Hypothesis testing

 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:

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, 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.

1. Determine degrees of freedom (df). df = number of classes - 1.

2. Consult calculator (on web) or Table 3.1 (textbook). Know how to use the table

A. 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)

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.

B. Chi square computation for a dihybrid ratio Detail on chi square for dihybrids. Review if you are not familiar with dihybrids and chi square calculation

Inheritance patterns

Inheritance patterns are a function of which genome the gene is found in

Nuclear ________

autosomal ________

sex-linked ________

Cytoplasmic ________

Mendelian genetic analysis

Mendelian genetic analysis

Qualitative/discontinuous variation vs. quantitative/continuous variation ________

• Histograms

Polymorphisms ________

 

Autosomal inheritance

Autosomal inheritance

A. 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

Mendelian genetic analysis: the "classical" approach to understanding the genetic basis of a difference in phenotype is to use progeny to understand the parents.

• If you use progeny to understand parents, then you make crosses between parents to generate progeny populations of different filial (F) generations: e.g. F1, F2, F3; backcross; doubled haploid; recombinant inbred, etc. 
• 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.
• Mendelian analysis is straightforward when one or two genes determine the trait.
• 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

Information on DNA sequence and/or gene expression may let you look directly at differences between individuals.

Example: Vrs1 sequences

B. 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 phenotype: VV vv

Vrs1 as genotype: Komatsuda et al. 2007. PNAS 104:1424-1429. Read (at least) abstract and introduction.

Principles of Komatsuda et al. Mutation as a source of new alleles; genes can encode transcription factors; the basis of Mendelian inheritance can be determined.

Brewing and Vrs1 Single genes may have big effects (sometimes due only to perception)

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

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.
Understand that DH = producing 2n plants form n gametes (male or female).

 

C. Dihybrid Model:

Segregation and independent assortment ________

Details on dihybrids: Review if you do not remember how to do 2-factor Punnet squares