Fall 2014 HORT6033 Molecular Plant Breeding INSTRUCTOR: AINONG SHI HORT6033 web site:...

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Fall 2014 HORT6033 Molecular Plant Breeding INSTRUCTOR: AINONG SHI HORT6033 web site: http://comp.uark.edu/~ashi/MB

Transcript of Fall 2014 HORT6033 Molecular Plant Breeding INSTRUCTOR: AINONG SHI HORT6033 web site:...

Page 1: Fall 2014 HORT6033 Molecular Plant Breeding INSTRUCTOR: AINONG SHI HORT6033 web site: ashi/MBashi/MB.

Fall 2014HORT6033

Molecular Plant BreedingINSTRUCTOR: AINONG SHI

HORT6033 web site: http://comp.uark.edu/~ashi/MB

Page 2: Fall 2014 HORT6033 Molecular Plant Breeding INSTRUCTOR: AINONG SHI HORT6033 web site: ashi/MBashi/MB.

Fall 2014 HORT6033Molecular Plant Breeding

Lecture 6 (09/10/2014)

I. Mendel genetics II.ExampleIII.Research ProjectIV.HomeworkV.Questions

Page 3: Fall 2014 HORT6033 Molecular Plant Breeding INSTRUCTOR: AINONG SHI HORT6033 web site: ashi/MBashi/MB.

Biology Junction

• Genetics is the study of genes.• Inheritance is how traits, or

characteristics, are passed on from generation to generation.

• Chromosomes are made up of genes, which are made up of DNA

• Arrangement of nucleotides in DNA• DNA RNA Proteins

• Genetic material (genes,chromosomes, DNA) is found inside the nucleus of a cell.

Mendel Genetics

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Gene – a unit of heredity; a section of DNA sequence encoding a single protein

Genome – the entire set of genes in an organism

Alleles – two genes that occupy the same position on homologous chromosomes and that cover the same trait (like ‘flavors’ of a trait), such as Sw-5 and sw-5 of tomato spotted wilt virus resistance.

Locus – a fixed location on a strand of DNA where a gene or one of its alleles is located.

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Genotype – the genetic makeup of an organisms,Such as Sw5Sw5Phenotype – the physical appearance

of an organism (Genotype + environment)Such as Tomato spotted wilt virus resistace

Monohybrid cross: a genetic cross involving a single pair of genes (one trait); parents differ by a single trait.P = Parental generationF1 = First filial generation; offspring from a genetic cross.F2 = Second filial generation of a genetic cross

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Homozygous – having identical genes (one from each parent) for a particular characteristic such as Sw-5Sw-5

Heterozygous – having two different genes for a particular characteristic, such as Sw-5sw-5.

Dominant – the allele of a gene that masks or suppresses the expression of an alternate allele; the trait appears in the heterozygous condition.

Recessive – an allele that is masked by a dominant allele; does not appear in the heterozygous condition, only in homozygous.

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Flower colorFor example, flower color:

P = purple (dominant)

p = white (recessive)

If you cross a homozygous Purple (PP) with a homozygous white (pp):

P P p p

P p ALL PURPLE (Pp)

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Cross the F1 generation:

P p P p

P P P p

P p

p p

P

p

P pGenotypes:1 PP2 Pp1 pp

Phenotypes: 3 Purple 1 White

Page 9: Fall 2014 HORT6033 Molecular Plant Breeding INSTRUCTOR: AINONG SHI HORT6033 web site: ashi/MBashi/MB.

1. Suppose:Pea has two flower colors: purple and white, controlled by a major gene ‘T1’ and the purple is dominant. Parent 1 is purple flower (T1T1) and Parent 2 white flower (t1t1). Questions:

(1) If we make a cross between the two parents, what is the color in F1?Color=

(2) If the F1 self-pollinated, what is the expected ratio between purple to white color plants in the F2 population?

Purple : white =

(3) If F1 is backcrossed to Parent 1, what is the expected ratio between purple to white color plants in BC1F1(P1) population?

Purple : white =

(4) If F1 is backcrossed to Parent 2, what is the expected ratio between purple to white color plants in BC1F1(P2) population?

Purple : white =

Page 10: Fall 2014 HORT6033 Molecular Plant Breeding INSTRUCTOR: AINONG SHI HORT6033 web site: ashi/MBashi/MB.

Public Talk for Mendel Genetics

http://www.slideshare.net/mazz4/genetics-a-d-29687785?next_slideshow=1

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F2AABB x aabb

male

AB(1-r)/2

ab(1-r)/2

Abr/2

aBr/2

Female

AB(1-r)/2

AABB(1-r)(1-r)/4

AaBb(1-r)(1-r)/4

AABbr(1-r)/4

AaBBr(1-r)/4

ab(1-r)/2

AaBb(1-r)(1-r)/4

aabb(1-r)(1-r)/4

Aabbr(1-r)/4

aaBbr(1-r)/4

Ab r/2

AABbr(1-r)/4

Aabbr(1-r)/4

AAbbr*r/4

AaBbr*r/4

aBr/2

AaBBr(1-r)/4

aaBbr(1-r)/4

AaBbr*r/4

aaBBr*r /4

Two Gene ModelP1 (AABB) x P2 (aabb) F1 (AaBa)

When r=0.5, two genes independent, i.e. two genes are located at different chromosomes;When r=0, two genes are located at same locus, i.e. same gene or no crossover

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F2

AAbb x aaBB†

Male Ab

(1-r)/2aB

(1-r)/2AB r/2

ab r/2

Female

Ab (1-r)/2

AAbb(1-r)(1-r)/4

R‡

AaBb(1-r)(1-r)/4

Seg=R

AABbr(1-r)/4

R

Aabbr(1-r)/4

Seg

aB(1-r)/2

AaBb(1-r)(1-r)/4

Seg=R

aaBB(1-r)(1-r)/4

R

AaBBr(1-r)/4

R

aaBbr(1-r)/4

Seg

AB r/2

AABbr(1-r)/4

R

AaBBr(1-r)/4

R

AABBr*r/4

R

AaBbr*r/4

Seg=R§

ab r/2

Aabbr(1-r)/4

Seg

aaBbr(1-r)/4

Seg‡

AaBbr*r/4Seg=R

aabbr*r /4

S‡

Genotypes and their frequencies postulated for the two allele A and B with a recombination frequency (r) in F2 and F2:3 population derived from a

cross P1(AAbb) x P2 (aaBB).

† A, presence of Rsv1 allele; B, presence of Rsv1-y allele; a, presence of rsv1 allele; b, presence of rsv1-y allele.‡ R, resistant; S, susceptible; Seg, segregating for resistant and susceptible.§ Since it is extremely difficult to distinguish F2:3 lines that are segregating for two closely genes (AaBb) from F2:3 lines

that are homogenous resistant due to limitation of population, lines with AaBb are grouped with All R class marked Seg=R in the table.

LINKED

Page 13: Fall 2014 HORT6033 Molecular Plant Breeding INSTRUCTOR: AINONG SHI HORT6033 web site: ashi/MBashi/MB.

Questions: (1) Write the Chi-square (χ2) formula using above count for testing M1 and T1 as a single allele:Answer: χ2 (M1) =

Answer: χ2 (T1) = (2) What is the recombination rate r between M1 and T1?

Answer: r =

13. Suppose: A SNP marker M1 [A/C] is linked to the pea color gene ‘T1’ with the recombination rate r. In the above 1-(4) BC1F1(P2) population, the genotypes and phenotypes and their count are below.

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14. If there are 100 individuals and the recombination rate r between M1 and T1 is 0.1 in above BC1F1(P2) population, what are the expected count for the four genotypes?ACT1t1 =CCt1t1 = ACt1t1 = CCT1t1 =

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Research Project Available in UAF Vegetable Breeding program I. Dandelion1. EST-SSR discovery and validation in dandelion2. EST-SNP discovery and validation in dandelion II. SpinachAssociation analysis and SNP markers for leaf traits in spinachAssociation analysis and SNP markers for mineral components in spinachSNP discovery and genetic diversity in spinachSSR discovery and genetic diversity in spinachSlowing-booting and QTL mapping in spinachQTL mapping and SNP marker identification for downy mildew resistance in spinachDevelop a SNP set for spinach variety determination III. Chicory and EndiveEST-SSR discovery and validation in chicory and endiveEST-SNP discovery and validation in chicory and endiveSNP discovery and genetic diversity in chicorySNP discovery and genetic diversity in endiveDevelop a SNP set for chicory and endive variety determination

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IV. Cowpea

1. EST-SSR discovery and validation in cowpea and genetic diversity2. EST-SNP discovery and validation in cowpea and genetic diversity3. SNP discovery from genotyping by sequencing in cowpea4. Genetic diversity analysis for world-wide cowpea germplasm5. Association analysis of low phosphorus efficiency in cowpea6. Association analysis of aphid tolerance in cowpea7. Association analysis of cowpea bacterial blight in cowpea8. Association analysis for morphological traits in cowpea9. Association analysis of salt tolerance in cowpea10. Association analysis of cowpea mosaic virus in cowpea11. Association analysis of iron deficiency chlorosis in cowpea12. Association analysis of seed size in cowpea13. Association analysis of maturity in cowpea14. Association analysis of seed ANTIOXIDNT content in cowpea15. Association analysis of seed protein content in cowpea16. Develop a SNP set for cowpea variety determination

Research Project Available in UAF Vegetable Breeding program