Mendelian Inheritance
Mendelian genetic analysis is based on discontinuous variation
Polymorphisms ________
Inheritance patterns are a function of which genome the gene is found in
Nuclear ________
autosomal ________
sex-linked ________
Cytoplasmic ________
Autosomal inheritance
The determinants of polymorphisms are the alleles at a locus
Many alleles are possible, but there are only two per locus in a diploid individual
The "classical" approach to understanding the genetic basis of a difference in phenotype is to use progeny to understand the parents. Information on DNA sequence and/or gene expression may let you look directly at differences between individuals. More
Point: Read and understand!
Monohybrid Model:
Segregation ________
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.
Vrs1 as genotype: Komatsuda et al. 2007. PNAS 104:1424-1429.
Point: Read (at least) abstract and introduction.
Principles of Komatsuda et al.
Point: Mutation as a source of new alleles; genes can encode transcription factors; the basis of Mendelian inheritance can be determined!
Brewing and Vrs1
Point: FYI!
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Generation: |
Parent 1 |
Parent 2 |
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Genotype: |
VV |
vv |
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Phenotype: |
Two-row |
Six-row |
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Generation: |
F1 |
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Genotype |
Vv |
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Phenotype: |
Two-row |
If the F1 is selfed to give an F2 generation, the genotypic ratio will be 1VV; 2Vv; 1vv and the phenotypic ratio will be 3 two-row: 1 six-row
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V |
v |
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VV |
Vv |
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v |
Vv |
vv |
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In a doubled haploid (DH) population there are no heterozygotes. Thus, the expected phenotypic and the genotypic ratios are the same 1 VV: 1 vv; 1 two-row: 1 six-row. The actual numbers in the Oregon Wolfe Barley population are 40 VV and 54 vv.
Point: Understand that DH = producing 2n plants form n gametes (male or female).
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V |
v |
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Dihybrid Model:
Segregation and independent assortment ________ More
Point: Review if you do not remember 2 factor Punnet squares
As demonstrated in the previous examples the expected ratios will be a function of the
- type of progeny
- number of loci involved
- degree of dominance
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 general formula is:
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Chi square = (O1 - E1)2/E1 +........+ (On - En)2/En |
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where O1 = number of observed members of the first class |
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E1 = number of expected members of the first class |
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On = number of observed members of the nth class |
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En = number of expected members of the nth class |
As deviations from hypothesized ratios are smaller, the chi square value approaches 0; there is a good fit. If there is a poor fit of observed data to the hypothesized ratio, the chi square value is larger. Generally speaking, 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.
1. Determine degrees of freedom (df). df = number of classes - 1.
2. Consult calculator (on web) or Table 3.1 (textbook).
Point: Know how to use the table
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 1:1 (expectation for 2 alleles at 1 locus in a doubled haploid population)
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Gametes |
V |
v |
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| DH genotypes |
VV |
vv |
| DH phenotypes |
Two-row |
Six-row |
| Number |
40 |
54 |
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Phenotype |
#Observed |
#Expected |
O - E |
(O - E)2/E |
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VV |
40 |
47 |
-7 |
1.04 |
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vv |
54 |
47 |
7 |
1.04 |
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Totals |
94 |
94 |
0 |
2.08 = chi square |
Prob( X^2 >= 2.08 | Degrees of Freedom=1 ) <= 0.14924. 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 (VV:vv genotypes) in the OWB population.
Chi square computation for a dihybrid ratio More
Point: Review if you are not familiar with dihybrids and chi square calculation
More ways to deal with Mendelian expectations
Cytoplasmic inheritance: usually maternal inheritance but there are examples of paternal inheritance in plants More
Point: Size and autonomy of organellar vs. nuclear genomes
Mitochondrial genomes ________
Chloroplast genomes ________
Text: 37 - 42 (segregation); 90 - 101 (independent assortment and Chi square test)
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