1. Generations and MendelX A B F1 F2 F3
F ~∞
% Heterozygosity x
100 50 25
~ 0

2. Generation Advance + = + = N n NN Nn nn + = + =
Note: At this point in the figure, the antipodals and synergids are deleted and only the
fertilized endosperm nuclei (now 3n) and fertilized egg (now 2n) are shown. Only the fertilized
egg is carried to the Punnett square.

3. X A B % Heterozygosity F1 100 x F2 50 F3 25 Gregor Mendel F ~∞ ~ 0
en.wikipedia.org

4. Selfing (assuming homozygous parents)
Crosses between parents generate progeny populations of different types
Filial (F) generations of selfing x
Selfing (assuming homozygous parents) X A B F1 F2 F3
F ~∞
% Heterozygosity x
100 50 25
~ 0

5. Each plant generation produces seed of the
next generation: The seed on an F1 plant is F2 seed; the seed
on an F2 plant is F3 seed, etc. X A B F1 F2 F3
F ~∞
% Heterozygosity x
100 50 25
~ 0

6. Backcross: The F1 crossed to either recurrent parent
The number of crosses to the recurrent parent and subsequent cycles of selfing are dictated by experiment/breeding goals
A X B
F1 X A (or B)
BC1
BC1 X A (B) x
BC2
BC2 X A (B)
BC3
~ ∞

7. Testcross: A backcross where the recurrent parent is recessive for the target gene(s)
A X B
F1 X A
BC1
BC1 X A x
BC2
BC2 X A
BC3
~ ∞

8. Recombinant inbred lines: A cross between two parents, with multiple F2 plants advanced to subsequent generations of selfing to generate
a population of inbred siblings X A B F1 F2 F3
F ~∞ x …. …. …. ….

9. Double chromosome number
Doubled Haploid: haploid gametes used to generate homozygous 2n plants
A X B F1
Gametes
Plants = F ∞
Double chromosome number …. …. …. ….

10. Making a cross Hermaphrodite (barley) Monoecious (maize)
Dioecious (hops)

11. The genetic status (degree of homozygosity) of the parents will determine which generation is appropriate for genetic analysis and the interpretation of the data (e.g. comparison of observed vs. expected phenotypes or genotypes).

12. The degree of homozygosity of the parents will likely be a function of their mating biology, e.g. cross vs. self-pollinated.

13. Expected and observed ratios in cross progeny will be a function of:
the degree of homozygosity of the parents
the generation studied
the degree of dominance
the degree of interaction between genes
the number of genes determining the trait

14. Mendelian analysis
Genetic analysis is straightforward when one or two genes determine the target trait + = n
n n
N N N
N N
N N n
N n + = N
N N
N N N N n NN Nn nn + = N
N N
n n n
n n
n n N
n N + = n
n n
n n n

15. Mendelian Inheritance
RR x OO
F1 x F2
Mendelian genetic analysis:
The "classical" approach to understanding the genetic basis of a trait.
Based on analysis of inheritance patterns in the progeny of a cross R R0 R RR R0 00
Gregor Mendel
en.wikipedia.org

16. Mendelian genetics and transgenic gene flow
Roundup Ready (RR) x Organic (OO)
F1: RO = Roundup Ready
F2: 1 RR: 2RO: 1OO = 3 Roundup Ready: 1 non-Roundup Ready R R0 R RR R0 00
Gregor Mendel
en.wikipedia.org

17. Mendelian Qualitative (discontinuous) variation Quantitative
Parent 1
Parent 2
Parent 1
Parent 2
The number of genes determining the trait and/or
The effects of the environment

18. Qualitative Quantitative variation
With just 3 unlinked loci: A B C – start to approach a quantitative distribution
Homozygous resistance alleles at each locus = 0% disease severity
Homozygous susceptibility alleles at each locus = 10% disease severity
Imagine if there were 10 genes and heterozygotes!
A locus
B locus
C locus
Value
- -
+ + 20 40 ++ 60

19. Quantitative Trait Loci
Genetic determinants of quantitative variation
Loci that determine, or are associated with loci that determine,
variation in a phenotype

20. Inheritance patterns for polymorphismsRR Rr rr
Nuclear genome
Autosomal = Biparental
Sex-linked = XX vs. XY X XX Y XY
Cytoplasmic genomes
Chloroplasts and mitochondria: ~ uniparental
~Maternal in angiosperms
~Paternal in gymnosperms Z

21. Cytoplasmic inheritance
Mitochondria
Chloroplast
Dombrowski et al. 1998
Biomedcentral.com

22. Cytoplasmic inheritance – origins, function, and important traits
Mitochondria
Once free-living bacteria – now endosymbionts
Respiration
~ 50 genes
Cross talk with nucleus – coevolution and horizontal transfer
Male sterility (cytoplasmic male sterility)
Chloroplast
Once free-living cyanobacteria– now endosymbionts
Photosynthesis
Cross talk with nucleus – coevolution and horizontal transfer
~ 100 genes
Variegation

23. Mendelian analysis of spike type in barley*
Phenotype 2-row
Vrs1 Vrs1
(Or Vrs1vrs1)
6-row
vrs1vrs1
*Autosomal trait

24. Genotype

25. vrs1 genotypes
Vrs1 phenotypes

26. “Six-rowed barley originated from a mutation in a………
homeobox gene”
Two-rowed is ancestral (wild type) Vrs1
Six-rowed in the mutant vrs1
Homeobox genes are transcription factors
The vrs1 (hox1) model:
In a two-row, the product of Vrs1 binds to another (unknown) gene (or genes) that determine the fertility of lateral florets
By preventing expression of this other gene, lateral florets are sterile and thus the inflorescence has two rows of lateral florets
In a 6-row (vrs1vrs1) there is a loss of function

27. X Transcription of Vrs1 Translation of Vrs1 Binding of Vrs1 to “Lat”*
No expression of Lat =2 -row
Vrs1
Lat X
*Lat a hypothetical gene

28. X X No transcription of vrs1 (or) No translation of vrs1
No binding of vrs1 to “Lat”*
“Lat” expresses and lateral florets
are fertile = 6-row
vrs1
Lat X X
*Lat a hypothetical gene

29. What mutations happened to Vrs1 to make it vrs1 (loss of function)?
Complete deletion of the gene ( - transcription, - translation so no protein)
Deletions of (or insertions into) key regions of the gene leading to - transcription and/or + transcription but – translation, or incorrect translation
Nucleotide changes leading to + transcription, but incorrect translation leading to non-functional protein

30. How many alleles are possible at a locus?
Only two per diploid individual, but many are possible in a population of individuals
New alleles arise through mutation
Some mutations have no discernible effect on phenotype
Different mutations in the same gene may lead to the same or different phenotypes

31. Double chromosome number
Doubled haploid generation advance
OWB-D X OWB-R F1
Gametes
Plants = F ∞
Double chromosome number …. …. …. ….

32. Double chromosome number
Doubled haploid generation advance
OWB-D X OWB-R F1
Gametes
Plants = F ∞
Double chromosome number …. …. …. ….

33. A quick overview of doubled haploids
Gametes (male or female)
Nud nud (n = 7)
F1 = Nud nud (2n = 2x = 14)
Induction/regeneration
Nud
nud
Mature plants/grain
Haploid plantlets (n = 7)
Nud Nud
nud nud
(2n = 2x = 14)
Chromosome doubling (spontaneous/induced)

34. Anther Culture DH production in barley
Harvest donor spikes
Apply stress conditions
Place anthers on induction medium
Sub-culturing for shoots and roots
Spontaneous doubling
Produce seed
Breeding/genetics forever

35. OWB dominant x OWB recessiveF1
Doubled haploids

36. Hypothesis Testing: Determining the “Goodness of Fit”
Expected and observed ratios in cross progeny will be a function of
the degree of homozygosity of the parents
the generation studied
the degree of dominance
the degree of interaction between genes
the number of genes determining the trait

37. Determining the “Goodness of Fit”
Hypothesis Testing:
Determining the “Goodness of Fit”
The Chi Square statistic tests "goodness of fit“; that is, how closely observed and predicted results agree
The degrees of freedom that are used for the test are a function of the number of classes
This is a test of a null hypothesis: “the observed ratio and expected ratios are not different”

38. The Chi square formula
Chi square = (O1 - E1)2/E (On - En)2/En
where O1 = number of observed members of the first class
E1 = number of expected members of the first class
On = number of observed members of the nth class
En = number of expected members of the nth class

39. The Chi square concept
As deviations from hypothesized ratios get smaller, the chi square value approaches 0; there is a good fit.
As deviations from hypothesized ratios get larger, the chi square value gets larger; there is a poor fit.
What determines good vs. poor?
The probability of observing a deviation as large, or larger, due to chance alone.
p values below 0.05 (i.e , 0.01, .005) lie in the area of rejection.

40. Interpreting the chi square statistic in terms of probability.
Determine degrees of freedom (df). df =  number of classes - 1.
2. Consult chi square table and/or calculator (on web)

42. Chi square computation for a
monohybrid ratio
The data: Number of kernel rows (Vrs-1/vrs-1) in barley (Hordeum vulgare).  For simplicity, vrs-1 is abbreviated as "v" in the following table.  Hypothesis is 1:1 (expectation for 2 alleles at 1 locus in a doubled haploid population).
Gametes V v
DH genotypes VV vv
DH phenotypes
Two-row
Six-row
Number 35 47

43. Chi square computation for a
monohybrid ratio
Phenotype
#Observed
#Expected
O - E
(O - E)2/E VV 35 41 -6
0.89 vv 47 6
Totals 82
1.75 =  chi square
Consult table (next slide): p-value (1 df) = 0.18
This chi square  is well within the realm of acceptance, so we conclude that there is indeed a 1:1 ratio of two-row: six-row phenotypes in the OWB population.

45. Chi square computation for dihybrid ratios
Be able to calculate chi-square tests for dihybrid F2, testcross and DH 
A X B
F1 X A
BC1
BC1 X A x
BC2
BC2 X A
BC3
~ ∞ X A B F1 F2 F3
F ~∞
% Heterozygosity x
100 50 25
~ 0