Flour Beetle :This exercise examines the components present in a typical image

BUGS: A calibration and measurement exercise on a couple of images.

Density Slice : How to calculate the area of objects using Density Slicing (A technique used to highlight a range of pixel values)

Shell Area

	Left Shell	Center Shell	Right Shell	Total
Area (mm. squared)	=1102	=696	=725	2523
% of Total	44%	28%	29%

 

Tree Ring Geometry: The Chaco Canyon Story

1. Did the tree grow slowly or rapidly during its first few years? How can you tell?The tree grew rapidly during the first few years - the growth rings are very wide.

2. How many years did the tree live? The tree was about 60 years old when it was felled.

Representative diameter and circumference results

Table 1:

Year	Diameter	Radius	Circumference (cm)	Area (cm-squared)
970	1.5	0.8	4.7	2.0
980	4.8	2.4	15.1	18.1
990	6.4	3.2	20.1	32.2
1000	7.9	4.0	24,8	50.2
1010	8.9	4.5	27.9	63.6
1020	10.3	5.2	32.3	84.9

3 - 5. Use the data in your table to make a graph of the change in the tree’s circumference versus time. Plot the year on the x-axis and the area on the y-axis.

6. Graph the change in the tree’s cross-sectional area versus time. Plot the year on the x-axis and the circumference on the y-axis.

7. Did the tree appear to grow at the same rate during each decade? No

Why might the tree have grown more during certain decades than others? The rate of growth changed over time. Factors affecting growth rate include (but may not be limited to) rainfall, temperature, and competition from neighboring trees.
8. During which decade did the tree grow most slowly? The tree grew most slowly during the decade 1000-1010 A.D.
9. During which decade did the tree’s circumference increase the most? The circumference increased the most from 970-980 A.D.
10. On average, how much did the tree’s diameter increase every year? (Divide the total diameter by the total number of years.) 1020 A.D. - 970 A.D. = 50 years. Growth of 10.3 cm - 3.4 cm. = 6.9 cm. - (6.9 cm/50 years = 0.14 cm/yea

 

General Science

 

Butterfly : A measuring exercise + a little background on butterflies

This is a monarch butterfly - by the way it is a male!

Wing Span (mm)	Missing Wing (mm)	% of Right Wing Missing
=46.5	46.5 - 37.7= 8.8 mm.	= 19%

Fibers: Calculating the porosity of different fabrics

1. Compare the LM view with the SEM view.

Are the images exactly the same magnification?

The second LM view is at a slightly higher magnification.

Which image gives you more detail?

The SEM image

Any idea why?

Higher resolution scanning the image with electrons (shorter) than longer light waves.

2. Which fabric weave (F1, or F2) would you want to be wearing during the Winter? Summer? Explain.

F2 Tighter weave with less air holes.

Table 1

Fabric	Thread	Count	Total Area	Open Area	Thread Area	% Open Area
	Vertical	Horizontal
F1	4	4	293760	136414	157346	46%
F2	6	5	298240	80217	218023	27%

Optional

TF1LM fibers run around 280 µ thick while SEMF1 fibers are close to 230 thick

 

It's Just a Phase: Observing the Moon's Cycles

Phase	Day	Area Visible (km-squared)	% Visible
Waxing Crescent	4	2,045,000	23
First Quarter	7	4,915,000	55
Waxing Gibbous	10	6,881,000	77
Full	14	8,934,000	100
Waning Gibbous	20	6,198,000	69
Third Quarter	22	4,257,000	48
Waning Crescent	26	2,414,000	27
New	28	0	0

 

1. Make a graph of the % Visible vs. Day

2. Describe how the % Visible changes through a lunar month.

Answers will vary. The %Visible changes most rapidly near the first and third quarters, and least rapidly near the full and new phases. Students may (correctly) attribute this phenomenon to the Moon's spherical shape combined with its relatively constant orbital and angular velocities. This cycle does change slightly due to factors such as libration and tidal forces, but most students will be unaware of this. It is reasonable for them to assume that the cycle is stable.

3. What percent of the moon is visible on day 5? ....on day 17?

Day 5 = 33%, Day 17 = 84%

 

Aral Sea : Environmental changes in the Aral Sea

1. What do you get for an increased length to the peninsula?

The penninsula changed from an initial length of � 7 km. to 20 km.

2. Aral Sea water area change.

1973 Area in square kilometers	1987 Area in square kilometers
= 14000 km	= 10500 km

Biology

Cheek Cell : A comparison between plant and animal cells.

Table 1

	Length µ	Area µ(squared)
Cell 1	= 51	= 1280
Cell 2	= 53	= 1320

1. Compare this length with your other two light microscope lengths.

2. What is the % difference between the light microscope length measurements and the SEM length measurement? (Show your calculations) Comment on possible reasons for the difference.

3. What is the most prominent cellular difference between these plant cells and the cheek cells.

Plant cells have a regular, rigid cell well while the cheek cells show an irregular membrane.

4. If this is a plant cell, what two structures are not present?

Vacuole and cell membrane.

5. Why is the copy portion selected in this manner?

This will allow for the thickness of the lines.

Table 2

	Length µ	Area µ(squared)
Cell	63 µ	2304 µ

Table 3

	Length µ	Area µ(squared)
Cell	140 µ	6170 µ
Nucleus	12µ	114 µ

6. Compare the nuclear area to the area of the entire cell as a percentage. How accurate a method is this for determining the size of a nucleus in the cell?

114 µ/6170 µ = 18.5% If the section/view you are looking at is midway through cell and nuceus then it is an accurate % of the two-dimensional cell. (A more accurate portrayal would be the nuclear volume as a % of the cellular volume.)

Elodea:The effect of osmosis and plasmolysis on the plant cell Elodea

1. Name three structures that are shown in this plant cell that you would not expect to find in an animal cell?

Cell wall, vacuole, chloroplasts

2. Why is the copy portion selected in this manner?

To account for the thickness of the line.

3. What % of the cell is occupied by the vacuole?

3411 - 1060 = 2351 sq.µ Possible volume of vacuole.

4. Can you see any problems with calculating the area of the cell vacuole in this manner?

You are assuming the vacuole is composed entirely of water and that all that water has exited. There is also a significant water content to the cyoplasm. Factors other than diffusion might be at work. (Many cell membranes have "carrier" proteins that "assist" water movement.

Table 1

	Perimeter µ	Area µ(squared)
Cell Wall	= 294	= 3411
Cell Membrane	= 147	= 1060

 

Bone Density : This exercise examines normal and osteoporotic bone.

Table 1

	Patient 1	Patient 2
	Density (Mean Gray Value)	Density (Mean Gray Value)
L1	156	136
L2	157	142
L3	168	151
L4	165	149

 

Compare the two X-Rays.

Based on your bone density results which person do you think is suffering from osteoporosis?

The Lumbar vertebrae in Patient 1 alll have a higher density than similar vertebrae in Patient 2.

 

Last updated: November 22, 2002

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