BSCI 207 HW2-Scaling S20 PDF

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BSCI 207 Spring, 2020 Homework #2: Scaling Please type your answers into this document and upload to ELMS by 11:59 P.M., Thursday, February 20 Also on ELMS: some time on Wed. (February 19) or Thursday (February 20) take a short timed quiz on Scaling You may work on this homework on your own or discuss it with others. If you work through questions with others, be sure to turn in your own work, at the level at which you understand it. Do not simply copy answers from or for someone else. Questions such as these may show up on quizzes or exams, so be sure to work to understand how to arrive at the answers. Note: Questions or statements in italics require a response from you. Problem 1. Fish scaling You have caught eight individuals of a particular fish species and measured the length, depth (vertical height), and thickness (side-to-side distance) of each. For each fish, approximate total surface area and volume of each fish were calculated, as shown in the table below. LENGTH (cm) 2 4 6 10 16 24 48 60

THICKNESS (cm) 0.5 1 1.5 2.5 4 6 12 15

VOLUME (cm3)

DEPTH (cm) 1 2 3 5 8 12 24 30

1 8 27 125 512 1728 13824 27000

AREA (cm2) 6 24 54 150 384 864 3456 5400

Use Excel (or a similar program) to fit power functions (scaling equations, Y = axb) to these data (a power function, which has the form y = axb, has a variable base raised to a fixed exponent). Exactly how to do this will depend on your version of Excel. But the procedure will be something like this: 1. Select the two columns of numbers you wish to plot (e.g., depth and volume). 2. Click on the Charts tab [or Insert tab] & select Scatter plot [X Y scatter] 3. Select the resulting plot and click on the Chart Layout tab & Trendline subtab [Add chart element, then Trendline] 4. Under Trendline Options, select Power [possibly under More trendline options] & under Options (just to the left) [or under Format Trendline] click on “Display equation on chart” and click OK. The equation of the power function fitted to your data should be displayed on your graph. 1a) (1/2 pt) What is the value of b for depth vs. volume (note that “A versus B” tells you that A is on the Y axis and B is on the X axis)? Paste your graph below.

b = 1/3 35 30 f(x) = x^0.33

depth (cm)

25 20 15 10 5 0 0

5000

10000

15000

20000

25000

30000

volume (cm^3)

1b) (1/2 pt) What is the value of b for area vs. volume? Paste your graph below. b = 2/3 6000 f(x) = 6 x^0.67

5000

area (cm^2)

4000 3000 2000 1000 0

0

5000

10000

15000

volume (cm^3)

20000

25000

30000

1c) (1/2 pt) What is the value of b for volume vs. area? Paste your graph below. b = 3/2 30000

volume (cm^3)

25000

f(x) = 0.07 x^1.5

20000 15000 10000 5000 0 0

1000

2000

3000

4000

5000

6000

area (cm^2)

1d) (1/2 pt) What is the value of b for thickness vs. area? Paste your graph below. b = 1/2 16 14

f(x) = 0.2 x^0.5

thickness (cm)

12 10 8 6 4 2 0

0

1000

2000

3000

4000

5000

6000

area (cm^2)

1e) (1/2 pt) Do your fish appear to be growing isometrically or allometrically? Explain. The fish seems to be growing isometrically because the scaling for each of the variables is growing as isometrically expected. For example, the b=1/3 for depth (cm) vs. volume (cm3) which indicates both variables are growing at the same rate since the change in depth=1 and volume=3 which produces 1/3. This reasoning can be applied for the other graphs as well. Problem 2. Dinosaur scaling

Theropod Dinosaur Species Staurikosaurus pricei Coelophysis bauri Coelophysis rhodesiensis

Body Mass (kg)

Body Length (m)

19

2.08

15.3 13

2.68 2.15

Elaphrosaurus bambergi

210

6.2

Liliensternus liliensterni

127

5.15

Dilophosaurus wetherilli Ceratosaurus nasicoris

283 524

6.03 5.69

Eustreptospondylus oxoniensis

218

4.63

Metriacanthosaurus? sp.

130

3.8

1010 1320

7.4 7.9

Gorgosaurus libratus

700

5.8

Gorgosaurus libratus

2500

8.6

Daspletosaurus torosus

2300

9

Tarbosaurus bataar

760

5.8

Allosaurus fragilis Allosaurus atrox

Tarbosaurus bataar

2100

7.7

Tyrannosaurus rex

5700

10.6

45 73

3.06 3.43

Velociraptor mongoliensis

15

2.07

Ornithomimus edmontonicus

110

3.3

Ornithomimus brevitertius Gallimimus bullatus

144 27

3.66 2.15

Gallimimus bullatus

440

6

Deinonychus antirrhopus Deinonychus antirrhopus

The dinosaur data above were plotted in Excel, as shown below

2a) (1/2 pt) Does one variable change faster relative to the other? If so, which one? Yes, mass seems to be changing faster than length. Log-transforming and plotting the data. As discussed in lecture, logarithms are very useful for manipulating and visualizing numbers that span large scales (for example, masses and lengths of mammals ranging in size from mice to elephants) and for analyzing functions that are multiplicative in nature. In principle, the exponents of any positive, non-zero number can be used as the base of a logarithm, but only two numbers are used regularly in biology: 10 and e (=2.71828..., which is known as the base of the natural logarithm). The function that transforms a standard number into a base 10 logarithm is typically abbreviated log10 and the one that transforms a standard number into a base-e or natural logarithm is abbreviated ln. We use log10 for this exercise, but ln would work just as well. Excel was used to log-transform the body mass and body length data and fit a linear trend line to it

2b) (1/2 pt) Compare the plot of the original data to the log-transformed data. Describe how the distribution of points on the graph has changed. The points on the log-transformed graph is more spread out and easier to see the individual points, whereas the original graph seemed to have the majority of the points all bunched up in an area. 2c) (1/2 pt) Based on these data, what is the scaling coefficient (b) for the relationship of length vs. mass in theropods? How do you know? The scaling coefficient is b = 0.2719 because it’s the scale factor (slope) for x in the equation given on the graph. 2d) (1/2 pt) Do theropod dinosaurs scale isometrically? If not, what does the observed deviation from isometry tell us about how theropod shape varies with overall size (mass)? No, they are not scaled isometrically. The expected value of b for the dinosaurs to be scaled isometrically should be b=0.33, but instead it’s b=0.27. Since the b value is lower than isometrically expected, this deviation tells us the mass is growing at a greater rate than the length.

2e) (1 pt) Using the equation for the line Excel fit to the log-transformed data, write the power function (=”scaling equation”) for untransformed data that describes the original data in the familiar form y = axb (let L =length and M= mass) and explain how you got your answer. y=1.0677x^0.2719, I got this answer by first using the scaling coefficient (b) as the (b) in the power function so 0.2719 would be the exponent. Then, I raised 10 to the power of the y-intercept from the transformed equation to “un-log” that value, which produced 1.0677—the a value.

Problem 3. Earthworm scaling

A. Let’s assume that earthworms are shaped essentially like cylinders of length L and radius R. Assume that growth in earthworms is isometric. What is the predicted value of the exponent (b, the scaling coefficient) for each of the following scaling equations (ignore the allometric coefficient, a). Remember: “A versus B” tells you that A is on the Y axis and B is on the X axis: (1/4 pts each) a) length (L) vs. mass (M): L = aMb b=1/3 b) cross-sectional area (A) vs. mass (M): A = aMb b=2/3 c) radius (R) vs. cross-sectional area (A): R = aAb b=1/2 d) volume (V) vs. radius (R): V = aRb b=3 e) mass (M) vs. total surface area (A): M = aAb b=3/2 f) cross-sectional area (A) vs. total surface area (A): A = aA b b=2

B. (1/2 pt) Here is a graph of actual measurements of body length (upper line) and of the diameters of two particular segments (15 and 50, lower lines) from a sample of resting individuals of the earthworm Lumbricus terrestris. What do these data suggest about whether the geometric scaling of growth in earthworms is isometric? Explain your answer. The geometric scaling of the earthworms are in fact isometric because the exponent for mass is about 1/3 (0.33) which is representative of the length (1) to mass (3) ratio.

Problem 4. Interpreting Allometric Plots W = Watt = Joules per second Explain why each of the following three statements about graph A is wrong: a) (1/2 pts) The metabolic rate of a bull is twice that of a goose. It seems that the metabolic rate of the bull is ~800, whereas the rate for the goose is ~20. With that being said, the metabolic rate of the bull is way more than twice that of the goose. b) (1/2 pts) The untransformed data plotted here provide an example of positive allometry between metabolic rate and mass. The example does not portray positive allometry because the slope is less than 1, which indicates negative allometry. c) (1/2 pts) The difference in mass between a rat and a guinea pig is about the same as between a dog and a cassowary. The difference in mass between a rat and a guinea pig is between a scale of 0.1 to 1, whereas the difference between a dog and a cassowary is between the scale of 10 to 100. This means although the points for the rat and guinea pig are about the same distance away from each other as the dog and

A

cassowary, since the scale for the dog/cassowary is wider (because it is log transformed), the difference in mass is most likely larger. Explain why each of the two following statements about graph B is wrong: d) (1/2 pt) The total metabolic rate of an elephant is much lower than the total metabolic rate of a mouse. Although the elephant’s BMR is lower than that of the mouse, that does not mean the total metabolic rate of an elephant is much lower than the total for a mouse. Since the elephant is significantly larger than the mouse, its total metabolic rate has to be greater than the total for the mouse. e) (1/2 pt) This is an example of negative allometry. This is more of an example of inverse allometry because the relationship between body mass and BMR is a negative one where the slope is less than 0.

B...