1. Flowers, seeds, generations, and “stems*”
*stems used loosely in this context to describe plant tissues capable of asexual reproduction
viachicago.wordpress.com
birdsandbloomsblog.com
tinyfarmblog.com

2. Flowers - sexual reproduction in Angiosperms
Advantages > disadvantages

3. Alternation of generations
The gametophytic (n) and sporophytic (2n) generations

4. Alternation of generations
The sporophytic generation may be diploid (2n = 2x) or polyploid (2n = _x)
VAVA
VAVAVBVB
VAVAVBVBVDVD
2n = 2x = 14
30,000 genes
2n = 4x = 28
60,000 genes
2n = 6x = 42
90,000 genes
1 pair homologous chromosomes
0 sets of homoeologous chromosomes AA
2 pairs of homologous chromosomes
2 sets of homoeologous chromosomes
AABB
3 pairs of homologous chromosomes
3 sets of homoeologous chromosomes
AABBDD
A A B B
A A
A A B B D D

5. Angiosperm reproductive organs and gamete formation Development of the female gametophyte
Reproductive structure: Ovule(s), style, stigma

6. Development of the female gametophyte
Source: arizonabeetlesbugsbirdsandmore.blogspot.com
Pollinator attraction: Petals, nectaries, etc.

7. Development of the female gametophyte
Megaspore mother cell (MMC) 
MMC undergoes meiosis
Of 4 megaspores produced 1 survives (most species)
Three post-meiotic mitoses 1 2 3

8. Development of the female gametophyte
The 8-nucleate embryo sac (1 egg, 2 synergids, 2 primary endosperm nuclei, 3 antipodals)
Source: yougems.reflectionsinfos.com
lima.ohio-state.edu

9. Development of the male gametophyte
Reproductive structures: Anthers; pollen within anthers

10. Development of the male gametophyte
Pollen mother cell (PMC) 
PMC undergoes meiosis
Meiosis gives a tetrad of microspores
Note, this is different than ♀
Meiosis 1
Meiosis 2

11. Development of the male gametophyte
The first mitosis gives vegetative and generative nuclei; at the second mitotic division, the generative nucleus gives 2 sperms.
mitosis
mitosis
mitosis
mitosis
mitosis
mitosis
mitosis
mitosis

12. The pollen pathway The stigma is the site of pollen recognition
Pollen germinates and the vegetative (tube nucleus) grows through the style to the ovule
The two sperm use the tube as conduit

13. Double fertilization
One sperm fertilizes the egg to give the 2n embryo, the other fertilizes the polar nuclei to give the 3n endosperm
antipodals ♀ ♀ 3n
endosperm
polar
nuclei ♀ ♀ ♂ ♀ ♂ ♀
egg 2n
embryo
synergids

14. A review… https://www.youtube.com/watch?v=bUjVHUf4d1I Nn sporophytic
n n n
n n
MMC Nn OR
MMC Nn n N n N
N N N
N N
N N N
PMC Nn n N Nn
sporophytic
generation
n n
N N n N n N n N
n n
N N

15. After double fertilization….
….there are at least four independent and genetically distinct generations coexisting in the seed:
maternal sporophyte diploid tissue
maternal gametophyte haploid tissue
offspring sporophyte diploid tissue
fusion of male (1) and female (2) gametophyte to form triploid tissue

16. A transition to Seeds + = + = 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.

18. A transition to Generation Advance+ = 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
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.

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

20. Backcross
A X B
F1 X A
BC1
BC1 X A x
BC2
BC2 X A
BC3
~ ∞

21. Double chromosome number
Doubled Haploid
A X B F1
Gametes
Plants = F ∞
Double chromosome number

22. 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).

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

24. 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

25. A transition to 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
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.

26. Mendelian Inheritance
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

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

28. Inheritance patterns for polymorphisms
Nuclear genome
Autosomal = Biparental
Sex-linked = XX vs. XY
Cytoplasmic genomes
Chloroplasts and mitochondria = ~ uniparental V v VV Vv vv

29. Cytoplasmic inheritance
Usually maternal inheritance in angiosperms but there are examples of paternal inheritance
Mitochondria
Chloroplast
Dombrowski et al. 1998
Biomedcentral.com

30. Mendelian analysis of spike type in barley
Phenotype 2-row
Vrs1 Vrs1
(Or Vrs1vrs1)
6-row
vrs1vrs1

31. Genotype

32. vrs1 genotypes
Vrs1 phenotypes

33. “Six-rowed barley originated from a mutation in a………
homeobox gene”
Two-rowed is ancestral (wild type)
Six-rowed in the mutant
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

34. X Transcription of Vrs1 Translation of Vrs1 Binding of Vrs1 to “Lat”
No expression of Lat =2 -row
Vrs1
Lat X

35. 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

36. 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

37. 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

38. Doubled haploid production using another culture
Pollen
V V v v
F1 = Vv V v
Induction
Regeneration
Plantlets
Mature plants
Chromosome Doubling
See the 3-D examples in class! VV VV vv vv VV VV vv vv

40. 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

41. 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”

42. 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

43. 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.

44. 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)

46. 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

47. 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.

49. Double chromosome number
Chi square computation for dihybrid ratios
Be able to calculate chi-square tests for dihybrid F2, testcross and DH (see link on webpage)  X A B F1 F2 F3
F ~∞ x
A X B F1
Gametes
Plants = F ∞
Double chromosome number

50. Doubled Haploid production using another culture – 2 loci (unlinked)
Pollen
VN Vn vN vn
F1 = VvNn VN Vn vN vn
Induction
Regeneration
Plantlets
Mature plants
Chromosome Doubling
See the 3-D examples in class!
VVNN
VVnn
vvNN
vvnn
VVNN
VVnn
vvNN
vvnn

51. A transition to Asexual Reproduction+ = 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
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.

52. Asexual reproduction: The clone is immortal
Example #1: Allium sativum
“As far as we know, garlic in cultivation throughout history has only been propagated asexually by way of vegetative cloves, bulbs, and bulbils (or topsets), not from seed. These asexually propagated, genetically distinct selections of garlic we cultivate are more generally called "clones". Yet this asexual lifestyle of cultivated garlic forgoes the possibility of combining traits proffered by interpollinating diverse parental stocks.”

53. Asexual reproduction: The clone is immortal
Example #2: Populus tremuloides
The world's heaviest living thing
1 clone in the Wasatch Mountains of Utah
47,000 stems of genetically identical aspen trees
Total weight: 6 million kilograms
Aspen is dioecious species - this clone is one big male

54. Asexual reproduction plus thus sexual reproduction option
Delicious
Red Delicious
Fuji

55. A transition to Apomixis+ = 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
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.

56. Seeds without sex: Apomixis
Apomixis = parthenogenesis: Development of an egg without fertilization
(Parthenocarpy = development of fruits (seedless) without fertilization)
Seed propagation of heterozygotes: genetically identical from generation to generation.
Obligate
Facultative 

57. Seeds without sex: Apomixis
Prevalence 400 species; 40 families; Common in Poaceae, Asteraceae, Rosaceae
Examples
Tripsacum
Poa pratensis
Pennisetum
Dandelion (Taraxacum spp)
Rubus

58. Seeds without sex: Apomixis
Mechanisms
no (or altered) meiosis to produce unreduced female gametophyte (embryo sac)
no fertilization – but do get autonomous embryo formation
may have autonomous endosperm development, or endosperm development may be triggered by fertilization. Most cases it is triggered by fertilization (pseudogamy = fertilization of central cell)  

59. Seeds without sex: Apomixis
Hand and Koltunow Genetics 197:

60. Seeds without sex: Apomixis
Genetic basis
Apomixis loci and candidate genes
Lots of breeding effort; little success
Epigenetics?

61. Seeds without sex: Apomixis
Economic implications of apomixis
Hybrid seed production? 
Transgenic apomict escape?
Evolutionary implications of apomixis
Obligate
Facultative

62. The Rubus armeniacus case study Himalayan (Armenian) blackberry

63. Is the Himalayan blackberry the perfect weed?
Attribute
Description
Flower
Hermaphroditic
Pollination biology
Self and outcross
Apomixis
Facultative
Seeds
Small and numerous
Vegetative propagation
Yes
Ploidy level
Polyploid
Protection
Thorns
Attraction
Tasty fruit
Was Luther Burbank the “father” of this
“perfect weed”?
promo.idahopotato.com