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

Flowers - sexual reproduction in Angiosperms Advantages > disadvantages

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

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

Angiosperm reproductive organs and gamete formation Development of the female gametophyte Reproductive structure: Ovule(s), style, stigma http://www.extension.org/pages/32204/stigma

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

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

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

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

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

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

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

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

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

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

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.

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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.

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

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

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

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).

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

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

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.

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

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

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

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

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

Genotype

vrs1 genotypes Vrs1 phenotypes

“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

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

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

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

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

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

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

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”

The Chi square formula Chi square = (O1 - E1)2/E1 +........+ (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

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. 0.025, 0.01, .005) lie in the area of rejection.

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)

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

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.

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

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

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.

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.” http://www.ars.usda.gov/Research/docs.htm?docid=5232

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 http://waynesword.palomar.edu/ww0601.htm#aspen

Asexual reproduction plus thus sexual reproduction option Delicious Red Delicious Fuji

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.

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 

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

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)  

Seeds without sex: Apomixis Hand and Koltunow. 2014. Genetics 197: 441-450.

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

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

The Rubus armeniacus case study Himalayan (Armenian) blackberry

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