Mendelian Inheritance

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Mendelian Inheritance

• Mendelian genetic analysis
• The "classical" approach to understanding
  thegenetic basis of a trait.
• Based on analysis of inheritance patterns in the
  progeny of a cross

R R
0 R0 R0
0 R0 R0
R 0
R RR R0
0 R0 00
Gregor Mendel
en.wikipedia.org
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Qualitative/discontinuous variation vs.
quantitative/ continuous variation The number of
genes determining the trait The effects of the
environment
RR beet
rr chard
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Polymorphisms A trait, a gene, a
nucleotide
vs.
Trait (Phenotype)
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A gene (genotype)
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A single nucleotide (genotype)
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• Inheritance patterns for polymorphisms
• Nuclear genome
• Autosomal Biparental
• Sex-linked XX vs. XY
• Cytoplasmic genomes
• Chloroplasts and mitochondria uniparental
• Details at end of this section

V v
V VV Vv
v Vv vv
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Delving into the nuclear genome Polymorphisms,
loci, and alleles Many alleles are possible,
but there are only two alleles per locus in a
diploid individual
V
V
The V locus
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Parent
Gametes
v
v
v
v
Homozygous Dominant
or
v
v
v
v
Heterozygous
or
v
v
v
v
Homozygous Recessive
or
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Crosses between parents generate progeny
populations of different sorts Filial (F)
generations of selfing e.g. F1, F2, F3
backcross doubled haploid recombinant inbred,
etc. 
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Crosses between parents generate progeny
populations of different types. Filial (F)
generations of selfing ( )
Selfing
Heterozygosity
F1
100
F2
50
25
F3
F 8
0
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Crosses between parents generate progeny
populations of different types.
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Crosses between parents generate progeny
populations of different types
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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).
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The degree of homozygosity of the parents will
likely be a function of their mating biology,
e.g. cross vs. self-pollinated. Mendelian
analysis is straightforward when one or two genes
determine the trait.
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• 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

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Monohybrid Model Segregation of alleles at a
locus Example Number of kernel rows in barley
(Hordeum vulgare). The VRS1 locus. Alleles are
Vrs1 and vrs1.  V and v for short. 2
row vs. 6 row V v
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Phenotype2-row V V 6-row v v
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Genotype
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Genotype Vrs1 sequences
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• Six-rowed barley originated from a mutation in
  a
• homeobox gene
• Two-rowed is ancestral (wild type)
• Homeobox genes are transcription factors they
  encode proteins that bind to other genes and thus
  regulate the expression of other genes
• The 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

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• Transcription of Vrs1
• Translation of Vrs1
• Binding of Vrs1 to Lat
• No expression of Lat
• 2-row

Vrs1
Lat
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1. No transcription of vrs1 (or)
2. No translation of vrs1
3. No binding of vrs1 to Lat
4. Lat expresses and lateral florets are fertile
   6-row

vrs1
Lat
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• What 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

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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

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All this happened to VRS1 in the past 10,000
years!
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2012 2-row, 6-row and your local barley
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Determining the inheritance of row type based on
phenotype
Crosses between parents generate progeny
populations of different filial (F) generations
e.g. F1, F2, F3 backcross doubled haploid
recombinant inbred, etc. 
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Doubled Haploid Production using anther culture
F1 (or other generation)
Genotype Vv Ww
Induction
Regeneration
Plantlets
Harvest Seed
Chromosome Doubling
VVww
vvWW
vvww
VVWW
VVWW
VVww
vvWW
vvww
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• 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

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• 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

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The general 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
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• 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.

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• 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)

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Chi square computation for a monohybrid ratio
Example 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 11 (expectation for 2 alleles at 1
locus in a doubled haploid population). The data
are for a SNP in HvHox1 (3_0897) from the Hb
population (n 82).  SNPs are assayed as
nucleotides but converted to "A" and "B" alleles
for each locus. In this example, the OWB-D allele
is A and the OWB-R allele is B. Reviewing the
sequence alignment, OWB-D Guanine (G)  and
OWB-R Adenine (A)
Gametes V v
     
DH genotypes VV vv
     
DH phenotypes Two-row Six-row
     
Number 35 47
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Genotype Vrs1 sequences
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Phenotype Observed Expected O - E (O - E)2/E
VV 35 41 -6 0.89
vv 47 41 6 0.89
Totals 82 82 0 1.75  chi square
p-value (1 df) 0.18. This chi square  is well
within the realm of acceptance, so we conclude
that there is indeed a 11 ratio of two-row
six-row phenotypes (VVvv genotypes) in the OWB
population. Be able to calculate chi-square
tests for monohybrid F2, monohybrid backcross
(including testcross) and DH.
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Chi square computation for dihybrid ratios See
online review if you are not familiar with
dihybrids and chi square calculation Be able to
calculate chi-square tests for dihybrid testcross
and DH. Know how many df you would use for F2
dihybrid. 
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• Cytoplasmic inheritance
• usually maternal inheritance but there are
  examples of paternal inheritance in plants
• Mitochondrial genomes
• Chloroplast genomes

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Mitochondrial genomes
Dombrowski et al. 1998
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Chloroplast genomes
Biomedcentral.com