Integrating Concepts in Biology
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Transcript of Integrating Concepts in Biology
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Integrating Concepts in Biology
PowerPoint Slides for Chapter 8:Evolution of Organisms
Section 8.1: What causes individual variation?
byA. Malcolm Campbell, Laurie J. Heyer, and
Chris Paradise
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Figure 8.1
Relationship between height of parents and offspring slope of one
best-fit line
Size of circles proportional to the number of comparisons
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Midparent ChildMidparent Squared
Midparent times Child Coef on m Coef on b RHS
70.5 61.7 4970.25 4349.85 4333017.5 63390 4318042
68.5 61.7 4692.25 4226.45 63390 928 63186.1
65.5 61.7 4290.25 4041.35
64.5 61.7 4160.25 3979.65 m= 0.646
64 61.7 4096 3948.8 b= 23.942
67.5 62.2 4556.25 4198.5
67.5 62.2 4556.25 4198.5
67.5 62.2 4556.25 4198.5
BME 8.1: How does linear regression work?
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Midparent ChildMidparent Squared
Midparent times Child Coef on m Coef on b RHS
70.5 61.7 4970.25 4349.85 4333017.5 63390 4318042
68.5 61.7 4692.25 4226.45 63390 928 63186.1
65.5 61.7 4290.25 4041.35
64.5 61.7 4160.25 3979.65 m= 0.646
64 61.7 4096 3948.8 b= 23.942
67.5 62.2 4556.25 4198.5
67.5 62.2 4556.25 4198.5
67.5 62.2 4556.25 4198.5
slope
y-intercept
BME 8.1: How does linear regression work?
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Coef on m Coef on b RHS
4333017.5 63390 4318042
63390 928 63186.1
m= 0.646
b= 23.942
BME 8.1: How does linear regression work?Consider:rm+sb=t um+vb=w
Solve for m and b:m=(sw-tv)/(su-rv) b=(tu-rw)/(su-rv)
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BME 8.1: How does linear regression work?Consider:rm+sb=t um+vb=w
Solve for m:m=(sw-tv)/(su-rv)
Σ(xi2) * m + Σ(xi) * b = Σ(xi * yi)
Σ(xi) * m + n*b = Σ(yi)
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(E) Coef on m
(F)Coef on b
(G)RHS
4333017.5 63390 4318042
63390 928 63186.1
m= 0.646
b= 23.942
BME 8.1: How does linear regression work?Consider:rm+sb=t um+vb=w
Solve for m:m=(sw-tv)/(su-rv)
Σ(xi2) * m + Σ(xi) * b = Σ(xi * yi)
Σ(xi) * m + n*b = Σ(yi)
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(E) Coef on m
(F)Coef on b
(G)RHS
4333017.5 63390 4318042
63390 928 63186.1
m= 0.646
b= 23.942
BME 8.1: How does linear regression work?Consider:rm+sb=t um+vb=w
Solve for m:m=(sw-tv)/(su-rv)
Excel formula for m:=((F2*G3)-(G2*F3))/((F2*E3)-(E2*F3))
=[(Σ(xi)*Σ(yi)) – (Σ(xi*yi)*(n)] / [(Σ(xi)*Σ(xi)) – (Σ(xi2)*(n))]
= ((s * w) – (t * v)) / ((s * u) – (r * v))
Σ(xi2) * m + Σ(xi) * b = Σ(xi * yi)
Σ(xi) * m + n*b = Σ(yi)
Σ(xi)
Σ(xi2)
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(E) Coef on m
(F)Coef on b
(G)RHS
4333017.5 63390 4318042
63390 928 63186.1
m= 0.646
b= 23.942
BME 8.1: How does linear regression work?Consider:rm+sb=t um+vb=w
Solve for b:b=(tu-rw)/(su-rv)
Σ(xi2) * m + Σ(xi) * b = Σ(xi * yi)
Σ(xi) * m + n*b = Σ(yi)
Σ(xi)
Σ(xi2)
Excel formula for b:=((G2*E3)-(E2*G3))/((F2*E3)-(E2*F3))
=[((Σ(xi*yi)*Σ(xi)) – (Σ(xi2)*Σ(yi)] / [(Σ(xi)*Σ(xi)) – (Σ(xi
2)*(n))]
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Figure 8.2
Mean blood pressures for rats in two colonies
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Figure 8.3
Portions of the β and α adducin subunit DNA sequences and corresponding amino acid sequence
position along the protein
letters correspond to different amino acids
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Figure 8.2
Mean blood pressures for rats in two colonies
αY/αY ; βR/βR
αF/αF ; β_/β_
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Figure 8.4
Q and R refer to the amino acid at position 529 of β subunit
Systolic BPs of the 3 combinations of 2 versions of the β adducin gene in rats from the low BP colony
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Figure 8.5
BP of 9 combinations of two versions of the α and β adducin genes in rats after two generations of breeding low and high blood pressure rats together.
BP of high BP parental strain BP of low BP parental strain
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a. Range of alleles contributed from male parent
F / Q F / R Y / Q Y / R
Range of alleles contributed from female parent
F / Q FF / QQ FF / QR FY / QQ FY / QRF / R FF / QR FF / RR FY / QR FY / RRY / Q FY / QQ FY / QR YY / QQ YY / QRY / R FY / QR FY / RR YY / QR YY / RR
b.Range of gene versions contributed from male parent
F / Q Y / R
Range of alleles contributed from female parent
F / Q FF / QQ FY / QR
Y / R FY / QR YY / RR
Table 8.1
Variation of combinations of the two adducin genesAll parents are heterozygous
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Table 8.2
Zinc and copper contamination and pH in soils surrounding a smelting operation in Pennsylvania
Species and site Zinc (ppm) Copper (ppm)
Sandwort – near 7,500 15
Sandwort – medium 3,344 10
Sandwort – far 975 2.5
Honeysuckle – near 5,875 3
Honeysuckle – far 40 1
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Figure 8.6
Stomata and hair densities of sandwort collected at two times and grown in controlled conditions
What do the results of the common garden experiment show?
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Figure 8.6
Stomata and hair densities of honeysuckle collected at two times and grown in controlled conditions
What do the results of the common garden experiment show?
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Figure 8.7
Responses of the acorn barnacle (Chthamalus anisopoma) to the snail predator (Acanthina).
bent and cone shell shapes
Spine used to pry barnacles
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Figure 8.7
Responses of the acorn barnacle (Chthamalus anisopoma) to the snail predator (Acanthina).
results of predator exclusion experiment
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Figure 8.7
Responses of the acorn barnacle (Chthamalus anisopoma) to the snail predator (Acanthina).
Survival of barnacles
without predator
with predator
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Integrating Concepts in Biology
PowerPoint Slides for Chapter 8:Evolution of Organisms
Section 8.2: How does selection act on individuals with variable characteristics?
byA. Malcolm Campbell, Laurie J. Heyer, and
Chris Paradise
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Poecilia reticulata
http://www.erdingtonaquatics.com/guppies.html
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Figure 8.8
Survival of guppies with different behavioral tendencies to inspect potential predators
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Figure 8.9
Mean number of predator inspections by bright and drab male guppies
Is there an effect of brightness or predation?
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Figure 8.10
Effects of presence of female and male color on predator inspection behavior
a. Predator inspections initiated by bright or drab males when females either present or absent.
b. Relationship between boldness and brightness of males.
c. Relationship between minimum distance of an approaching predator and brightness of males.
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Figure 8.11
Preferences of female guppies in choice tests
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BME 8.2: Do females really prefer bold males? When a predator was present,
14 / 20 female guppies preferred the bright male to the drab male when the bright male was the bold one
16 / 20 preferred the drab male to the bright male when the drab male was the bold one.
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BME 8.2: Do females really prefer bold males? • p-value of preference test = probability of getting
at least the observed number of guppies choosing one color and behavior combination.
• Here, the p-value = probability of at least 14 of 20 guppies making this choice.
• BME IQ 8.2a leads to estimate of this probability. • In BME IQ 8.2e, you computed the probability of
exactly 14 heads in 20 tosses of a fair coin • =(1/2)20 x number of arrangements in which there
are exactly 14 heads (20 choose 14). • =38,760 x (1/2)20 = 0.037. • Probability of at least 14 heads in 20 tosses,
repeat for situations of 15 – 20 heads, and sum. • http://
stattrek.com/online-calculator/binomial.aspx
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Summary of guppy experiments• Bold males selected against in first experiment. • In presence of females, bright more likely than drab guppies to
swim towards, and inspect, potential predators. • Bright guppies more likely to flee before predator gets too close. • Females preferred to mate w/ bold males when predator was nearby
and bright males when predator absent; appearance = boldness? • Bold males inspections may signal to predator that it has been
spotted. Bold individuals more aware?• Bold, healthy males may contribute more advantageous genes. • Phenotypes remain in population when providing advantage to
possessor; phenotypes selected against reduce ability of individual to survive or reproduce.
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Figure 8.12
Spatial pattern of flower color along two transects that cut through a ravine of desert snow (Linanthus parryae)
Position of ravine dividing populations
How does the frequency of the white allele change across the ravine?
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Figure 8.12
Estimates of frequencies for four allozymes along each transect
How does the spatial pattern in flower color compare to the allele frequencies?
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Figure 8.12
Estimates of frequencies for four allozymes along each transect
No spatial pattern evident in these allozymes – note the slope of each line
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Figure 8.12
Spatial pattern of flower color along two transects that cut through a ravine, and estimates of frequencies for four allozymes along each
Position of ravine dividing populations
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Figure 8.13
Mean seed production (± 1 s.e.) for blue- and white-flowered desert snow plants from transplant plots
blue side white side
Do you detect evidence for natural selection?
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Figure 8.14
Variation in plant cover on two sides of a ravine where desert snow grows
Asterisks indicate that the cover for a species was statistically different on the two sides.
What do the patterns suggest to you?
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Figure 8.14
Differences in soil composition along a transect that crossed the ravine
Asterisks indicate that the variable was statistically different on the two sides.
What do the patterns suggest to you?
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Integrating Concepts in Biology
Chapter 8: Evolution of Organisms
Section 8.3: Can you observe descent with modification?
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Figure 8.15
Evolutionary tree of major groups of plant species.
To what group of species are flowering plants most closely related?
The species that existed at branch point “b” was the common ancestor of what groups of species?
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Figure 8.16
Morphology and pollen packet placement of orchids
Orchid flowers often have a lip
The pollen packet lies at the end of the erect anther.
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http://www.anu.edu.au/BoZo/orchid_pollination/
Meliorchis caribea. This orchidpollinarium, carried by a worker stingless bee, is preserved in amber and represents the first definitive fossil record for the family Orchidaceae. (scale bar, 1,000 mm).
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Table 8.3
Phenotypic characters for several orchid genera.
Specimen a b c d e f g h i j k l m n
Altensteinia 0 0 0 0 1 0 0 1 0 1 0 0 2 1
Gomphichis 1 0 0 0 0 0 0 0 0 1 0 0 1 1
Goodyera 0 1 0 0 1 0 0 1 0 1 1 1 0 1
Microchilus 0 1 0 0 1 0 0 1 0 0 1 1 0 1
Meliorchis 0 1 1 0 1 0 0 1 0 0 1 1 0 1
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Table 8.3
Phenotypic characters for several orchid genera.
Specimen a b c d e f g h i j k l m n
Altensteinia 0 0 0 0 1 0 0 1 0 1 0 0 2 1
Gomphichis 1 0 0 0 0 0 0 0 0 1 0 0 1 1
Goodyera 0 1 0 0 1 0 0 1 0 1 1 1 0 1
Microchilus 0 1 0 0 1 0 0 1 0 0 1 1 0 1
Meliorchis 0 1 1 0 1 0 0 1 0 0 1 1 0 1
Living orchids
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Table 8.3
Phenotypic characters for several orchid genera.
Specimen a b c d e f g h i j k l m n
Altensteinia 0 0 0 0 1 0 0 1 0 1 0 0 2 1
Gomphichis 1 0 0 0 0 0 0 0 0 1 0 0 1 1
Goodyera 0 1 0 0 1 0 0 1 0 1 1 1 0 1
Microchilus 0 1 0 0 1 0 0 1 0 0 1 1 0 1
Meliorchis 0 1 1 0 1 0 0 1 0 0 1 1 0 1
The fossilized orchid
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Table 8.3
Phenotypic characters for several orchid genera.
Specimen a b c d e f g h i j k l m n
Altensteinia 0 0 0 0 1 0 0 1 0 1 0 0 2 1
Gomphichis 1 0 0 0 0 0 0 0 0 1 0 0 1 1
Goodyera 0 1 0 0 1 0 0 1 0 1 1 1 0 1
Microchilus 0 1 0 0 1 0 0 1 0 0 1 1 0 1
Meliorchis 0 1 1 0 1 0 0 1 0 0 1 1 0 1
Phenotypes denoted by letters
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Table 8.3
Phenotypic characters for several orchid genera.
Specimen a b c d e f g h i j k l m n
Altensteinia 0 0 0 0 1 0 0 1 0 1 0 0 2 1
Gomphichis 1 0 0 0 0 0 0 0 0 1 0 0 1 1
Goodyera 0 1 0 0 1 0 0 1 0 1 1 1 0 1
Microchilus 0 1 0 0 1 0 0 1 0 0 1 1 0 1
Meliorchis 0 1 1 0 1 0 0 1 0 0 1 1 0 1
Particular form of a phenotype coded as a number
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Figure BME 8.3.1 Three rooted trees (b, c, d) consistent with the same unrooted tree (a).
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Figure BME 8.3.1 Three rooted trees (b, c, d) consistent with the same unrooted tree (a).
Unrooted tree
Rooted trees
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Figure BME 8.3.1 Three rooted trees (b, c, d) consistent with the same unrooted tree (a).
Tree b is rooted at point B, species 1 (blue)
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Figure BME 8.3.1 Three rooted trees (b, c, d) consistent with the same unrooted tree (a).
Tree c is rooted at point C
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Figure BME 8.3.1 Three rooted trees (b, c, d) consistent with the same unrooted tree (a).
Tree d is rooted at point D, species 3 (yellow)
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BME 8.3 Possible unrooted trees for 5 species, with species 3 in middle position
1
43
2
5
1
53
2
4
1
23
4
5
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BME 8.3 Best unrooted trees for 5 species
1
23
4
5
1
24
3
5
1
25
3
4
Required single mutation for b, k, and l occurs here
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Table 8.3
Phenotypic characters for several orchid genera.
Specimen a b c d e f g h i j k l m n
Altensteinia 0 0 0 0 1 0 0 1 0 1 0 0 2 1
Gomphichis 1 0 0 0 0 0 0 0 0 1 0 0 1 1
Goodyera 0 1 0 0 1 0 0 1 0 1 1 1 0 1
Microchilus 0 1 0 0 1 0 0 1 0 0 1 1 0 1
Meliorchis 0 1 1 0 1 0 0 1 0 0 1 1 0 1
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BME 8.3 Best unrooted trees for 5 species
1
23
4
5
1
24
3
5
1
25
3
4
Only tree #1 is optimal for j
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Table 8.3
Phenotypic characters for several orchid genera.
Specimen a b c d e f g h i j k l m n
Altensteinia 0 0 0 0 1 0 0 1 0 1 0 0 2 1
Gomphichis 1 0 0 0 0 0 0 0 0 1 0 0 1 1
Goodyera 0 1 0 0 1 0 0 1 0 1 1 1 0 1
Microchilus 0 1 0 0 1 0 0 1 0 0 1 1 0 1
Meliorchis 0 1 1 0 1 0 0 1 0 0 1 1 0 1
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Figure 8.17
Evolutionary tree of the orchid family
Estimate of earliest existence of orchids, range can be determined from two scales, top and bottom
Arrow heads indicate estimated ages of small subfamilies, A, B, and C
End of Cretaceous, 65 MYA
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Figure 8.17
Evolutionary tree of the orchid family
Youngest age estimated from a fossil orchid, 15 MYA
Oldest age estimated from a fossil orchid, 20 MYA
Fossil orchid, M. caribea, showing range of estimated fossil dates
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Figure 8.17
Evolutionary tree of the orchid family
Size of shaded area of a subfamily proportional to number of species in subfamily. E is most diverse
Small black scale bar represents thickness of 50 genera of orchids
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Figure 8.18
Microhabitats of the forest canopy and the distribution of orchid species (open bars) and numbers of individuals (black bars) in the microhabitats.
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Figure 8.19
Frequency distributions of plants living on trees (maroon) and on the ground (gold)
Orchids
Non-orchid plants
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Figure 8.20
Mean number of pollinators / species for orchids in five subfamilies Subfamilies same as in 8.17, D
and E most diverse (# of species above bar)
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Figure 8.21
Evolutionary relationships among birds, bats, and insects
Simplified cartoon of evolutionary relationships.
Real tree is much more complex.
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Figure 8.22
Evolutionary tree showing relationship among several vertebrates
Evolution of wings (W) occurred twice
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Figure 8.23
The fossil bat Icaronycteris index and skeleton from a living species
Bones trapped in rock
Skeleton from living species
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Figure 8.24
Elongated fifth digit (metacarpal bone) of bats compared to a combined index of body size for living and fossil bats
Fossil bats
Living bats
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Two hypotheses for selection of long digits
What are they?
Jumping hypothesis
In neither case did they have muscles or behavior necessary for true flight
Gliding hypothesis
bat- or squirrel-size creature that lived in Mongolia about 125 million years agohttp://palaeosbios.blogspot.com/2006/12/mammals-linked-to-earlier-flight.html
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Figure 8.25
Percentage of growing 3rd -5th forelimb digits composed of proliferating and differentiating /
elongating cells in mice and bats
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Figure 8.26
Percentage of cells in elongation and differentiation zone of metacarpals vs. length of the bat 5th
metacarpal as development proceeds.
18 1920
21
22Developmental
stages
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Bone Morphogenetic Protein 2Experiment ResultExpression levels of BMP2
31% higher in bat forelimb metacarpals than mice
Bat forelimb cell cultures with BMP2+
239 μm longer than control metacarpals
Bat forelimb cell cultures with BMP2 inhibitor
183 μm shorter than control metacarpals
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Integrating Concepts in Biology
Chapter 8:Evolution of Organisms
Section 8.4 Can you observe evolution in your lifetime?
byA. Malcolm Campbell, Laurie J. Heyer, and
Chris Paradise
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Figure 8.27
DDT and mosquitoesDDT
Incidence of malaria in Italy after a DDT spray campaign began
Anopheles gambiae female taking a blood meal.
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Figure 8.28
Relationship between incidence of malaria and DDT use in India, 1969-77
What do you conclude from what occurred in India?
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Malaria Transmission
MAlaria Risk in Africa
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Table 8.4
Mortality of two populations of Anopheles gambiae when exposed to a series of DDT concentrations
Mortality rate Gambian population
Tanzanian population
50% 0.017 0.33990% 0.040 1.160
Are Anopheles gambiae from Gambia more susceptible to DDT than those from Tanzania?
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Figure 8.29
Results from extraction of GST from two populations of Anopheles gambiae
Mass of each variant of GST
What is the significance of different amounts and activities of enzyme?
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Table 8.5
Time in minutes to 50 and 90% mortality of A. gambiae when exposed to permethrin and DDT
Pesticide / Mortality
rate
Permethrin-exposed Kenyan
pop.
Permethrin-exposed Kenyan pop. w/ PBO
DDT-exposed Tanzanian pop.
Laboratory control pop.
Permethrin
50% 32 25 nd 1790% 98 79 nd 36
DDT 50% 57 nd 145 <1590% 115 nd >250 41
Permethrin-resistant population somewhat resistant to DDT
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ELSI 8.1
ELSI Integrating Questions1. How do the proper and improper uses of pesticides
and antibiotics inhibit or facilitate the evolution of resistance?
2. What are the costs and benefits to society and ecological systems of pesticide use?
3. What are the ethical and social considerations to be made when choosing whether to apply a pesticide or prescribe an antibiotic?
When is there too much of a good thing? When chemicals are overused