Session 1 - Artificial Selection in Brassica rapa; Mendelian Genetics in Zea mays; Chi-Square Analysis Flashcards Preview

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Flashcards in Session 1 - Artificial Selection in Brassica rapa; Mendelian Genetics in Zea mays; Chi-Square Analysis Deck (13)
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1
Q

Natural selection

A

A process in which certain variants within a given population have traits that result in the production of more offspring

2
Q

Brassica rapa

A

A very fast growing plant from the mustard family; source of canola (rapeseed) oil.

3
Q

What trait is examined in Brassica rapa?

A

Trichome number - trichomes are small epidermal “hairs”

4
Q

Why are the seeds considered “wild-type”

A

Because no selection has been applied for traits

5
Q

What was observed under the microscope?

A

Suspension of (yellow) corn pollen (male gametes) stained with iodine (which in the presence of starch, stains dark purple)

6
Q

What are the phenotypes, genotypes, and ratios of the gametes observed?

A

starchy (Wx) and waxy (wx); starchy is dark purple and waxy is yellow. Ratio of phenotype was 1:1

7
Q

What was observed in the monohybrid cross?

A

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.

8
Q

What was observed in the dihybrid cross?

A

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].

9
Q

How does one the modified phenotypic ratio of 9:7 in a dihybrid cross?

A

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

10
Q

How does one the modified phenotypic ratio of 13:3 in a dihybrid cross?

A

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

11
Q

How does one the modified phenotypic ratio of 9:3:4 in a dihybrid cross?

A

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

12
Q

What does the chi-square test “do”?

A

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.

13
Q

How do you interpret the “P value” and derive a conclusion from the results?

A

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.