Generations and Mendel X A B F1 F2 F3 F ~∞ % Heterozygosity x 100 50 25 ~ 0

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.

X A B % Heterozygosity F1 100 x F2 50 F3 25 Gregor Mendel F ~∞ ~ 0 en.wikipedia.org https://keen101.wordpress.com/2015/08/19/pea-breeding-resources/ http://www.interactive-biology.com/3813/mendels-pea-plant-experiment-the-root-of-all-in-genetics/

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

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 http://www.johnnyseeds.com/flowers/sunflowers/tall-sunflowers/royal-hybrid-1121-sunflower-f1-sunflower-seed-2603.html

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 ~ ∞

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 ~ ∞

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 …. …. …. ….

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

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

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

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

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

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 https://www.genuity.com/specialty/Pages/Roundup-Ready-Sugarbeets.aspx https://tilth.org/ R R0 R RR R0 00 Gregor Mendel en.wikipedia.org

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

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

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

Inheritance patterns for polymorphisms RR 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

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

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 http://www.plantcell.org/content/11/4/571

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

Genotype

vrs1 genotypes Vrs1 phenotypes

“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

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

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

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

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

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

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)

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

OWB dominant x OWB recessive F1 Doubled haploids

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.

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