Genotypic variation and qualitative inheritance
Importance: There is phenotypic diversity obvious to the eye (e.g. two-row vs. six-row) and there is the underlying DNA-level diversity. There are a number of approaches to characterizing DNA sequence diversity – from molecular markers to whole genome sequencing. No matter what the method is that is used to characterize sequence diversity, the inheritance of DNA level polymorphisms can be approached with the same tools that are used to study phenotypes showing Mendelian inheritance.
Learning objectives:
- Learn how single nucleotide polymorphisms (SNPs) can lead to phenotypic differences.
- Understand the principles of characterizing SNPs genotyping with the Illumina platform.
- Given SNP data, be able to test hypotheses regarding allele segregation and interpret the results.
The case study: VRS1 cloning and allelic diversity
Cloning VRS1 (HvHox1)
- The paper. Komatsuda et al. 2007. PNAS 104:1424-1429
- The principles of Komatsuda et al.
Characterizing allelic variation in multiple genotypes
- The paper. Cuesta-Marcos et al. 2010. BMC Genomics.11:707
- Summary of VRS1 alleles - See Table 1 in Cuesta-Marcos et al. 2010
- VRS1 (and other) accession numbers
Inheritance of VRS1 in the OWB
Genotypic data
- Parents
How VRS1 was genotyped in the OWB using Illumina SNPs
- The principles of Illumina SNP genotyping
- The application of Illumina SNP genotyping to the OWBs
- The germplasm: Cistue et al. 2011. Theor Appl Genet. 122:1399–1410
Note: The final population size reported by Cistue et al. 2011 is n = 93 + 2 parents. In the application example n = 96 because the 2 parents and one extra doubled haploid line were included in the genotyping. Details on population size are available online.
The SNPs targeted within HvHox1
- 12_30897 and 12_30900
- SNP nomenclature 12_30897; 12 = BOPA-2; 3 = POPA 3; 0897 = sequential numbering
SNP genotype data in the OWB
- As SNPs (12_30897 = T/C and 12_30900 = G/C) or as “A” and “B”
- Translating nucleotides to A and B
- A and B data (file online at http://wheat.pw.usda.gov/ggpages/maps/OWB/
Locus | Observed A | Observed B |
---|---|---|
12-30897 | 45 | 48 |
The Χ2 test
Locus | Χ2 value | P value |
---|---|---|
12-30897 | 0.097 | 0.95 |
Conclusion: Accept the null hypothesis. Alleles (T and C) at HvHox1 (VRS1) segregate 1:1, as expected in a doubled haploid population.
Exercises:
This question addresses “What happens when your data do not give the results you expected.”
1. Go to the Cistue et al. 2011 data set and download the data for locus 3_0106.
A. Based on the data you retrieve, fill in the following table.
Locus name | # Observed A allele | # Expected A allele | # Observed B allele | # Expected B allele | X2 value | P value |
---|---|---|---|---|---|---|
3_0106 |
B. Based on these results do you, do you accept or reject your null hypothesis?
C. Cistue et al. 2011 address the issue of segregation distortion. Do the results of you chi square test provide evidence for segregation distortion at this locus? Briefly defend your answer.
D. Possible causes of segregation distortion include sampling error and selection. Based on the proximity of 3_0106 to the Zeo locus and the information presented by Cistue et al. 2011, what do you think caused segregation distortion in this case: sampling error or selection? Briefly defend your answer.
E. Give an example of how a plant breeder could cause segregation distortion during population development to work in her/his favor.
This question addresses “Making decisions about how to obtain data”.
2. VRS1 can be scored as a phenotype (2-row vs. 6-row) or as a genotype (HvHox1 alleles). Go to the OWB image gallery and score the inflorescence type for the anther culture (AC) set. You may score the phenotypes as 2 or 6, but you will need to convert your scores to A and B, following the convention of the OWB population where the OWB-D allele = A and the OWB-R allele = B. Keep track of how long it takes you to obtain the 2 vs. 6 (A vs. B) data.
A. How long did it take you to obtain the inflorescence phenotype data?
B. Did you make the same allele calls using genotype and phenotype data? Briefly explain the basis of your answer.
C. Considering the time it took you to obtain the phenotype data vs. the cost of obtaining the genotype data, which technique would you use to test for the inheritance of 2-row vs. 6-row in a new and different population of 94 doubled haploid lines? In order to estimate the cost of obtaining the genotype data, assume an average of $5.00 per individual/ per allele call. This average figure is based on $1.00 for a single allele-specific assay/individual and $10.00 for genome-wide data on thousands of SNPs/individual.
This material was developed by Pat Hayes (OSU), Shiaoman Chao (USDA-ARS) and Alfonso Cuesta-Marcos (OSU)
Version 1.1. January 12, 2012