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BIO 2C03 > Chapter 19 > Flashcards

Chapter 19 Flashcards

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Biology 101 flashcards
1
Q

Discontinuous variation

A

variation that produces discrete sharply distinguishable phenotypic categories, allowing for consistent phenotype ratios (3:1 or 9:3:3:1) - in single gene traits

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2
Q

Continuous variation

A

variation distributed across a range of values in an uninterrupted continuum - in polygenic and multifactorial traits

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3
Q

Human height is…

A

a multifactorial trait with continuous variation

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4
Q

In addition to genetics, human height is affected by ____________ and ____________ factors

A

developmental and environmental factors - nutrition, exercise, rest

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5
Q

In regard to adult human height, there are more than ____ genetic variants

A

700

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6
Q

Genetic Potential

A

is transmitted by parents to offspring, which can be attained if the offspring grows and develops under ideal conditions

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7
Q

Different genes contribute to polygenic traits differently, Major Genes have ______ influence while Modifier (Minor) Genes have _____ influence

A

greater, less

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8
Q

_______ is a major gene that influences eye color while ______ is a major gene that influences both eye color and skin tone

A

HERC2, OCA2

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9
Q

Additive Genes

A

when multiple genes make incremental contributions to a phenotypic value - in polygenic traits where no major gene effects are observed

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10
Q

The alleles of each ________________ can be assigned their own quantitative values of contribution

A

additive gene

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11
Q

Multiple-gene hypothesis

A

the hypothesis that alleles of each of the contributing genes obeyed the principles of segregation and independent assortment and had an additive effect in the production of phenotypic variation

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12
Q

To test the multiple-gene hypothesis, Nilsson-Ehle used two additive genes (A + B) that both had two alleles A1, A2, B1, and B2. Alleles ___ and ____ add an equal unit of color to the phenotype while ____ and ____ add no color. The more “__________” alleles a genotype has, the darker the color of what kernels and vice versa.

A

A1 and B1, A2 and B2, number 1

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13
Q

The number of phenotypes follows the rule “_______” where ______________

A

2n+1, n = the number of genes

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14
Q

The more additive genes there are that contribute to a phenotype, the ______ phenotypes there are, causing ______ demarcation between categories

A

more, less

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15
Q

In 1916, East used Nicotiana longiflora corolla length to test the multiple-gene hypothesis. He took pure-breeding long and short corolla lines and measured progeny. His two conclusions were…

A

(1) the intermediate corolla length in progeny results from segregation of alleles of multiple genes
(2) phenotypic expression is also influenced by environmental factors

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16
Q

T or F: when there is no gene-environment interaction, F2 phenotypic distribution is discontinuous with a predictable 1:2:1 ratio. With moderate gene-environment interaction, there’s a range of phenotypic values with F2 progeny having some overlap. Finally, substantial gene-environment interaction results in F2 progeny that have significant overlap

A

True

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17
Q

Phenotype = ……

A

P (phenotype) = G (genotype) + E (environment) + GEI (genotype-environment interactions)

18
Q

GEI (genotype-environment interactions) is estimated …

A

only when several genotypes are assayed in several defined environments

19
Q

Threshold Trait

A

a polygenic/multifactorial trait that has continuous distribution but can be divided into distinct categories (like affected or unaffected)

20
Q

Genetic Liability

A

the idea that certain alleles can push the phenotype toward the affected end of a continuous distribution, each individual’s risk of having the affected phenotype is the result of genotype or in the case of multifactorial phenotypes the genotype and nongenetic influences

21
Q

Threshold of Genetic Liability

A

the point when an individual will show the affected phenotype - environmental factors can contribute to reaching threshold

22
Q

Variance (s^2)

A

a measure of the spread of distribution around the mean
s^2 = sum [(x,i - x,bar)^2] / df

23
Q

Phenotypic Variance = ……

A

VP (phenotypic variance) = VG (genetic variance - genotypic contribution) + VE (environmental variance - environmental contribution)

24
Q

In parental pure-breeding lines, VP = VE since both parental lines are genetically uniform (VG = 0). In F1, VP = ________ and in F2, VP = _________

A

VP = VE, VP = VE + VG

25
Q

Genetic Variance (VG) has three kinds of allelic effects…

*AID

A

(1) Additive Variance (VA): added effects of all alleles contributing to the trait
(2) Dominance Variance (VD): contributions due to heterozygous individuals not intermediate between the two homozygous states
(3) Interactive Variance (VI): epistatic interactions between alleles of genes

26
Q

Heritability

A

the proportion of phenotypic variation that is due to genetic variation

27
Q

T or F: a high heritability value indicates most of the observed phenotypic variation is due to environmental influences while a low heritability value indicates that most of the observed phenotypic variation is due to genetic variation

A

False, it’s opposite.

28
Q

The two measures of heritability are __________________ and ____________.

A

Broad-sense heritability, Narrow-sense heritability

29
Q

Regarding the two measures of heritability, what are their respective equations and what do they estimate?

A

Broad-Sense (H^2 = VG / VP): estimates the proportion of phenotypic variation that is due to total genetic variation
Narrow-Sense (h^2 = VA / VP): estimates the proportion of phenotypic variation that is due to additive genetic variation

30
Q

Cave Fish case study: if you know the F1 fish are genetically uniform and that VP(F1) = 0.057cm and VP(F2) = 0.563cm, what is H^2?

A

H^2 = VG / VP
Since F1 is genetically uniform, VP(F1) = VE(F1) = 0.057cm
Since VE(F1) = VE(F2), VE(F2) = 0.057cm
VP(F2) = VG(F2) + VE(F2)
VG(F2) = 0.563cm - 0.057cm = 0.506cm
H^2 (F2) = 0.506cm / 0.563cm = 0.899
Conclusion: roughly 90% of phenotypic variation in eye size between these two populations of blind or sighted cave fish is due to genetic variation

31
Q

Since identical twins share all their alleles, VP(mz) = _____ but because fraternal twins share half their alleles, VP(dz) = ____________. This means VG = 2 [VP(dz) - VP(mz)]

A

VP(mz) = VE
VP(dz) = VE + 1/2(VG)

32
Q

The three caveats of Broad Sense Heritability are (1) Stronger ______, (2) Greater _______, and (3) Greater ___________.

A

(1) Stronger shared maternal effects in identical twins than in fraternal twins
(2) Greater similarity of treatment by parents of MZ than DZ twins
(3) Greater similarity of interactions between genes and environment in MZ than DZ twins

33
Q

What does a high narrow-sense heritability value correlate to?

A

A greater degree of response to selection (artificial or natural selection)

34
Q

Four attributes of heritability are central to its meaning, what are they?

A

(1) Heritability is a measure of the degree to which genetic differences contribute to phenotypic variation of a trait
(2) Heritability values are only accurate for the environment and population in which they are measured
(3) Heritability for a given trait in a population can change
(4) High heritability does not mean a trait is not being influenced by environmental factors at all

35
Q

Quantitative trait loci (QTLs) are genes…

A

that contribute to phenotypic variation in quantitative traits

36
Q

QTL Mapping is when you..

A

map QTLs to chromosome regions/linkage groups

37
Q

T or F: a chromosome region likely to contain a QTL is identified by frequent co-occurrence of an SNP with a particular phenotype

A

True

38
Q

There are two QTL Mapping Strategies: __________ and _________

A

Genetic Crosses, GWAS (genome-wide association study)

39
Q

Using genetic crosses to map QTLs, these are the four main steps

A

(1) construct genetic cross between parental strains w/ different phenotypes
(2) determine DNA markers that differ between parental strains
(3) obtain phenotype and genotype of all progeny
(4) identify associations between phenotype and genotype at individual loci

40
Q

Using GWAS to map QTLs, these are the five main steps

A

(1) select a natural population
(2) obtain their whole genome sequences
(3) compares “cases” to “controls” to assess their genotypic difference at each sequence variant
(4) nucleotide frequencies showing significant differences between “cases” and “controls” are candidate QTLs
(5) need to correct for “false discovery rate” (FDR) to reduce false positives

41
Q

Regarding selection differential (S) and response to selection (R), (h^2) = ________. Ms is ________ and M is _________. Ms - M = selection differential (S). The predicted average for the next generation is ______.

A

h^2 = R/S
Ms = mean phenotype selected for mating
M = mean phenotype in parental generation
M + R = predicted average for next generation

BIO 2C03 (10 decks)

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