Session 1 - Artificial Selection in Brassica rapa; Mendelian Genetics in Zea mays; Chi-Square Analysis Flashcards
Natural selection
A process in which certain variants within a given population have traits that result in the production of more offspring
Brassica rapa
A very fast growing plant from the mustard family; source of canola (rapeseed) oil.
What trait is examined in Brassica rapa?
Trichome number - trichomes are small epidermal “hairs”
Why are the seeds considered “wild-type”
Because no selection has been applied for traits
What was observed under the microscope?
Suspension of (yellow) corn pollen (male gametes) stained with iodine (which in the presence of starch, stains dark purple)
What are the phenotypes, genotypes, and ratios of the gametes observed?
starchy (Wx) and waxy (wx); starchy is dark purple and waxy is yellow. Ratio of phenotype was 1:1
What was observed in the monohybrid cross?
A cross between two F1 heterozygous individuals for the Color (C) gene. There were red and white kernels, with a phenotype ratio of 3:1 red to white.
What was observed in the dihybrid cross?
A cross between individuals heterozygous for 2 independent genes. One for kernel color (purple [P] or yellow [p]), the other for sweet (wrinkled) [s] or starchy (smooth) [S].
How does one the modified phenotypic ratio of 9:7 in a dihybrid cross?
Two genes involved in a pathway–functional products from both are required for expression (one or more recessive allelic pair would result in the mutant):
Precursor–Gene A–>Intermed.–Gene B–>Final Product
This epistatic interaction is called complementary gene action. You get 9 functional to 7 nonfunctional
How does one the modified phenotypic ratio of 13:3 in a dihybrid cross?
Certain genes have the ability to suppression the expression of a gene at a 2nd locus. Both synthesis and suppression of gene product are dominant traits; present of the dominant suppression allele leads to no product even with the synthesis dominant allele present. This masking is termed dominant suppression epistasis–13 nonfunctional to 3 functional
How does one the modified phenotypic ratio of 9:3:4 in a dihybrid cross?
Two genes involved in a pathway – these genes are not linked and thus 3 phenotypes are possible:
Phenotype 1–Gene A–>Phenotype 2–Gene B–>Phenotype 3
Note to get the third phenotype, dominant alleles of both genes must be present; if only A => 2; if only B => 1; if all recessive => 1.
In the example, the “a” (recessive) allele is epistatic on B and b. Because a recessive allele is epistatic, this is a case of recessive epistasis. 9 A&B functional : 3 A functional : 4 only B functional or all nonfunctional
What does the chi-square test “do”?
It allows us to test how well experimental data fits theoretical expectations by determining if it’s reasonable to attribute deviations from the model to chance.
How do you interpret the “P value” and derive a conclusion from the results?
A P value can be interpreted as “the probability of obtaining a chance deviation from theoretical expectation >= that observed (the chi-square value)”
If the P value >= 0.05, we can conclude that the data fit the model reasonably well and that the results provide no statistically compelling argument against the hypothesis/model.
