Integrating Concepts in Biology

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

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

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?

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?

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)


Solve for m:m=(sw-tv)/(su-rv)

Σ(xi2) * m + Σ(xi) * b = Σ(xi * yi)

Σ(xi) * m + n*b = Σ(yi)

(E) Coef on m

(F)Coef on b

(G)RHS

4333017.5 63390 4318042

63390 928 63186.1

m= 0.646

b= 23.942



Σ(xi2) * m + Σ(xi) * b = Σ(xi * yi)

Σ(xi) * m + n*b = Σ(yi)

(E) Coef on m

(F)Coef on b

(G)RHS

4333017.5 63390 4318042

63390 928 63186.1

m= 0.646

b= 23.942



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)

(E) Coef on m

(F)Coef on b

(G)RHS

4333017.5 63390 4318042

63390 928 63186.1

m= 0.646

b= 23.942


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))]

Figure 8.2

Mean blood pressures for rats in two colonies

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

Figure 8.2

Mean blood pressures for rats in two colonies

αY/αY ; βR/βR

αF/αF ; β_/β_

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

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

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

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

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?

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?

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

Figure 8.7


results of predator exclusion experiment

Figure 8.7


Survival of barnacles

without predator

with predator


PowerPoint Slides for Chapter 8:Evolution of Organisms

Section 8.2: How does selection act on individuals with variable characteristics?


Chris Paradise

Poecilia reticulata

http://www.erdingtonaquatics.com/guppies.html

Figure 8.8

Survival of guppies with different behavioral tendencies to inspect potential predators

Figure 8.9

Mean number of predator inspections by bright and drab male guppies

Is there an effect of brightness or predation?

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.

Figure 8.11

Preferences of female guppies in choice tests

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.

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

http://stattrek.com/online-calculator/binomial.aspx

http://stattrek.com/online-calculator/binomial.aspx

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.

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?

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?

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

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

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?

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?

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?


Chapter 8: Evolution of Organisms

Section 8.3: Can you observe descent with modification?

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?

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.

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).

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

Table 8.3



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

Table 8.3



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

Table 8.3



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

Table 8.3



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

Figure BME 8.3.1 Three rooted trees (b, c, d) consistent with the same unrooted tree (a).


Unrooted tree

Rooted trees


Tree b is rooted at point B, species 1 (blue)


Tree c is rooted at point C


Tree d is rooted at point D, species 3 (yellow)

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

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

Table 8.3



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

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

Table 8.3



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

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

Figure 8.17


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

Figure 8.17


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

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.

Figure 8.19

Frequency distributions of plants living on trees (maroon) and on the ground (gold)

Orchids

Non-orchid plants

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)

Figure 8.21

Evolutionary relationships among birds, bats, and insects

Simplified cartoon of evolutionary relationships.

Real tree is much more complex.

Figure 8.22

Evolutionary tree showing relationship among several vertebrates

Evolution of wings (W) occurred twice

Figure 8.23

The fossil bat Icaronycteris index and skeleton from a living species

Bones trapped in rock

Skeleton from living species

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

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

http://palaeosbios.blogspot.com/2006/12/mammals-linked-to-earlier-flight.html





Figure 8.25

Percentage of growing 3rd -5th forelimb digits composed of proliferating and differentiating /

elongating cells in mice and bats

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

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


Chapter 8:Evolution of Organisms

Section 8.4 Can you observe evolution in your lifetime?


Chris Paradise

Figure 8.27

DDT and mosquitoesDDT

Incidence of malaria in Italy after a DDT spray campaign began

Anopheles gambiae female taking a blood meal.

Figure 8.28

Relationship between incidence of malaria and DDT use in India, 1969-77

What do you conclude from what occurred in India?

Malaria Transmission

MAlaria Risk in Africa

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?

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?

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

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

Integrating Concepts in Biology

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