Generations and Mendel X - PowerPoint Presentation

Slide1

Generations and Mendel

X

A

B

F

1

F

2

F

3

F

~

∞

% Heterozygosity

x

x

x

100

50

25

~ 0

Slide2

Generation

Advance

N

n

N

NN

Nn

n

Nn

nn+

=NN NN N NN NN N NN N NN N

+=nn nN N NN NN N NN N nN n

+=nn nn n nn nn n nn n nn n

+=NN Nn n nn nn n nn n Nn 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 fertilizedegg is carried to the Punnett square.

Slide3

Gregor

Mendel

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/

X

A

B

F

1

F

2

F

3

F

~

∞

% Heterozygosity

x

x

x

100

50

25

~ 0

Slide4

Crosses between parents generate progeny populations of different types

Filial (F) generations of selfing

Selfing (assuming homozygous parents)

X

A

B

F

1

F

2

F

3

F

~

∞

% Heterozygosity

x

x

x

100

50

25

~ 0

x

Slide5

X

A

B

F

1

F

2

F

3

F

~

∞

% Heterozygosity

x

x

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.

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

Slide6

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

F

1

X A (or B)

BC1

BC1 X A (B)

BC1

x

x

x

BC2

x

x

BC2 X A (B)

BC3

~ ∞

~ ∞

Slide7

Testcross:

A backcross where the recurrent parent is recessive for the target gene(s)

A X B

F

1

X A

BC1

BC1 X A

BC1

x

x

x

BC2

x

x

BC2 X A

BC3

~ ∞

~ ∞

Slide8

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

F

1

F

2

F

3

F

~

∞

x

x

x

….

….

….

….

Slide9

Doubled Haploid:

haploid gametes used to generate homozygous 2n plants

A X B

F

1

Gametes

Plants = F ∞

Double chromosome number

….

….

….

….

Slide10

Making a cross

Hermaphrodite (barley)

Monoecious (maize)

Dioecious (hops)

Slide11

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).

Slide12

The degree of

homozygosity

of the parents

will likely be a function of their mating biology, e.g. cross vs. self-pollinated.

Slide13

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

Slide14

Mendelian

analysis

N

n

N

NN

Nn

n

Nnnn

+=NN NN N NN NN N NN N NN N

+=nn nN N NN NN N NN N nN n

+=nn nn n nn nn n nn n nn n

+=NN Nn n nn nn n nn n Nn N

Genetic analysis is straightforward when one or two genes determine the target trait

Slide15

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

R

0

R0

R0

0R0R0

R0RRRR00R000

Gregor Mendel

en.wikipedia.orgRR x OO F1 x F2

Slide16

R

R

0

R0

R0

0

R0R0

R

0

RRRR00R000

Gregor Mendel

en.wikipedia.orgRoundup Ready (RR) x Organic (OO)F1: RO = Roundup ReadyF2: 1 RR: 2RO: 1OO = 3 Roundup Ready: 1 non-Roundup Ready

Mendelian genetics and transgenic gene flow https://tilth.org/https://www.genuity.com/specialty/Pages/Roundup-Ready-Sugarbeets.aspx

Slide17

Mendelian

Qualitative

(discontinuous)

variation

Quantitative

(continuous)

variation

The number of genes determining the trait and/or

The effects of the environment

Parent 1

Parent 2Parent 1Parent 2

Slide18

Qualitative Quantitative variation

With just 3 unlinked loci: A B C – start to approach a quantitative distribution

If homozygous for resistance alleles (-) at a locus = 0% contribution to disease severity

Each susceptibility allele (+) at each locus = 10% disease severity

Imagine if there were 10 genes and heterozygotes!

A locus

B locus

C locus Value - -

- -- -

0- - - - + + 20- -+ + - - 20 + +- - - - 20+ + + + - -40+ + - - + + 40- - + + + +40 +++ ++ + 60

Slide19

Quantitative Trait Loci

Genetic determinants of quantitative variation

Loci that determine, or are associated with loci that determine,

variation in a phenotype

Slide20

Nuclear genome

Autosomal =

Biparental

Sex-linked = XX vs. XY

R

O

R

RRRrORrrrInheritance patterns for polymorphismsXXXXXXXYXY

XYCytoplasmic genomesChloroplasts and mitochondria: ~ uniparental~Maternal in angiosperms~Paternal in gymnosperms Z

ZZ

Slide21

Mitochondria

Chloroplast

Cytoplasmic inheritance

Biomedcentral.com

Dombrowski

et al. 1998

Slide22

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

Slide23

Phenotype

2-row

Vrs1

Vrs1

(Or

Vrs1vrs1)

6-row

vrs1vrs1

Mendelian analysis of spike type in barley*

*Autosomal trait

Slide24

Genotype

Slide25

vrs1

genotypes

Vrs1

phenotypes

Slide26

“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 factorsThe vrs1 (hox1

) model: In a two-row, the product of Vrs1 binds to another (unknown) gene (or genes) that determine the fertility of lateral floretsBy preventing expression of this other gene, lateral florets are sterile and thus the inflorescence has two rows of lateral floretsIn a 6-row (vrs1vrs1) there is a loss of function

Slide27

Vrs1

Lat

X

Transcription of

Vrs1

Translation of

Vrs1

Binding of

Vrs1

to

“Lat”*

No expression of

Lat

=2 -row

*

Lat

a hypothetical gene

Slide28

vrs1

Lat

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

X*Lat a hypothetical gene

Slide29

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

Slide30

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

Slide31

OWB-D X OWB-R

F

1

Gametes

Plants = F ∞

Double chromosome number

….

….

….

….

Doubled haploid generation advance

Slide32

A quick overview of doubled haploids

F

1

=

Nud

nud

(2n = 2x = 14)

Gametes (male or female)

Nud

nud (n = 7)

Induction/regeneration

Nud

nud

Haploid plantlets (n = 7)

Chromosome doubling (spontaneous/induced)

Mature plants/grain

Nud

Nud

nud

nud

(2n = 2x = 14)

Slide33

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

Slide34

OWB dominant x OWB recessive

F1

Doubled haploids

Slide35

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

Slide36

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

”

Slide37

Chi square = (O

1

- E

1

)

2

/E1 +........+ (On - E

n)2/E

n

where O1 = number of observed members of the first classE1 = number of expected members of the first class

On = number of observed members of the nth classEn = number of expected members of the nth

class

The Chi square formula

Slide38

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.

The Chi square concept

Slide39

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)

Slide40

Slide41

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 VVvv

   

DH phenotypesTwo-rowSix-row

   

Number 3547

Chi square computation for a monohybrid ratio

Slide42

Phenotype

#Observed

#Expected

O - E

(O - E)

2

/E

VV

35

41

-6

0.89

vv4741

60.89Totals

828201.75 =  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 a monohybrid ratio

Slide43

Slide44

Chi square computation for dihybrid ratios

Be able to calculate chi-square tests for dihybrid F2, testcross and DH 

X

A

B

F

1

F

2

F

3

F

~

∞

% Heterozygosity

x

x

x

100

50

25

~ 0

A X B

F

1

X A

BC1

BC1 X A

BC1

x

x

x

BC2

x

x

BC2 X A

BC3

~ ∞

~ ∞