lb/acre;
B2 = 26.0 lb/acre;
F1 = 38.5 lb/acre
From these family means, estimate the expected value of P1, P2, and F2, based on the
additive/dominance model of inheritance [6 points].
P1 =
P2 =
F2 =
9
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].
10