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General Genetics > Test 4: Genetic Engineering, Population Genetics, Quantitative, Micro and Macroevolution > Flashcards

Test 4: Genetic Engineering, Population Genetics, Quantitative, Micro and Macroevolution Flashcards

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Biology 101 flashcards
1
Q

Recombinant DNA technology

A
  • Recombining DNA segments from different species in the test tube
  • Reintroducing it back into a living organism
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2
Q

2 Discoveries that made Recombinant DNA technology possible

A

1) plasmids

2) restriction endonucleases

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3
Q

Plasmids

A
  • bacterial extrachromosomal DNA

- small circular DNA molecules w/ around 2700 bps

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4
Q

Restriction endonucleases

A
  • naturally-occuring enzymes in bactera
  • a defense mechanism against invading virus DNA
  • bacteria with these enzymes were able to restrict infection
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5
Q

Difference between Endonuclease & Exonuclease

A

Endonuclease: breaks an internal phosphodiester linkage, fragmentation of the DNA molecule

Exonuclease: breaks the terminal phosphodiester linkage again and again

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6
Q

E.Coli RNA polymerase

A
  • recognizes and binds the promoter to initiate transcription
  • does not recognize or initiate transcription at the promoter from another species
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7
Q

Type I diabetes (juvenile onset)

A

cause: lack of insulin (a protein hormone)
original treatment: bovine or porcine insulin
problem: allergies, side-effects to animal protein
recombinant source: engineered bacterial cells harboring the human insulin gene grown in bioreactors

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8
Q

Dwarfism

A

cause: lack of growth hormone (a protein)
original treatment: human GH isolated from pituitary glands of cadavers
problem: possible co-isolation of infectious agent
recombinant source: engineered bacterial cells harboring the human GH gene grown in bioreactors

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9
Q

Hemophilia

A

cause: lack of clotting factor VIII (a protein
original treatment: human factor VIII isolated from donated blood
problem: possible co-isolation of HIV
recombinant source: engineered mammalian cells harboring the human GH gene grown in tissue culture

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10
Q

Genetic Engineering in Plants

A
  • does not usually involve the simple modification of a zygote
  • scientists employed the help of bacteria that practice recombinant DNA technology
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11
Q

CRISPR-Cas9 components

A
  • dCas9
  • sgRNA 1
  • sgRNA 2
  • donor DNA
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12
Q

HDR

A
  • homology directed repair (HDR)

- if ligase repairs before HDR, paired nickases will act again

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13
Q

BT toxin

A

-protein that kills lepidopterans
-extremely species specific: onlu Lepidopterans (moths, caterpillars, butterflies)
-create genetically modified crop plants that harbor
the BT-toxin gene

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14
Q

Why was SCIDS chosen?

A

1) single gene trait
2) no complex regulation
3) human gene had been isolated and cloned
4) affected organ system easily accessible to genetic modification

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15
Q

Population

A
  • A group of interbreeding individuals

- same species, same range = matings produce fertile offspring

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16
Q

Hardy-Weinberg Equilibrium

A
  • p2 + 2pq +q2= 1

- if =1, in Hardy-Weinberg Equilibrium and allele frequencies will stay the same generation to generation

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17
Q

Polymorphisms

A

-Many forms (of a trait) in the population

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18
Q

2 Ways for a restriction fragment to differ in length between two chromosomes

A

1) missing or extra RE recognition site

2) more or less DNA between RE recognition sites

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19
Q

VNTRs

A

(1/2)n(n+1)

n= number of alleles in the population

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20
Q

Quantitative trait inheritance

A

-continuous variation between extremes in phenotype

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21
Q

Quantitative traits

A

-are polygenic (controlled by two or more genes)

number of phenotypic classes = 2n+1 (n = number of genes)

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22
Q

Sample mean

A

X = sigma(Xi)/n

Xi = each value
n = number of values in sample
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23
Q

Sample variance

A

s^2 = sigma(Xi - X)^2/(n-1)

-need to know X first (sample mean)

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24
Q

establish purebreeding line with smallest value…

A

… homozygous noncontributing alleles at all genes

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25
Q

establish purebreeding line with largest value…

A

… homozygous contributing alleles at all genes

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26
Q

Heritability

A

[S^2(F2) - S^2(parental)]/(S^2(F2)

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27
Q

F1

X = 65
s^2 = 4
s = 2

F2

X = 65
s^2 = 25
s = 5

Determine environmental s^2, genotypic s^2, and total s^2, then determine heritability

A

environmental s^2 = 4
total s^2 = 25
genotypic s^2 = 21

21/25 = 0.84 heritability

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28
Q

Darwin’s Postulates

A

1) Individuals within species are variable (VARIABLE)
2) Some of these variations are passed on to offspring (HEREDITY)
3) In every generation, more offspring are produced than can survive
4) Survival and reproduction are not random: The individuals that survive and go on to reproduce, or who reproduce the most, are those with variations that fir the environment best. They are naturally selected. (SELECTION)

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29
Q

Microevolution

A

-Allele frequency changes in populations (small incremental changes)

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30
Q

Macroevolution

A
  • speciation

- large differences as a result of accumulated microevolutionary changes over time

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31
Q

Allele frequencies will stay the same to generation if… (Hardy-Weinberg Requirements)

A

1) mating is random
2) no migration
3) no mutation
4) infinite population size (no sampling error)
5) no selection

32
Q

Nonrandom mating

A

-Also called assortative mating

positive - like with like (inbreeding)
negative - unlike with unlike (outbreeding)

phenotypic or genotypic
complete or partial

33
Q

Inbreeding decreases…
Inbreeding increases…
Outbreeding increases…
Outbreeding decreases…

A

heterozygosity
homozygosity
heterozygosity
homozygosity

34
Q

Inbreeding coefficient (F)

A

F = 1 - (observed frequency of heterozygotes / expected frequency of heterozygotes if HWE)

F = 1 - (Het0/2pq)
-if F > 0 = inbreeding
if F < 0 = outbreeding
F will be 0 = HWE

35
Q

Migration

A

p(new) = mp(migrant) + (1 - m)p(native)

m = migration rate (fraction of new population composed of migrants)

36
Q

300 butterflies from the continent are swept to an island where there were 1200 butterflies. The value of p on the continent is 0.4 and the value of p on the island was 0.6. What is the value of p after admixture?

A

p(new) = (300/1500)0.4 + (1-300/1500)0.6

= 0.56

37
Q

Migration can change allele frequencies, but does not lead to _________

A

-adaptations

  • but it can affect:
    1) direction of allele frequency change, therefore direction of evolution
38
Q

Mutation

A

p(n) = p(0) * e^(-nu)

-pn is the frequency of the allele after n generations
n is generations
-p0 is the frequency of the allele now
-e is the base of the natural log
-u is the mutation rate in alleles (gametes) per generation

39
Q

Mutation alone does not have a significant effect on allele frequencies, but it is the source of new ____

A

-alleles

40
Q

Random genetic drift

A
  • an allele’s frequency is free to drift over many generations
  • the amount of drift is a function of population size
  • does not lead to adaptations
41
Q

Population size

A
  • random chance events applied to reproduction

- the expected is determined by what he frequency was in the previous generation

42
Q

Founder Effect

A
  • A population size effect
  • small sample of large population founds a new population
  • small sample may not reflect the same allele frequencies as larger population
43
Q

Selection

A
  • differential reproductive success
  • not all phenotypes have the same nummber of offspring
  • is a function of the environment
44
Q

Fitness

A

-selection’s relationship to the environment

1) probability of survival
2) rate of reproduction

45
Q

Types of Selection

A

1) zygotic selection - survival
2) sexual selection - finding a mate
3) fecundity selection - gamete production
4) gametic selection - gamete quality

46
Q

Fitness (w)

A

w = 1 - s
-where s is the selection coefficient (fraction by which fitness is reduced relative to the phenotype with: maximum probability of survival and maximum rate of reproduction

47
Q

Direction selection

A

AA Aa aa
1 1 1 - s

-one homozygote has the highest fitness, other homozygote has lowest fitness

48
Q

Heterozygote advantage

A

AA Aa aa
1-s 1 1-s

-heterozygote has the highest fitness

49
Q

Homozygote advantage

A

AA Aa aa
1 1-s 1
1 1-s 1-s

-heterozygote has the lowest fitness

50
Q

Steps of speciation

A

1) elimination or reduction in gene flow (1 pop becomes 2)
2) Divergence (microevolution + other factors until…)
3) Separate species

51
Q

Biological Species Concept (BSC)

A

-If two individuals (from separate intermating populations) mate and produce fertile offspring, they are considered same species otherwise, they are separate species

52
Q

1) Elimination or reduction in gene flow: Allopatric speciation

A
  • different range
  • pop 1 and pop 2 are separated by geographical barrier

1) Dispersal - population migrates to island
2) vicariance - continent separates into island and creates new pop on island

53
Q

1) Elimination or reduction in gene flow: Sympatric speciation

A

-same range

-ex: Hawthorn trees are native to North America
Apple trees introduced to NA 350 years ago
some Hawthorn maggot flies began mating on apple fruit
2 populations in the same range mate different times of year apple vs Hawthorn fruit ripening time

54
Q

1) Elimination or reduction in gene flow: Parapatric speciation

A

-nearby range (overlapping)

55
Q

2) Divergence by microevolution: mutation

A
  • same mutations do not occur in both populations

- different alleles appear in each population independently

56
Q

2) Divergence by microevolution: selection

A
  • if environment is different:
  • different selection pressures occur
  • under directional selection pattern different alleles become fixed or lost
57
Q

2) Divergence by microevolution: drift

A

-different alleles fixed or lost in each population

58
Q

2) Divergence by microevolution: migration

A
  • both populations are not affected to the same degree

- different alleles fixed or lost

59
Q

2) Divergence by microevolution: nonrandom mating

A

-reinforces isolation, and decreases gene flow further

60
Q

Divergence Microevolution Summary

A
  • Microevolution does not just change allele frequencies
  • Over time, alleles can either become fixed, lost or created
  • Genetic distance between two different populations increases with time
61
Q

Inversions

A
  • inversion heterozygotes have reduced fertility (fecundity selection)
  • inversion is passed on to half the viable offspring

-inversion gamete + inversion gamete = inversion homozygote
no loop at meiosis, no reduced fertility

62
Q

Translocations

A

-translocation heterozygotes reduced fertility regardless of CO events and pass on translocation to progeny

63
Q

Polyploidy

A

-instant speciation

64
Q

Social biology: Cooperative, Selfish, Altruistic, Spiteful

A

Cooperative: actor + recipient benefit
Spiteful: actor + recipient harmed
Altruistic: actor harmed, recipient benefits
Selfish: actor benefits, recipient harmed

65
Q

Coefficient of relatedness

A
  • r
  • fraction of alleles shared by a relative
  • or probability of finding the same allele at any given gene in a relative
66
Q

r for full-siblings

A

r = 1/2

67
Q

r for cousins

A

r = 1/8

68
Q

r for aunt or uncle

A

r = 1/4

69
Q

Hamilton’s Rule

A
  • if Br - C > 0 then alleles that cause altruistic behavior will increase in frequency by natural selection
  • B = benefit (increase in fitness to relative)
  • C = cost (decrease in fitness to actor)
  • r = coefficient of relatedness
  • same as Br > C
70
Q

Kin selection in Belding’s Ground Squirrels

A

-Pairie dogs will be more likely to undergo altruistic behavior (warning call) when they have kin at home or offspring

71
Q

Kin selection in Hymenoptera

A
  • females are diploid, males are haploid (haplodiploidy)

- siblings r = 3/4, progeny r = 1/2

72
Q

Reciprocal altruism

A

-natural selection will favor alleles that cause altruistic (fitness decreasing) acts to nonkin if equally valuable (fitness increasing) acts are returned

73
Q

Prisoner’s Dilemma

A
  • T = 5 points you defect, opponent cooperates
  • R = 3 points both cooperate
  • P = 1 point both defect
  • S = 0 points you cooperate, opponent defects
  • T > R > P > S
  • R > (S + T)/2
74
Q

TFT

A
  • tit for tat
  • start by cooperating
  • repeat opponents last decision
  • never defects first
  • immediately retaliates for defection
  • does not hold a grudge
75
Q

ESS

A
  • evolutionarily stable strategy
  • a behavior strategy is evolutionary stable if:
  • population of individuals practicing it cannot be exploited by a rare form practicing another strategy

General Genetics (11 decks)

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