ʼ
1

TAS2R38
GLO

Leonardo da Vinici' s drawing
Vitruvian Man shows how the proportions of the human body fit perf e ctly into a circle or a square. This diagram by Leonardo da Vinci is an illustration ofVitruvius' theory .
According to Vitruvius's theory the distance from fingertip to fingertip (ann span) should b e equal to the distance from head to heel (height). In this activity you will explore the legitimacy ofVitruvius' theory by developing a hypothesis regarding the Vitruvian Man. ,
A hypothesis is a possible explanation for a set of observations or an answer to a scientific question. A hypothesis is useful only if it can be tested. Testable hypothesis are generally written in a fonnalized format using an iflthen statement.
• If my car does not start because the battery is dead, then when I replace the old battery with a new one it will start.
• If increasing physical activity causes a person to bum calories and lose weight , then I should lose weight if! run 2 miles a day.
Formalized hypotheses contain both a dependent and an independent variable. The ind e pendent variable is the one that you change and the depend e nt variable is the one you observe and measure to collect data. Using the if then fommt
Temperature is the independent variable because I change it and movement is the dependent variable because it is the one that is observed to look for change . Using the think about what results are expected.
, if! change temperature , then what will happen to movement. if then format forces the scientist to
PURPOSE
In this activity you will devise and test a hypothesis regarding Vitruvius' theory on human proportions.
In Part II , you will devise and test a hypothesis concerning the relationship between foot and arm span length s .
2

•
MATERIALS
14 metric measuring tapes
PROCEDURE
PART I
1. Write an if-then hypothesis based on Vitruvius' theory relating arm span and height. Record your hypothesis on the student answer page.
2. Working with a partner, measure your arm span by standing against a flat surface and spreading your arms out as far as possible. Have your partner measure the distance from the longest finger on one hand to the tip of the longest finger on the other hand. Record your measurements in Data Table 1.
3. Repeat step two on your partner.
4. Remove your shoes and have your partner measure your height as you stand against a flat surface.
Measure the distance from the top of your head to the floor. Record your measurements in Data
Table 1.
5. Repeat step 4 on your partner.
6. Calculate the difference between your arm span and your height (arm span-height). Record your calculations in Data Table 1.
7 . Gather data from 10 additional students in the classroom. Record the student's name, sex and data in Data Table 2.
PART II
1. Some people have observed that the length of their foot is the same as the length oftheir forearm.
Others disagree saying there is no relationship between the two. You have been assigned to investigate this phenomenon. As a good scientist, you know that first thing you need to do is write a hypothesis. Is there a direct relationship between the length of a person's foot and the length of their forearm? Write an if-then hypothesis for this relationship on the student answer page in the space labeled Hypothesis #2.
2. Collect foot-forearm data from five people and record the measurements in Data Table 3.
1
3

Name ______________ __ ______ _
Period ____________ __________
HYPOTHESIS #1
HYPOTHESIS #2
DATA AND OBSERVATIONS
Measured in cm
Data Table 1: Lab Partner Data
Arm Span Height
Your measurements
Your Partners measurements
Difference
Person
M/F
Data Table 2: Class Data in em
Arm Span Height Difference
I
4

AP BIOLOGY
HEREDITY
ACTIVITY #6
NAME_____________________
DATE___________HOUR_____
M&M S
– A C
S
A
I NTRODUCTION
Have you ever wondered why the package of M&Ms you just bought never seems to have enough of your favorite color? Or, why is it that you always seem to get a package with mostly brown M&Ms? What’s going on at the Mars Company? Is the number of the different colors of M&Ms in a package really different from one package to the next? Or, does the Mars Company do something to ensure that each package gets a certain number of each M&M color? The following information is from the Mars Company web site.
13% brown, 14% yellow,
13% red, 24% blue,
20% orange, & 16% green
23% brown, 15% yellow,
12% red, 23% blue,
23% orange, & 15% green
Source: http://us.mms.com/us/about/products/milkchocolate/ http://us.mms.com/us/about/products/peanut/
One way we can determine if the Mars Company is true to its claims is to sample a package of M&Ms and complete a statistical test known as a “goodness to fit” test.
This type of statistical test allows us to determine if any differences between our observed measurements (count of each color from the M&M package) and our expected (claims posted at the M&M web page) are due to chance or some other reason (the Mars Company sorters aren’t doing a good job of putting the correct number of M&Ms in each package.) The goodness of fit test we will be using is called a chi-square analysis. This test is generally used when dealing with discrete data (i.e. count data or discontinuous data.) We will be calculating a statistical value and using a table to determine the probability that the differences between observed data and expected data is due to chance alone.
Heredity Activity
5

The formula for calculating chi-square ( χ 2 ) is :
2
( o
e
) 2 e
5.
6.
3.
4.
That is chi-square is the sum of the squared difference between observed (o) and expected (e) data divided by the expected data in all possible categories.
H
YPOTHESIS
:
If the Mars Company sorters are working properly then any difference between the color percentage in an actual package of M&Ms and the color percentage posted on the web site should be due to random chance.
P
ROCEDURE
:
1.
Wash your hands. You will be handling food you may want to munch on later.
2.
Put paper towels down on your desk. You will be counting the M&Ms on your desk.
Open the bag of M&Ms.
Do NOT eat any of the M&Ms right now. Separate the M&Ms into color categories and count the number of each color.
Record your M&M color totals in the data table.
Determine the total number of M&Ms in your package and record this number in the data table.
7.
Calculate the expected number of M&Ms in your package by multiplying the total number of M&Ms in the package by the color percent listed on page 1 of the activity. For example, if your package contains 500 M&Ms and you want to find the expected number of red M&Ms you will need to multiply 500 by
20% (500 x 0.20). Record your calculations in the data table.
8.
Calculate the difference between the observed and expected numbers for each M&M color. Record your calculations in the data table.
9.
Square the difference between the observed and expected. Record your calculations in the data table.
Heredity Activity
6

10.
Divide the square of the difference by the expected. Record your calculations in the data table.
11.
Total all the answers from step 10 to determine the chi-square (
Record the chi-square ( χ 2 ) in the data table.
χ 2 ) value.
D
ATA
T
ABLE
Observed
(o)
Expected (e)
Difference
(o-e)
Difference squared (d 2 d 2 = (o-e) 2
) d 2 /e
Σ (d 2 /e) = χ 2
12.
Now you must determine the probability that the difference between the observed and expected values (as summarized by the calculated value of chisquare) occurred simply by chance. To do this you will need to compare the calculated value of chi-square with the appropriate value from the Chi-
Square Distribution Table on the next page. Examine the table. Note the term “degrees of freedom.” For this statistical test the degrees of freedom is equal to the number of classes (color categories) minus one. Complete the following to determine the degrees of freedom for the M&M analysis:
# of color categories degrees of freedom
13.
The reason why it is important to consider degrees of freedom is that the value of the chi-square statistic is calculated as the sum of the squared differences for all classes. The natural increase in the value of chi-square with an increase in classes must be taken into account. Scan across the row corresponding to 5 degrees of freedom. Values of the chi-square are given for several different probabilities ranging from 0.95 on the left to 0.001 on the right. Note that the chi-square increases as the probability increases.
Notice that a chi-square value of 1.63 would be expected by chance in 95%
Heredity Activity
7

(0.95) of the cases, whereas one of 12.59 would be expected in 5% (0.05) of the cases. Use the chi-square value calculated and recorded on the data table to determine the probability for the M&M analysis. If the exact chisquare value is not listed in the table estimate the probability. Record your answer below.
Chi-square value =
Probability =
C
HI
-S
QUARE
D
ISTRIBUTION
T
ABLE
Accept Hypothesis Reject Hypothesis
Degrees of
Freedom
1
5
6
7
2
3
4
8
9
10
0.95 0.90 0.80 0.70 0.50 0.30 0.20 0.10
0.004 0.02 0.06 0.15 0.46 1.07 1.64 2.71
0.05
3.84
0.01 0.001
6.64 10.83
0.10 0.21 0.45 0.71 1.39 2.41 3.22 4.60 5.99 9.21 13.82
0.35 0.58 1.01 1.42 2.37 3.66 4.64 6.25 7.82 11.34 16.27
0.71 1.06 1.65 2.20 3.36 4.88 5.99 7.78 9.49 13.38 18.47
1.14 1.61 2.34 3.00 4.35 6.06 7.29 9.24 11.07 15.09 20.52
1.63 2.20 3.07 3.83 5.35 7.23 8.56 10.64 12.59 16.81 22.46
2.17 2.83 3.82 4.67 6.35 8.38 9.80 12.02 14.07 18.48 24.32
2.73 3.49 4.59 5.53 7.34 9.52 11.03 13.36 15.51 20.09 26.12
3.32 4.17 5.38 6.39 8.34 10.66 12.24 14.68 16.92 21.67 27.88
3.94 4.86 6.18 7.27 9.34 11.78 13.44 15.99 18.31 23.21 29.59
14.
Scientists, in general, are willing to accept a hypothesis if the probability that the difference between the observed and expected results is greater than 5%
(0.05). If the probability determined in question 13 is greater than 5%
(0.05) then any differences between the observed color counts and the claims of the Mars Company (posted at their web site) is due to chance alone. Five percent! That’s not much … but it’s good enough for scientists!
If, however, the probability determined in question 13 is less than 5% (0.05) then any differences between the observed color counts and the claims of the
Mars Company is not due to chance. Some other factor caused the differences. Based on your results, should you accept or reject the hypothesis? Explain your answer.
_____________________________________________________________
_____________________________________________________________
Heredity Activity
8

Plain Peanut Crispy Minis
Peanut
Butter
Almond
Red 13% 12% 17% 12% 10% 10%
Green 16% 15% 16% 12% 20% 20%
Blue 24% 23% 17% 25% 20% 20%
Questions:
1. What is the chi-square value for your data?
2. What is the critical value for your data?
3. What is the p-value for chi-square statistic?
4. Given your p-value, create a statement that describes the “goodness of fit” of your data.
5. Was you null hypothesis (H
O
) accepted or rejected?
9

15-8939
TEACHER’S MANUAL
10

Student Instructions
15-8939
Name
Date
In this laboratory you will
• observe the inheritance of traits over three generations of plants.
• recognize contrasting phenotypes.
• propose and test models for the inheritance of the phenotypes.
• make predictions about the inheritance of the phenotypes.
• collect and analyze data from F
2 seedlings.
• compare predicted results with results obtained from actual data.
• modify predictions on the basis of new data.
P
1
P
2
F
1
Wisconsin Fast Plants TM , Brassica rapa , are widely used to study inheritance. There are several mutant types available for study, and they can be crossed by transferring pollen from the flower of one plant to the flower of another plant. You will germinate Wisconsin Fast Plants seeds collected from three generations of plants and make observations that will help you determine the pattern of inheritance.
In sexual reproduction, two gametes fuse, bringing together homologous chromosomes from both parents. One chromosome of the homologous pair is inherited from the male parent and the other is inherited from the female parent. Since genes (the hereditary units) are located on the chromosomes, there are two copies of each gene. For example, Drosophila (fruit flies) have a gene for eye color, and there are two forms or alleles of this gene, Se , a dominant allele for red, and se , a recessive allele for sepia (brown). If both alleles on the chromosome pair are for red ( Se/Se ) the phenotype (appearance or form that is observed) is red eyes and the genotype is homozygous . If the two alleles are different
( Se/se ), the phenotype is red eyes and the genotype is heterozygous . If the two alleles are for sepia
( se/se ), the phenotype is sepia eyes and the genotype is homozygous recessive.
A cross involving one gene (one set of alleles) is a monohybrid cross. Monohybrid crosses usually involve parents that have one set of contrasting phenotypes. An example in corn would be crossing a red kernel corn with a yellow kernel corn. A cross between parents that differ in two separate genes (two separate sets of alleles) is a dihybrid cross. Dihybrid crosses usually involve parents that have two sets of contrasting phenotypes. An example of a dihybrid cross in corn would be crossing red, starchy kernel corn with yellow, sugary kernel corn.
Here are some symbols that you will find useful in this investigation.
F
2 maternal parent or “mother” paternal parent or “father” the first generation offspring that result from crossing of the P
1
Short for first filial , a word that refers to offspring. and P
2
; the children, so to speak. second-generation offspring, the result of crossing two F
1 plants
© 2 0 0 5 C a r o l i n a B i o l o g i c a l S u p p l y C o m p a n y
11
S-1

Germinating the P
1 and F
1
Seeds
Use a straightedge and pencil to draw a line through the center of a circle of filter paper, dividing the filter in half. Label one side P
1 and the other F
1
. Add your name or group number. Place the filter in the bottom of a petri dish and flood the paper with water until it is completely soaked. Pour off any excess water not absorbed by the filter. Place five P
1 seeds on the P
1 side of the filter and five F
1 seeds on the other side. Place the lid on the dish and leave it under the light bank as indicated by your teacher. You will examine the dish in 3 days, after the seeds have germinated.
Observing the P
1 and F
1
Seedlings
Observe the P
1 and F
1 seedlings in your petri dish. Record any consistent difference or differences between these two sets of seedlings. Compare your seedlings to those of other groups to be certain that all groups are seeing the same differences.
What phenotypic differences do you notice between the P
1 and F
1 plants?
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
Forming a Hypothesis and Making Predictions:
Based on your observations and general knowledge of genetics, state a hypothesis (or model) of how you think these phenotypes are inherited. Be as complete as possible and cite evidence for or give a reason for each part of your hypothesis. Your hypothesis should include whether the phenotypes are inherited through a single set of alleles (monohybrid) or two sets of alleles (dihybrid). Include a one- or two-word designation for each phenotype and a symbol for the allele that gives rise to each phenotype. Indicate which allele of each pair is dominant and explain how you know.
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
___________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
© 2 0 0 5 C a r o l i n a B i o l o g i c a l S u p p l y C o m p a n y
12
S-2

Each parenr p l anr
(P, and
P ,) was homozygous for the alleles inv olved in the cr os s that you are studying. Given [his informacion , use your hypmhesis to predicr the following for each prediction):
(give your reas o ning
I. the phenotype of the P,:
Z . the genotype of the F,:
3 . the phenotype (5) of the F,:
Te sting Hyp o th esis and Predicti ons
Your teacher wi ll now exp l ain wha t mat e ri als are a va ilable to you . Outline a procedure whereby you ca n use these materia l s to tes t your hy pot h es i s. Incl u de methods a nd expected ou tcomes. Li st a t l e ast two poss i b le outcomes t ha t cou l d cause you to r e je c t your hypo thes i s or rnodi/y it .
Now set up yo ur test.
Observing
F l
Seedlings
Three days after se t t in g up your test, r etr i eve your petr i dis h and inspect t he F, seed lin gs.
Has this te s t confi rme d your hypothes i s! Expl ain your answer.
Based on the res u lts of yo u r test, should you
I. accept your hypothesis without modification! If so , sta t e your argument (or accepting your hypothes i s.
Z. r eject your hypothesis! If so, develop a n ew , detailed hyp o thesis to r eplace your o l d one.
3 . m odi/Y your hypothesis? I f so , explain i n detail your modification.
©2005 Carolina B iolo g ic al Supply Company
13
5 3

(or How Selection Affects the Hardy-Weinberg Equilibrium)
Introduction:
Understanding natural selection can be confusing and difficult. People often think that animals consciously adapt to their environments (that the peppered moth can change its color, the giraffe can permanently stretch its neck, the polar bear can turn itself white) so that they can better survive in these environments.
In this laboratory you will use fish crackers to help further your understanding of natural selection and the role of genetics and gene frequencies in evolution.
Background: Facts about the 'Fish'
1.
These little fish are the natural prey of the larger fish-eating sharks ( Carcharodon sapiens )- YOU!
2.
Fish come with two phenotypes: gold and brown: a.
Gold is a recessive trait (f); these fish are palatable and are easy to catch. b.
Brown is a dominant trait (F); these fish taste salty, are sneaky and hard to catch.
3.
You, the terrible fish-eating shark ( Carcharodon sapiens ), much prefer to eat the palatable gold fish; you eat ONLY gold fish unless none are available in which case you resort to eating brown fish in order to stay alive.
4.
New fish are born every 'year'; the birth rate equals the death rate. You simulate births by reaching into the container of 'spare fish' and selecting randomly.
5.
Since the gold trait is recessive, the gold fish are homozygous recessive (ff). Because the brown trait is dominant, the brown fish are either homozygous or heterozygous dominant (FF or Ff).
Hardy-Weinberg:
G. H. Hardy, an English mathematician, and W.R. Weinberg, a German physician, independently worked out the effects of random mating in successive generations on the frequencies of alleles in a population.
This is important for biologists because it is the basis of hypothetical stability from which real change can be measured.
For fish crackers, you assume that in the total population, you have the following genotypes, FF, Ff, and ff. You also assume that mating is random so that ff could mate with ff, Ff, or FF; or Ff could mate with ff,
Ff, or FF, etc. In addition, you assume that for the gold and brown traits there are only two alleles in the population - F and f. If you counted all the alleles for these traits, the frequency of 'f' alleles plus the frequency of 'F' alleles would add up to 1.
14

The Hardy-Weinberg equation states that: p + q = 1 p 2 + 2pq + q 2 = 1
This means that the frequency of pp (or FF) individuals plus the frequency of pq (or Ff) individuals plus the frequency of qq (ff) individuals equals 1. The pq is multiplied by 2 because there are two ways to get that combination. You can get F from the male and f from the female OR f from the male and F from female.
If you know that you have 16% recessive fish (ff), then your qq or q 2 value is .16 and q = the square root of .16 or .4; thus the frequency of your f allele is .4 and since the sum of the f and F alleles must be 1, the frequency of your F allele must be .6 Using Hardy Weinberg, you can assume that in your population you have .36 FF (.6 x .6) and .48 Ff (2 x .4 x .6) as well as the original .16 ff that you counted.
Procedure for SELECTION:
1.
Get a random population of 10 fish from the 'ocean.'
2.
Count gold and brown fish and record in your chart; you can calculate frequencies later.
3.
Eat three (3) gold fish; if you do not have three (3) gold fish, fill in the missing number by eating brown fish.
4.
Add three (3) fish from the 'ocean.' (One fish for each one that died.) Be random. Do NOT use artificial selection.
5.
Record the number of gold and brown fish.
6.
Again eat three (3) fish, all gold if possible.
7.
Add three (3) randomly selected fish, one for each death.
8.
Count and record.
9.
Repeat steps 6, 7, and 8 until all of the generations are filled.
10.
Fill in the class results on your chart.
11.
Fill Table 1 and calculate frequencies, prepare your graph, and answer the questions.
15

Generation
1
2
3
7
8
9
4
5
6
10
Table 1. Pair Data: SELECTION
Gold Brown p q p 2
Generation
1
5
6
7
2
3
4
8
9
10
Table 2. Class Data: SELECTION
Gold Brown p q p 2
16
2pq
2pq q 2 q 2

Procedure for NO SELECTION:
1.
Get a random population of 10 fish from the 'ocean.'
2.
Count gold and brown fish and record in your chart; you can calculate frequencies later.
3.
Without looking/choosing (close your eyes), pull three (3) fish RANOMLY from your pool of ten
(10).
4.
Add three (3) fish from the 'ocean.' (One fish for each one that died.) Be random. Do NOT use artificial selection.
5.
Record the number of gold and brown fish.
6.
Again, w ithout looking/choosing (close your eyes), pull three (3) fish RANOMLY from your pool of ten (10). Add three (3) randomly selected fish, one for each death.
7.
Count and record.
8.
Repeat steps 6, 7, and 8 until all of the generations are filled.
9.
Fill in the class results on your chart.
10.
Fill Table 2 and calculate frequencies, prepare your graph, and answer the questions.
17

Generation
1
2
3
4
7
8
9
5
6
10
Table 3. Pair Data: NO SELECTION
Gold Brown p q p 2
Generation
1
2
6
7
8
3
4
5
9
10
Table 4. Class Data: NO SELECTION
Gold Brown p q p 2 2pq
2pq q 2 q 2
18

Analysis:
1.
Prepare a graph of your data and the class results. On the 'x' axis put generations 1-10 and on the 'y' axis put frequency (0-1). Plot both the q and p for your data and for the class data. Use one color for your data and another color for class data. What generalizations would you make about your results? How do they compare to the class results?
2.
According to Hardy-Weinberg, what conditions would have to exist for the gene frequencies to stay the same over time?
3.
Why is it important to collect class data?
4.
Explain which phenotype is NOT favorable to the fish and why?
5.
What happens to the genotypic frequencies from generation 1 to generation 10?
6.
What process is occurring when there is a change in genotypic frequencies over a long period of time?
7.
What would happen if it were more advantageous to be heterozygous (Ff)? Would there still be homozygous fish? Explain.
8.
What happens to the recessive genes over successive generations and why?
9.
Why doesn't the recessive gene disappear from the population?
10.
Explain what would happen if selective pressure changed and the recessive gene was selected for.
For Further Investigation:
Design an experiment to show how one of the following affects allele frequencies over several generations: a.
migration b.
isolation c.
no selection d.
no random mating e.
very small population f.
mutations
Taken and modified from:
The Woodrow Wilson National Fellowship Foundation http://www.woodrow.org/teachers/bi/1994/fishfreq.html
19

Based on: D. W. Goldsmith (2003) American Biology Teacher 9: 679-682
Imagine a race through the woods with eight runners (A-H). All the runners start at a single starting line but at various places the course forks, and runners are free to choose either path. To help keep track of the race, each runner carries a card that gets stamped at check-in stations distributed at various places in the course. The following rules apply:
• Runners are obligated to collect a stamp from each check-in station they pass
• Each check-in station uses only one stamp
• No two check-in stations use the same stamp (this rule will be dropped later)
All eight runners finish the race carrying their cards, but they each cross a different finishing line. The aim is to use the information in the cards collected from the eight runners to reconstruct the course and the placement of check-in stations.
Draw a map of the course. Labeling the starting point, each check-in station (with the symbol of the stamp it uses), and each finishing line (with the letter of the runner who emerges there).
When you are done, copy it onto the overhead transparency provided.
What principles did you used in generating your map?
Compare your map to those generated by other groups. How are they similar? How do they differ?
Consider the analogy between the great clade race and biological evolution. What are the biological analogs of: a) The runners b) The check-in stations c) The race course
Discuss the following questions: a) Can you infer whether a particular path went right or left? b) Can you infer whether particular segments of the course (between one branch point and the next) are longer or shorter? c) When there were two check-in stations on a segment, can you figure out which came first? d) In the biological case, when is it valid to assume that each “runner” runs the same total distance? e) Is it reasonable to expect that the check in stations will be distributed evenly across the course? f) In the biological case, is it valid to assume that the “runners” receive a blank card at the start and that they always gain, but never lose, stamps?
20

Clade Race Two. This is the same as the original clade race with two important exceptions:
1) The runners all the get the same card at the beginning of the race, but it is not necessarily empty – it could have some stamps already added
2) The check-in stations either add a stamp or erase a stamp of a particular shape. To make the challenge easier, only one station on the course acts on each shape : if one station erases a square then no other station either erases or adds square; if one station adds circles, no other station either adds or erases circles, etc.
You are given the eight runners’ cards (“Clade race 2a”) and are told that at the first split in the course, runners A and G went one way, whereas everybody else went the other way.
Reconstruct the course.
If you had not been told that A and G split at the first branch-point, would this challenge have been possible?
Clade Race Three. For simplicity, this case is like the original clade race in that you are told that runners start with a blank card. However, we will now relax the assumption that only one station adds a given shape and will also allow that stations might erase particular preexisting shape. It will remain true, however, that a given station will act exactly the same on all runners that pass . For example, it could be that in the whole course two stations add squares and one station removes squares, but a given station that adds squares will do so to all runners that pass. In evolutionary biology the technical term to describe cases where characteristics are either gained in parallel (convergence) or gained and later lost (reversal) is “homoplasy.” Use the cards
(‘clade race 3a’) and make your best guess as to the correct course.
In the previous cases there was only one possible solution. Is that true this time? Why has homoplasy confused things?
Suppose several alternative courses were offered to explain these data. How would you decide which was most likely to be true? Can you think of a general criterion for picking among alternative trees? Can you use it to decide which of the three attached trees is best and which is worst?
Using the method you propose, will you always identify the correct tree? Will you know if you have identified the correct tree?
21

A
A
A
D
G
D
G
E
G
E
F
E
F
C
F
C
B
C
B
A H
B
H
H
22

Name ______________________________
An important part of any modern course in organismal biology is to understand cladistics. This is perhaps not the most obvious or easily understood concepts in biology, so we will work into it gently. We will use a modification of a published exercise designed for non-majors (David W.
Goldsmith. 2003. The great clade race: presenting cladistic thinking to biology majors and general science students. The American Biology Teacher 65: 679-682). For this exercise, work with one partner to carry out each part of this project.
Our warm-up exercise will involve a group of eight cards with human names on them. All of the names consist of three letters. The names are not to be used in the classification as the names are artificial common names. You may not imply gender or other characteristics from the names.
Rather, using only the graphical symbols on the cards, classify the cards. You may make as many groups as you want, but each group of cards should show or possess a particular theme. In the remaining space, discuss and defend your classification.
The groups we have defined and their membership:
The rules separating the individuals into these groups:
/8
Koning, Ross E. 1994. Cladistics Seminar. Plant Information Website .
http://plantphys.info/organismal/labdoc/cladisticslab.doc
23

The cards you have classified are the products of a fictitious cross-country meet carried out in a forest. This race was governed by the following rules:
1. All runners enter the woods at a single entrance at the south end of the forest. But there are several trails within the forest. (REK modification)
2. Runners may not stop part of the way down a trail or retrace their path.
3. When a trail branches, it only branches into two new trails, never three or more.
4. Once trails have branched apart, they do not reunite later in the race.
5. Check-in stations are located along the straight sections of the trail, never at a branch point. Runners must have their card punched at each station. The workers at the stations work quickly so the sequence of punches means nothing; only the shape of the punch out matters. (REK edit)
6. All runners must complete the race and emerge out the north edge of the forest through one of several different exits. (REK modification)
7. Using the punch-cards turned in by the runners, draw a map of the trails, check-in stations, and exits used by each of the runners. Put a hash-mark across the trail where each station is located.
Finish
Start
/20
24

Oh No! We have received news that a ninth runner has just finished the race…coming in very late! You will receive a ninth card. Remap your race with the information included on this ninth entry. Perhaps there are multiple explanations? Beneath the map showing the whole race and particularly highlighting the possible pathways and stations for this last runner, write a few sentences to explain your best guess about how all of this fits together.
Finish
Start
Explanation:
/7
25

Your project with the cross-country meet is very much like cladistic analysis of organisms. Give answers below that translate the meet into cladistics ideas:
Interpret the rules about how the race operates:
1. All runners enter the woods at a single entrance at the south end of the forest. But there are several trails within the forest.
2. Runners may not stop part of the way down a trail or retrace their path.
3. When a trail branches, it only branches into two new trails, never three or more.
4. Once trails have branched apart, they do not reunite later in the race.
5. Check-in stations are located along the straight sections of the trail, never at a branch point. Runners must have their card punched at each station. The workers at the stations work quickly so the sequence of punches means nothing; only the shape of the punch out matters.
6. All runners must complete the race and emerge out the north edge of the forest through one of several different exits.
7. Using the punch-cards turned in by the runners, draw a map of the trails, check-in stations, and exits used by each of the runners.
What do biologists call the runners who finished the race? ______________________________
Biologically, what do we call the group of all runners who passed by one station but then traveled in various pathways from that station?_______________
What do biologists call a group of just some of the runners who passed by one station? ________________
When a pathway branches in the woods, what do cladists call that evolutionary event?___________________________________
When there are multiple stations along a particular pathway, what would cladists call that evolutionary event?________________________________
/12
26

168 Chapter J5
Ellercise 8 (Module 15.131
, ,..
' , .. ,. ".: _ ' . ,. '.: .. ::'. '.' :.
Cladistic analysis seeks to clarify evolutionary and t a x ono mic relationship s by finding clades, groups o f o r genisms made up o f an ancestor an d all its descendants. Thi s simpli· fie d phylogenetic tre e u s e s cl adistic analysis (bas e d on anatomy, but backed up by mo le c · ular data) to recon s tru c t the relationship s among f ou r groups of plants and th eir closest rel a ti ves, the green a lgae . Read M o dul e 15.l3, examine the trees in the module and below, and then answer t h e followin g qu es tions. This e xercise is rather difficult, so teke you r time .
Green a lg a e Mo ss e s Ferns Con il e r s Fl owe rin g pl a nt s
/ / '1
)
1. Whjch four groups of organisms above m ake up the in g r o up7
2. Whi c h organisms cons titut e the o utgroup ?
3. Are analogous or homologous fea t ures used in cladi s ti c analys i s?
4 . Whkh characters are W1ique to a lineage o f organisms, shared derived characte r s or s hared primitive characters? Which are more useful in differenti<=l~ing among (s e parating ou ' t) distin c llin e age s ?
5. What i s a shared primitivE' cha racter common to aJl pla nt s ?
6. Whet i s a share d derived charecter common to all p la nt s?
7. What is a share d primitive charac ter common to all plants v v ith seeds?
8. What is a shared derived charecter common to all plants with s e eds'
9. Which charact e r s are most useful in deciding \vhether an organ i s m i s in the out g r o up o r t he in gro up , shared primitive cha ra c t ers or sha r ed derived charac ters ?
10. If we are inte re sted i n focusin g on all plant s that have vasculClr tissues, \· .... hich g roup s on the phylogeneti c tree constitute the outg r oup? The ingroup?
11. WllClt i s the name o f a t;1 xonomic group consisting of an ancestor t1lld ;:'I I! its d escend.mt
s ?
1 2. What other organisms Me in the clad e tha t includ es th e first pi<'lllts with s e eds?
1 3. Na tHe or desc rib t' nin e differ en t c l.1d
es s h own on the
~ ""h)'l o gt?n
eti l tree abo \' l' .
27

28

29

30

31

32

33

34

35

36

37

38

•   Uptake of foreign DNA , often a circular plasmid
GFP
Amp Resistance
•   A circular piece of autonomously replicating DNA ori bla

Originally evolved by bacteria

May express antibiotic resistance gene or
be modified to express proteins of interest
39

ori bla araC pGLO
GFP
Transmission electron micrograph
Agarose Gel
Graphic
–   Beta Lactamase
•   Ampicillin resistance
–   Green Fluorescent
Protein
•   Aequorea victoria jellyfish gene
–   araC regulator protein
•   Regulates GFP transcription ori bla pGLO araC
GFP
Genomic
Bacterial cell
Plasmid DNA
Genomic DNA
40

ori bla araC pGLO
GFP
LacI
lac Operon
Z Y A
Effector
(Lactose) araC ara Operon
B A D
Effector
(Arabinose)
LacI Z Y A araC B A D
RNA Polymerase
Z Y A araC
RNA Polymerase
B A D araC araC
ara Operon
B A D
Effector
(Arabinose) araC ara GFP Operon
GFP Gene
Effector
(Arabinose)
B A D araC GFP Gene araC B
RNA Polymerase
A D araC
RNA Polymerase
GFP Gene
41

•   Electroporation
–   Electrical shock makes cell membranes permeable to DNA
•   Calcium Chloride/Heat Shock
–   Chemically-competent cells uptake DNA after heat shock
•   Suspend colonies in Transformation
Solution
•   Add loop of pGLO plasmid DNA to “+” tube
•   Incubate tubes 10 minutes on ice
•   Heat shock 50 seconds at 42 o C
•   Incubate 2 minutes on ice
•   Incubate 10 minutes with nutrient broth
•   Plate
Ca ++
O
   Transformation solution = CaCl
2
Ca ++
O P O
O
CH
2
O
Sugar
Base
Positive charge of Ca +2 ions shields negative charge of DNA phosphates
Ca ++
O
O
P O
O
CH
2
O
Sugar
Base
OH
42

   Incubation on ice slows fluid cell membranes
   Heat-shock increases permeability of cell membrane
   Nutrient broth incubation allows beta lactamase expression
Cell wall
GFP
Beta lactamase
(ampicillin resistance)
•   Luria-Bertani (LB) broth
•   Medium that contains nutrients for bacterial growth and gene expression
–   carbohydrates
–   amino acids
–   nucleotides
–   salts
–   vitamins
43

+
-
•   Using a bulb pipet, transfer
250uL of transformation solution, from your microfuge tubes labeled ‘TS,” to each of your tubes labeled (+) & (-)
44

Using a sterile loop, pick up one single colony from your starter plate and add it to the + tube by placing the loop into the solution in the tube and swirling it to remove the cells.
Repeat for the - tube
Place both tubes on ice
Examine the pGlo plasmid solution with the UV light. Note your observations.
Immerse a new sterile loop into the plasmid DNA stock tube. Withdraw a loop full. Mix the loop into the + micro tube
Return to the ice for 10 minutes
45

Heat shock your cells in the 42˚C waterbath for 50 seconds. The movement from ice to hot water and back again must be rapid.
Incubate on ice for 2 minutes
Add 250ul of luria broth to each tube
Incubate at room temp for 10 minutes
During this time label your 4 agar plates
Spread 100uL of suspension onto corresponding plates
•   Follow protocol
•   On which plates will colonies grow?
•   Which colonies will glow ?
46

Making
Observations &
Data Collection
•   TE = total # of cells on agar plate
amount of DNA spread on plate
1.
Determine # of cells growing
2.
Determine amount of DNA spread (ug)
3.
10uL at 0.03ug/uL
4.
Volume spread and total volume in t.t.
(0.2)x0.3 = 0.06 ug (pGLO)
5. Use of scientific method
“Spin-Off” Lessons & Activities
•   Spin-off activities that tie into any science curriculum
•   Chromatography to purify GFP from pGLO-transformed bacteria (kit available from Bio-
Rad)
•   Other lessons include the use of genetic engineering, recombinant
DNA technology, biochemistry, and ethics in science
47

Ships at room temperature. Store DNA in the refrigerator
(4°C) or freezer (–20°C) within 4 weeks of arrival.
Duplication of any part of this document is permitted for classroom use only.
For Technical Service Call Your Local Bio-Rad Office, or in the U.S., Call 1-800-4BIORAD (1-800-424-6723)
48

One of the basic tools of modern biotechnology is DNA splicing: cutting DNA and linking it to other DNA molecules. The basic concept behind DNA splicing is to remove a functional DNA fragment — let’s say a gene — from one organism and to combine it with the DNA of another organism in order to study how the gene works. The desired result of gene splicing is for the recipient organism to carry out the genetic instructions provided by its newly acquired gene. For example, certain plants can be given the genes for resistance to pests or disease, and in a few cases to date, functional genes have been given to people with nonfunctional genes, such as those who have a genetic disease like cystic fibrosis.
This activity may be used to simulate the real world application of gene splicing.
You may suggest to your students that the DNA they are working with represents a chromosome that has been cut into many fragments. Of the fragments that are produced, one particular fragment may represent a specific gene. This imaginary gene can code for any number of traits, but before it can be given to a recipient organism, your students must first identify the gene by its size using agarose gel electrophoresis.
Restriction Enzymes
The ability to cut and paste, or cleave and ligate, a functional piece of DNA predictably and precisely is what enables biotechnologists to recombine DNA molecules. This is termed recombinant DNA technology. The first step in DNA splicing is to locate a specific gene of interest on a chromosome. A restriction enzyme is then used to cut out the targeted gene from the rest of the chromosome. This same enzyme is also used to cut the DNA of the recipient into which the fragment will be inserted.
Restriction enzymes are biomolecules that cut DNA at specific sites. Restriction enzymes, also known as endonucleases, recognize specific sequences of DNA base pairs and cut, or chemically separate, DNA at that specific arrangement of base pairs.
They were first identified in and isolated from bacteria that use them as a natural defense mechanism to cut up the invading DNA of bacteriophages — viruses that infect bacteria. Any foreign DNA encountering a restriction enzyme will be digested, or cut into many fragments, and rendered ineffective. These enzymes in bacteria make up the first biological immune system. There are thousands of restriction enzymes and each is named after the bacterium from which it is isolated. For example:
Eco RI = The first restriction enzyme isolated from Escherichia coli bacteria
Hin dIII = The third restriction enzyme isolated from Haemophilus influenzae bacteria
Pst I = The first restriction enzyme isolated from Providencia stuartii bacteria
49

Each restriction enzyme recognizes a specific nucleotide sequence in the DNA, called a restriction site, and cuts the DNA molecule at only that specific sequence.
Many restriction enzymes leave a short length of unpaired bases, called a “sticky” end, at the DNA site where they cut, whereas other restriction enzymes make a cut across both strands creating double stranded DNA fragments with “blunt” ends. In general, restriction sites are palindromic, meaning they read the same sequence of bases forwards and backwards on the opposite DNA strand.
For example, here is a list of enzymes and the sites where they cut:
Eco RI
✄
G A-A-T-T-C
Hin dIII
Pst I
C-T-T-A-A G
✄
✄
A A-G-C-T-T
T-T-C-G-A A
✄
✄
C-T-G-C-A G
G A-C-G-T-C
✄
Lambda Phage DNA
Lambda DNA comes from a bacterial virus, or bacteriophage, which attacks bacteria by inserting its nucleic acid into the host bacterial cell. Lambda is a lytic bacteriophage, or phage, that replicates rapidly inside host cells until the cells burst and release more phages to carry out the same infection process in other bacterial host cells. Bacteriophage lambda is harmless to man and other eukaryotic organisms, and therefore makes an excellent source of DNA for experimental study.
In this investigation, students observe the effects of three restriction enzymes on lambda genomic DNA. Since the lambda genome is significantly large, with approximately 48,000 base pairs, each restriction enzyme will cut the DNA several times and generate restriction fragments of different sizes. In this kit, three separate samples of lambda DNA have been precut using the three different restriction enzymes, and one sample remains undigested. Each sample produces DNA fragments whose size can be estimated when run on an agarose gel using electrophoresis.
Lambda Phage Genome
This diagram represents bacteriophage lambda genomic DNA, showing the locations of important gene clusters (Ausubel et al. 1998). Arrows mark the sites where the restriction enzyme Hin dIII cuts the DNA, and the numbers indicate the number of base pairs in each fragment.
bacterial chromosome
Integration into
Early gene expression
DNA synthesis
Head
Tail host
23,130 2,027 2,322
48,502 base pairs
9,416 564 125 6557
Bacteriophage lambda consists primarily of a head, which contains the genomic DNA, and a tail that is involved in phage attachment to bacterial cells.
4,361
5
50

Electrophoretic Analysis of Restriction Fragments
The three-dimensional structure or shape of a restriction enzyme allows it to fit perfectly in the groove formed by the two strands of a DNA molecule. When attached to the DNA, the enzyme slides along the double helix until it recognizes a specific sequence of base pairs which signals the enzyme to stop sliding. The enzyme then chemically separates, or cuts, the DNA molecule at that site — called a restriction site. In this way, a restriction enzyme acts like molecular scissors, making cuts at the specific sequence of base pairs that it recognizes.
If a specific restriction site occurs in more than one location on a DNA molecule, a restriction enzyme will make a cut at each of those sites, resulting in multiple fragments of DNA. Therefore, if a given piece of linear DNA is cut with a restriction enzyme whose specific recognition sequence is found at five different locations on the DNA molecule, the result will be six fragments of different lengths.
The length of each fragment will depend upon the location of restriction sites on the DNA molecule.
A DNA fragment that has been cut with restriction enzymes can be separated using a process known as agarose gel electrophoresis . The term electrophoresis means to carry with electricity . Agarose gel electrophoresis separates DNA fragments by size. DNA fragments are loaded into an agarose gel slab, which is placed into a chamber filled with a conductive buffer solution. A direct current is passed between wire electrodes at each end of the chamber. Since DNA fragments are negatively charged, they will be drawn toward the positive pole
(anode) when placed in an electric field. The matrix of the agarose gel acts as a molecular sieve, or a matrix of holes, through which smaller DNA fragments can move more easily than larger ones. Therefore, the rate at which a DNA fragment migrates through the gel is inversely proportional to its size in base pairs. Over a period of time smaller DNA fragments will travel farther than larger ones.
Fragments of the same size stay together and migrate in single bands of DNA.
These bands will be seen in the gel after the DNA is stained.
An analogous situation is one where all the desks and chairs in the classroom have been randomly pushed together. An individual student can wind his/her way through the maze quickly and with little difficulty, whereas a string of four students would require more time and have difficulty working their way through the maze.
Visualizing DNA Restriction Fragments
DNA is colorless so DNA fragments in the gel cannot be seen during electrophoresis. A blue loading dye, containing two blue dyes, is added to the DNA solution. The loading dye does not stain the DNA but make it easier to load the gels and monitor the progress of the DNA electrophoresis. The dye fronts migrate toward the positive end of the gel, just like the DNA fragments. The “faster” dye co-migrates with DNA fragments of approximately 500 bp, while the “slower” dye co-migrates with DNA fragments approximately 5 kb in size. Staining the DNA pinpoints its location on the gel. When the gel is immersed in Fast Blast DNA stain
(diluted to 1x for overnight staining or 100x for quick staining), the stain molecules attach to the DNA molecules trapped in the agarose gel. When the bands are visible, your students can compare the DNA restriction patterns of the different samples of DNA.
51

The DNA pattern that will be obtained by your students following electrophoresis of DNA samples that have been digested using three different restriction digestion enzymes is shown in Figure 1. By convention, the lanes are numbered from the top left. Notice that each restriction enzyme produces a unique banding pattern in each lane. The relative size of fragments contained in each band can be determined by measuring how far each band has traveled from its origin. Since the fragment sizes are known for the Hind III digest, this sample will function as a DNA standard or marker.
1 2 3 4
Fig. 1. Electrophoresis of lambda DNA digested using three different restriction enzymes. Lane 1 contains uncut lambda DNA. Lane 2 contains lambda DNA digested by Pst I. Lane 3 contains lambda
DNA digested by Eco RI. Lane 4 contains lambda DNA digested by Hind III.
7
52

Student Workstations . Materials and supplies that should be present at each student workstation prior to beginning each laboratory experiment are listed below.
The components provided in this kit are sufficient for eight student workstations (4 students per workstation).
Teacher’s (Common) Workstation . A list of materials, supplies, and equipment that should be present at a common location, which can be accessed by all student groups, is also listed below. It is up to the discretion of the teacher as to whether students should access common buffer solutions and equipment, or whether the teacher should aliquot solutions and operate equipment.
( ✔ ) Quantity per Station
LESSON 1
Student Workstations
Agarose gel electrophoresis system
(Electrophoresis chamber, gel tray,
8-well comb)
Laboratory tape (not Scotch tape)
Permanent marker
1
1 roll
1
❏
❏
❏
❏
Teacher’s Workstation
Molten 1% agarose in 1x TAE 40–50 ml/gel ❏
LESSON 2
Student Workstation
Electrophoresis power supply
Micropipet, 2–20 µl
Pipet tips, 2–200 µl
Empty micro test tubes (4 colors)
Foam micro test tube holder
Permanent marker
Gel support film (if applicable)
Fast Blast DNA stain (1x or 100x)
1
1
20
1
1
1
1 sheet ❏
120 ml per 2 stations ❏
Gel staining tray 1 per 2 stations
Large containers for destaining (if applicable) 1–3 per 2 stations
❏
❏
❏
❏
❏
❏
❏
❏
Teacher’s workstation
Four stock DNA samples
Uncut lambda DNA (L)
Hind III lambda digest (H)
Eco RI lambda digest (E)
Pst I lambda digest (P)
Sample loading dye
Electrophoresis buffer (1x TAE)
LESSON 3
Student Workstation
Gel support film (if applicable)
Millimeter ruler
Semilog graph paper
Teacher’s workstation
None required
1 vial
1 vial
1 vial
1 vial
1 vial
275 ml/gel
1 sheet
1
1
❏
❏
❏
❏
❏
❏
❏
❏
❏
53

Lesson 1 Sample Preparation
1. Obtain one of each colored micro test tube for each team and label each as follows: yellow, L = lambda DNA violet, P = Pst I lambda digest green, E = Eco RI lamgda digest orange, H = Hind III lambda digest
2. Using a fresh tip for each sample, pipet
10 µ l of DNA sample from each stock tube and transfer to the corresponding colored micro test tube.
3. Add 2 µ l of sample loading dye to each tube. Mix the contents by flicking the tube with your finger.
4.
Optional : Heat the DNA samples at
65°C for 5 minutes.
5. Pulse-spin the tubes in the centrifuge to bring all of the liquid to the bottom or tap them gently on the benchtop.
6. You have two options:
Option one: Put the DNA samples into the refrigerator and run the agarose gel during the next class.
Option two: Run the agarose gel the same day. Proceed directly to step 3 below.
Lesson 2 Agarose Gel Electrophoresis
1. Remove the DNA samples from the refrigerator (if applicable).
L P E
Centrifuge
H
2. Pulse-spin the tubes in the centrifuge to bring all of the liquid to the bottom or tap them gently on the benchtop.
3. Remove the agarose gel from the refrigerator (if applicable), remove the plastic wrap, and place the gel in the electrophoresis chamber. Fill the electrophoresis chamber and cover the gel with approximately 275 ml of
1 x buffer.
4. Check that the wells of the agarose gels are near the black (–) electrode and the bottom edge of the gel is near the red
(+) electrode.
–
Centrifuge
+
54
Tap
Tap

5. Load 10 µl of each sample into separate wells in the gel chamber in the following order:
Lane 1: L (yellow tube)
Lane 2: P (violet tube)
Lane 3: E (green tube)
Lane 4: H (orange tube)
6. Place the lid on the electrophoresis chamber carefully. Connect the electrical leads into the power supply, red to red and black to black.
7. Turn on the power and run the gel at 100 V for 30 minutes.
Visualization of DNA Fragments
1. When the electrophoresis run is complete, turn off the power and remove the top of the chamber. Carefully remove the gel and tray from the gel box. Be careful — the gel is very slippery. Slide the gel into the staining tray.
2. You have two options for staining your gel:
Option one : Quick staining (requires
12–15 minutes) a. Add 120 ml of 100x Fast Blast stain into a staining tray (2 gels per tray).
b. Stain the gels for 2 minutes with gentle agitation. Save the used stain for future use.
c. Transfer the gels into a large washing container and rinse with warm
(40–55°) tap water for approximately
10 seconds.
d. Destain by washing twice in warm tap water for 5 minutes each with gentle shaking for best results.
e. Record results.
f. Trim away any unloaded lanes.
g. Air-dry the gel on gel support film and tape the dried gel into your laboratory notebook.
Option two : Overnight staining a. Add 120 ml of 1x Fast Blast DNA stain to the staining tray (2 gels per tray).
b. Let the gels stain overnight, with gentle shaking for best results. No destaining is required.
c. Pour off the water into a waste beaker.
d. Record results.
e. Trim away any unloaded lanes.
f. Air-dry the gel on gel support film and tape the dried gel into your laboratory notebook.
55
–
+

Consideration 2. How Can Fragments of DNA Be Separated From One Another?
Agarose gel electrophoresis is a procedure used to separate DNA fragments based on their sizes. DNA is a molecule that contains many negative electrical charges. Scientists have used this fact to design a method that can be used to separate pieces of DNA. A solution containing a mixture of DNA fragments of variable sizes is placed into a small well formed in an agarose gel that has a texture similar to gelatin. An electric current causes the negatively-charged DNA molecules to move towards the positive electrode.
Imagine the gel as a strainer with tiny pores that allow small particles to move through it very quickly. The larger the size of the particles, however, the slower they are strained through the gel. After a period of exposure to the electrical current, the DNA fragments will sort themselves out by size. Fragments that are the same size will tend to move together through the gel and form bands.
A piece of DNA is cut into four fragments as shown in the diagram.
A solution containing the four fragments is placed in a well in an agarose gel. Using the information given above, draw (to the right) how you think the fragments might be separated. Label each fragment with its corresponding letter.
• Have your teacher check your diagram before you proceed.
• Where would the larger fragments, those with the greater number of base pairs, be located, toward the top of the gel or the bottom? Why?
• Suppose you had 500 pieces of each of the four fragments, how would the gel appear?
• If it were possible to weigh each of the fragments, which one would be the heaviest? Why?
• Complete this rule for the movement of DNA fragments through an agarose gel.
The larger the DNA fragment, the …
56

This diagram represents the bacteriophage lambda genomic DNA, showing the locations of important gene clusters (Ausubel et al. 1998). Arrows mark the sites where the restriction enzyme Hin dIII cuts the DNA, and the numbers indicate the number of base pairs in each fragment.
bacterial chromosome
Integration into
Early gene expression
DNA synthesis
Head
Tail host
23,130 2,027 2,322
48,502 base pairs
9,416
➜ ➜
564 125 6557
Bacteriophage lambda consists primarily of a head, which contains the genomic DNA, and a tail that is involved in phage attachment to bacterial cells.
4,361
• How many fragments were produced by the restriction enzyme Hin dIII?
On the gel diagram at the right, show how you believe these fragments will sort out during electrophoresis.
• Label each fragment with its correct number of base pairs.
57

Restriction Analysis Standard Curve
100,000
60,000
40,000
20,000
10,000
6,000
4,000
2,000
1,000
600
400
200
100
10 15 20
Distance traveled, mm
25 30 35
In this experiment, band 2 of Pst I migrated 23.5 mm (A). From the 23.5 mm mark on the X-axis, read up to the standard line; when you intersect your standard curve, mark the spot with a shaded circle (B). Follow the intersect point over to the Y-axis and determine where the graph line meets the Y-axis to read the approximate size of the fragment (C). Band 2 of Pst I is approximately 4,500 bp.
Repeat this procedure for all unknown fragments in the linear range.
5
58

(Numbers in parentheses are from the data from the gel comparison of bands.)
• When this data table has been completed, describe what you have done to determine DNA fragment sizes in this investigation. Use no more than two sentences.
The first determination of size involved the approximation of unknown
DNA band size by comparison to the migration of known DNA samples directly on the agarose gel. The second determination more accurately determined unknown DNA size by plotting a standard curve from known
DNA bands, and then using the curve to determine the sizes of unknown samples.
• Explain how you think you could make your DNA size estimation more accurate.
Drawing two standard curves, rather than one, would make the size estimation more accurate. One curve could be drawn for the larger data points (bands 1, 2, and 3) and a second curve could be drawn for the smaller points (4, 5, and 6). Estimation of unknown fragment sizes could then be made from the most appropriate curve.
• Compare the two methods — direct gel examination and semilog graph — of determining the fragment size in base pairs. Which method seems to be more accurate? Explain your answer.
Both methods have advantages and disadvantages. With the gel examination method, it is possible to estimate sizes over the entire range of the gel, in particular for extremely large fragments. Because large fragments are outside of the linear range of the standard curve, you cannot accurately estimate sizes from the curves, but you can estimate the sizes from the gel.
Use of the semilog graph standard curve is very accurate within the linear range. The logarithmic cycles on the graph paper allow you to accurately estimate sizes of fragments, such as band 3 of the Eco RI lane, which migrated between standard band points. It is harder to estimate these intermediate sizes directly on the gel.
59

Diffusion & osmosis: Teacher’s GuiDe
Kit # 3674-04
This lab addresses the properties of osmosis and diffusion and their function in maintaining homeostasis in the cell. Students use two phospholipid bilayer models to simulate the movement of water and nutrients across a cell membrane and observe osmosis in living tissue. In Part 1, students calculate the surface area-to-volume ratios of differently-sized cuboidal cell models.
In Part 2, the movement of molecules across a membrane is simulated using dialysis tubing and solutions of varying composition. In Part 3, students directly observe osmosis in a living specimen.
In all parts of this lab, after performing a guided activity, students are then directed to design their own experiments, to further develop their understanding of the topics explored. The students’ understanding of these exercises will allow them to explain how cell size and shape affect rates of diffusion, as well as pose scientific questions about the selective permeability properties of cell membranes.
©2012, Ward’s Natural Science
All Rights Reserved, Printed in the U.S.A.
US: www.wardsci.com
60
250-7454 v.1/12
Page 1

Diffusion & osmosis: Teacher’s GuiDe
Kit # 3674-04
‹ The student is able to use calculated surface area-to-volume ratios to predict which cell(s) might eliminate wastes or procure nutrients faster by diffusion (2A3 & SP 2.2).
‹ The student is able to explain how cell size and shape affect the overall rate of nutrient intake and the rate of waste elimination (2A3 & SP 2.2).
‹ The student is able to use representations and models to pose scientific questions about the properties of cell membranes and selective permeability based on molecular structure (2B1 &
SP 4.2, SP 4.3, SP 4.4).
Part 1: Diffusion and Osmosis
Part 2: Modeling Osmosis
Part 3: Osmosis in Living Plant Cells
Analyzing Results and Class Discussion
Structured Inquiry: 5 minutes
Guided Inquiry: 45 minutes
Open Inquiry: Will vary, depending on students’ experimental designs
Structured Inquiry: 45 minutes
Guided Inquiry: 45 minutes
Open Inquiry: Will vary, depending on students’ experimental designs
Structured Inquiry: 45 minutes
Guided Inquiry: 45 minutes
Open Inquiry: Will vary, depending on students’ experimental designs
45 minutes
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Diffusion & osmosis: Teacher’s GuiDe
Kit # 3674-04 objeCtIves
‹
‹
‹
Use calculated surface area-tovolume ratios to predict which cell(s) might eliminate wastes or procure nutrients faster by diffusion.
Explain how cell size and shape affect the overall rate of nutrient intake and the rate of waste elimination.
Use representations and models to pose scientific questions about the properties of cell membranes and selective permeability based on molecular structure.
Figure 1:
Plant cell in hypertonic solution.
Net movement of H O
The cell is plasmolyzed.
Why are cells so small? Most cells grow, but upon reaching a certain size, a cell will divide becoming two smaller cells. This is how multicellular organisms, like humans, grow. But why do cells stop growing once they reach a certain size? Why does a cell divide and multiply rather than simply growing bigger? One possible answer can be found in the relationship between cell size and the diffusion of substances across the cell membrane.
The absorption of nutrients, excretion of cellular wastes, and the exchange of respiratory gases are life processes which depend upon efficient transport of substances into, out of, and throughout living cells. Diffusion is one of the most common and efficient means by which substances are passively transported between cells and their aqueous environment. Diffusion is the movement of a substance
(liquid or gas) along a concentration gradient from high to low concentration. Diffusion is vital to many life functions of a cell.
Diffusion allows the transport of vital nutrients and compounds without the expenditure of energy.
The cell membrane is the selectively permeable barrier whose total surface area is important to regulating the substances that diffuse into or out of the cell. Small, neutrally charged molecules such as oxygen, carbon dioxide, and glucose can pass freely through the membrane, while the diffusion of other materials is restricted. Materials that cannot diffuse across the membrane or need to be transported against a diffusion gradient can be actively transported across the membrane with the expenditure of energy. Osmosis is a special kind of diffusion that occurs as water is separated by a selectively permeable membrane with different solute concentrations on either side of the membrane. During osmosis, water moves from regions of low solute concentration to regions of high solute concentration without the expenditure of energy.
Organisms rarely exist in environments with solute concentrations that match their cytoplasm; there are usually more or fewer dissolved particles in one of two compared solutions separated by a membrane, such as a cell and the media in which it exists. A hypertonic solution is a solution in which the solute concentration is higher outside of the cell; therefore, water will flow to the external environment, causing the cell to shrink. Hypotonic solutions consist of a low concentration of solutes outside the cell; therefore, water will flow into the cell, causing cellular expansion. In the case of plant cells with cell walls , expansion is restricted, so pressure builds. This pressure is called turgor pressure.
(continued on next page)
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Diffusion & osmosis: Teacher’s GuiDe
Figure 2
Plant cell in hypotonic solution.
Net movement of H O
The cell is turgid.
Figure 3
Plant cell in isotonic solution.
Net movement of H O
Net movement of H O
Figure 4
Water Potential in a Tube.
Selective
Permeable
Membrane
Kit # 3674-04
(continued)
Isotonic solutions, on the other hand, are solutions in which the solute and solvent concentrations are at equilibrium: there is no net flow of materials across the selectively permeable membrane.
Only a solute’s relative concentration, or water potential ( y ), affects the rate of osmosis. Water potential consists of two components
– pressure potential ( y p s
), the relative concentration of solutes within the two solutions.
) the exertion of pressure on a solution; and solute (or osmotic) potential ( y
Water Potential ( y ) = Pressure Potential ( y p
) + Solute Potential ( y s
)
Water moves from an area of high water potential (free energy) to an area of lower water potential. For example, in Figure 4 the water initially enters the tube because there is a negative solute potential in the sugar–water solution. However, the force of gravity begins to exert pressure on the rising column; when the force of gravity, pressure potential, equals the solute potential, the sugar-water solution in the column stops rising. The water potential is at zero and dynamic equilibrium has been established. The pressure potential can be determined from the height of the column. With the water potential and the pressure known, solute potential can be experimentally determined.
The solute potential can be calculated as yy s
= – i CRT, where: i = ionization constant (number of ions Na + Cl – = 2, sucrose = 1)
C = Molar concentration
R = pressure constant (0.0831 Liter bars/mole, ° Kelvin)
T = Temperature in Kelvin (273 + temp C) bar = measure of pressure
1 bar = 1 atmosphere at sea level
Water
Sugar
Molecule
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Diffusion & osmosis: Teacher’s GuiDe notes
©2012, Ward’s Natural Science
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Kit # 3674-04
q q q q q q q q q
Celery stick soaked in water\
Celery stick soaked in saltwater
3 Phenolphthalein agar cubes: 3 x 3 cm, 2 x 2 cm, and 1 x 1 cm
1 Plastic knife
1 Plastic spoon
1 Plastic cup
1 Vinylite white plastic ruler, 6” metric system
1 White vinegar, 100 mL
1 Timer
1. Observe the celery stick that was soaked in water. Record your observations.
2. Break the celery stick that was soaked in water. Record your observations.
3. Observe the celery stick that was soaked in saltwater. Record your observations.
4. Break the celery stick that was soaked in saltwater. Record your observations.
‹ The agar cubes have been prepared with 1% phenolphthalein, which is a pH indicator. The chart below indicates a color scale of pH for phenolphthalein. The blocks are pink because the agar blocks were soaked in 0.01 % sodium hydroxide.
Phenolphthalein Color Indicator
Color
Colorless
Pink to Red pH
0 - 8.2
8.2 – 12.0
Acid or Base
Acidic or slightly neutral
Basic
1. Obtain agar cubes in a plastic cup from your teacher.
Be careful not to scratch any surface of the cubes.
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Diffusion & osmosis: Teacher’s GuiDe
Kit # 3674-04 forMUlAs
All formulas for calculations are listed below:
‹ Surface Area =
Length x width x # of sides
‹ Volume =
Length x width x height
‹ Surface Area Volume Ratio =
Surface Area
Volume
‹ Extent of Diffusion =
Total Cube Volume –
Volume of cube that has not changed color x 100
Total Cube Volume
©2012, Ward’s Natural Science
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(continued)
2. Using the metric ruler, measure the dimensions of each agar cube and record the measurements in your lab notebook.
3. Place the three cubes carefully into a plastic cup. Add white vinegar (acidic solution) until the cubes are submerged. Using a plastic spoon, keep the cubes submerged for 10 minutes turning them frequently.
‹ Be careful not to scratch any surface of the cubes.
‹ Be sure to start the timer once the cubes are submerged.
4. As the cubes soak, calculate the surface area, volume, and surface area to volume ratio for each agar cube. Record this data in a table similar to the one below.
Block
#
Start 1
Start 2
Start 3
End 1
End 2
End 3
Length
(cm)
Width
(cm)
Height
(cm)
Surface area (cm 2 )
Volume
(cm 3 or mL)
5. After 10 minutes, use the spoon to remove the agar cubes and carefully blot them on dry paper towel. For more accurate measures of diffusion, use a knife to cut the cubes.
6. Using a metric ruler, measure the distance in centimeters (cm) that the white vinegar diffused into each cube. (Distance from surface)
7. Calculate the rate of diffusion for each cube in centimeters per minute (cm/min.).
8. Calculate the volume of the portion of each cube which has not changed color (in other words, the portion of the cube that is still pink).
9. Calculate the extent of diffusion into each cube as a percent of the total volume.
10. Graph the rate of diffusion relative to cell volume and surface area.
11. Graph the extent of diffusion relative to cell volume and surface area.
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Diffusion & osmosis: Teacher’s GuiDe
Kit # 3674-04 eXPerIMent desIgn tIPs
The College Board encourages peer review of student investigations through both formal and informal presentation with feedback and discussion. Assessment questions similar to those on the AP exam might resemble the following ques tions, which also might arise in peer review:
‹
‹
‹
‹
Explain the purpose of a procedural step.
Identify the independent variables and the dependent variables in an experiment.
What results would you expect to see in the control group? The experimental group?
How does XXXX concept account for YYYY findings?
• Describe a method to determine XXXX.
What questions occurred to you as you investigated diffusion in agar blocks and the flexibility of the celery sticks? Design an experiment to investigate one of your questions. Questions may involve examining diffusion in different shapes of agar blocks, the effect of temperature on rates of diffusion, the amount of time it takes to make crisp celery limp, the effect of salt concentration on celery limpness, or the effect of other solutes on celery limpness.
Before starting your experiment, have your teacher check over your experiment design and initial your design for approval. Once your design is approved, investigate your hypothesis. Be sure to record all observations and data in your laboratory sheet or notebook.
Use the following steps when designing your experiment.
1. Define the question or testable hypothesis.
2. Describe the background information. Include previous experiments.
3. Describe the experimental design with controls, variables, and observations.
4. Describe the possible results and how they would be interpreted.
5. List the materials and methods to be used.
6. Note potential safety issues.
After the plan is approved by your teacher:
7. The step by step procedure should be documented in the lab notebook. This includes recording the calculations of concentrations, etc. as well as the actual weights and volumes used.
8. The results should be recorded (including drawings, photos, data print outs).
9. The analysis of results should be recorded.
10. Draw conclusions based on how the results compared to the predictions.
11. Limitations of the conclusions should be discussed, including thoughts about improving the experimental design, statistical significance and uncontrolled variables.
12. Further study direction should be considered.
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Diffusion & osmosis: Teacher’s GuiDe
Kit # 3674-04
ProCedUre tIPs
‹ Record all data IMMEDIATELY in your laboratory notebook.
‹ Wash your hands before handling the dialysis tubing, and keep physical contact with the tubing to a minimum.
‹ Remember to label your model cells. Record the pairs in your laboratory notebook in their respective order of your lab set-up.
‹ If you choose to tie off the end of the dialysis tubing with string, tie two knots, about 1/4” apart, to prevent leaking.
In this lab activity, you will construct and simulate model cells in an external environment, to relate solutes passing through a semipermeable membrane in hypertonic, hypotonic, and isotonic solutions.
q q q q q q q q q q
1
1
1
5
Roll of String
Balance
Graduated Cylinder
Disposable beaker, 1000 mL
7 ft. Piece of Dialysis Tubing, 20 cm
250 mL 1M Sucrose Solution
250 mL 1M Sodium Chloride (Salt)
250 mL 1M Glucose Solution
250 mL 5% Albumin Solution (Protein)
500 mL Distilled OR Tap Water
The pores in dialysis tubing allow some molecules to freely diffuse across the membrane and some to be restricted. In this lab, you will use dialysis tubing as a model cell membrane.
1. Obtain five pieces of pre-soaked dialysis tubing from the beaker of water. Tie a tight knot in one end of each piece of tubing, or use a piece of string to tie off the end.
2. Measure and pour 10 mL of each of the four prepared solutions into a separate graduated cylinder. The solutions are salt, glucose, sucrose, and protein.
3. Open the tubing by rubbing the untied end between your fingers.
Pour 10 mL of prepared solution into the tubing. Carefully tie a knot in the open end to form a closed cell membrane (similar to a bag). Be sure to leave enough space in the bag for expansion.
Minimize air enclosed in the tubing.
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Diffusion & osmosis: Teacher’s GuiDe
Kit # 3674-04
Cell 1 (protein)
Cell 2 (sucrose)
Cell 3 (water)
Cell 4 (glucose)
Cell 5 (salt) notes
(continued)
4. Fill a beaker about 100 mL of the solutions to be paired with your model cell. Use either water or salt. (See sample data table below for parings.)
Paired
Extra-Cellular
Solution (in cup)
Salt
Water
Water
Salt
Water
Start time
0
Cell Weight (g)
After 30 minutes
% Change
Final Mass - Initial Mass
Initial Mass x 100
5. Repeat Steps 3 and 4 for the remaining four cells.
‹ Remember to clean the graduated cylinder between solutions.
6. Determine the initial weight of each cell and record in a table similar to the one shown below.
7. Completely immerse the model cells in their pairing solutions in the beaker or cup. Start your timer.
8. Given what you know about solute concentration, predict whether each “cell” volume will grow, shrink, or remain constant. Record your predictions in your laboratory notebook.
9. Allow the “cells” to soak for 30 minutes. Record any observations in your laboratory notebook.
10. When 30 minutes has passed, remove the model cells from the solution, pat dry, and determine the final weight of each of the model cells. Record the final weights and any additional observations.
11. Calculate the percent change in weight and record your results in your laboratory notebook.
‹ Do not discard any of your solutions from this part of the lab activity as they will also be used in Parts 2B, 2C, and Part 3 of this investigation.
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CELL PROCESSES: CELLULAR RESPIRATION – TEAChER’S GUIdE
Kit #36-7406
Living organisms must metabolize compounds derived from food to produce energy for maintenance, growth and reproduction. Cellular respiration is a process that produces energy by metabolizing glucose in the presence of oxygen (O
2
). In this lab, students will measure the rate of oxygen consumption related to cellular respiration. This will be achieved through the construction and utilization of a microrespirometer. Students will compare the results obtained using germinating seeds versus a non-germinating control, acrylic beads. Students will then design their own experiments to investigate the effects of various factors on the rate of cellular respiration.
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CELL PROCESSES: CELLULAR RESPIRATION – TEAChER’S GUIdE
Kit #36-7406
ProCedUre tIPs
‹
‹
When performing this lab activity, all data should be recorded in a lab notebook. You will need to construct your own data tables, where appropriate, in order to accurately capture the data from the investigation.
If directed to do so by your teacher, this part of the lab may be done at the same time as
Part 2 of the lab.
1. Fill a 100 mL graduated cylinder with 50 mL water. Add 10 germinating peas and take a reading of the displaced water. This is the volume of the germinating peas. Record the volume in the space below, or in your lab notebook. Decant the water, remove the peas and place them on a paper towel; pat the peas dry and set aside.
Volume of germinating peas for vial 1 _____________
2. Refill the graduated cylinder with 50 mL water. Add beads until the water level is the same as that of the germinating peas. Be sure to get the water level as close as possible to that of the germinating peas. If you go over, pour out the contents of the graduated cylinder and start again. Record the volume in the space below. Decant the water, remove the beads and place them on a paper towel; pat the beads dry and set aside.
Volume of beads for vial 2 ______________
3. Obtain two vials with steel washers on the bottoms (to prevent floating). Number the vials 1 and 2 with a glass marking pen.
Place an absorbent cotton ball in each of the vials and push each down to the bottom using a pipet or pencil tip. Be sure to use the absorbent cotton balls and NOT the non-absorbent rayon.
Potassium hydroxide is corrosive.
Without getting liquid on the sides of the respirometers, use a pipet to add 1 mL 15% potassium hydroxide (KOH) to the cotton.
4. Add a piece of non-absorbent rayon that is slightly smaller than that of the cotton ball and place it on top of the KOH-soaked cotton. Do not tamp down this layer.
5. Add the germinating peas to vial 1 and the acrylic beads to vial 2.
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CELL PROCESSES: CELLULAR RESPIRATION – TEAChER’S GUIdE
Kit #36-7406
ProCedUre tIPs
‹ The petroleum jelly is used to create a seal around the pipet where it enters the stopper if necessary (see note on page
16). It should not be necessary to use the petroleum jelly as a lubricant for inserting the pipet into the stopper.
Figure 2
(ContInUed)
6. Insert the non-tapered end of the pyrex graduated pipet into the wide end of a stopper so that the tapered end of the pipet points away from the stopper and so that the pipet extends just beyond the bottom of the stopper (see Figure 2).
7. Firmly insert the stopper into the vial. The seal that has been created between the stopper and the vial should be sufficient to prevent the pipet from easily moving up and down in the stopper. Place a washer over the pipet tip and guide it down the pipet until it rests on the stopper. Repeat this entire step for the other vial. The respirometers should look like those shown in
Figure 3 below.
8. Place a thermometer and vials 1 and 2 in the 20 °C waterbath with the pipet tips resting on the edge of the tray as shown in
Figure 4. Allow the respirometers to equilibrate for 10 minutes.
(continued on next page)
Figure 3
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Germinating peas Beads
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CELL PROCESSES: CELLULAR RESPIRATION – TEAChER’S GUIdE notes
Kit #36-7406
(ContInUed)
9. Add one drop of food coloring to the exposed tip of each respirometer and wait one minute. Turn each of the respirometers so that the graduation marks on the pipets are facing up. Carefully shift the two respirometers until the seed container is completely immersed in the waterbath. Do not touch the respirometers once the experiment has started! Let the respirometers equilibrate for another 5 minutes before proceeding.
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Figure 4
(continued on next page)
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CELL PROCESSES: CELLULAR RESPIRATION – TEAChER’S GUIdE
Kit #36-7406 note
It is normal for a small amount of water to enter the pipets when they are first immersed and for a small amount of food coloring to enter the water. However, if a pipet begins to fill with water, that respirometer has a leak that should be repaired immediately in the following manner:
‹
‹
Remove the vial from the water and remove the stopper assembly.
Blot the end of the pipet on a paper towel to remove all liquid.
‹
‹
Reassemble the respirometer in the same manner as in Steps
9 and 10 of this procedure. Be sure to firmly insert the stopper to prevent leaks. Petroleum jelly can be used to seal the outside of where the pipet enters the stopper if it is loose.
Submerge the vial portion of the respirometer and add one drop of food coloring to the tip. Carefully submerge the tip of the respirometer in the same manner as previously mentioned.
note
The corrected difference is being used because this procedure is very sensitive and may be influenced by factors such as an increase in ambient temperature or varying barometric pressure from passing weather.
(ContInUed)
10. Read all of the respirometers to the nearest 0.01 mL and take the temperature of the waterbath to ensure temperature stability.
Copy Table 1 into your laboratory notebook or sheet. Record the initial readings of volume (mL) and the temperature of the waterbath (°C) in Table 1.
gerMInAtIng PeAs ACrylIC beAds
Temp Time Reading Diff.
—
Corr.
Diff.
—
Reading Diff.
— 0
5
10
15
20
25
30
11. Take additional readings every 5 minutes for 30 minutes, and record the readings and temperature in Table 1.
12. When all of the readings have been taken, calculate the difference and the corrected difference for each result and record each value in Table 1.
Difference = (initial reading at time 0) – (reading at time X)
Corrected difference = (initial pea reading at time – pea seed reading at time X) – (initial bead reading at time 0 – bead reading a time X)
Graph your results from the corrected difference column in
Table 1 for the germinating peas and beads. Plot the time in minutes.
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CELL PROCESSES: CELLULAR RESPIRATION – TEAChER’S GUIdE notes
Kit #36-7406
Adapt the experiment in Part 1 to test the effect of altering an abiotic condition on the rate of respiration. This part of the experiment may be run in parallel with Part 1.
Suggestions for conditions to alter include: temperature, light, and amount oxygen (starting volume of air in respirometer). Alternative seed types are also included (black-eyed peas and kidney beans that will not be germinating if not soaked days ahead of time).
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CELL PROCESSES: CELLULAR RESPIRATION – TEAChER’S GUIdE
Kit #36-7406 exPerIMent desIgn tIPs
The College Board encourages peer review of student investigations through both formal and informal presentation with feedback and discussion. Assessment questions similar to those on the AP exam might resemble the following questions, which also might arise in peer review:
‹
‹
‹
‹
‹
Explain the purpose of a procedural step.
Identify the independent variables and the dependent variables in an experiment.
What results would you expect to see in the control group?
The experimental group?
How does XXXX concept account for YYYY findings?
Describe a method to determine
XXXX.
What questions occurred to you as you completed the structured and guided inquiry? Now that you are familiar with the use of a respirometer, design an experiment to investigate one of your questions.
Possible questions could involve: Do rates of respiration differ in different seed types or sizes? Do rates of respiration differ in different organism types (like insects)? Can the respirometer be adapted to measure respiration in an aquatic organism?
Before starting your experiment, plan your investigation in your lab notebook. Have your teacher check over and initial your experiment design. Once your design is approved, investigate your hypothesis.
Be sure to record all observations and data in your laboratory sheet or notebook.
Use the following steps when designing your experiment.
1. Define the question or testable hypothesis.
2. Describe the background information. Include previous experiments.
3. Describe the experimental design with controls, variables, and observations.
4. Describe the possible results and how they would be interpreted.
5. List the materials and methods to be used.
6. Note potential safety issues.
After the plan is approved by your teacher:
7. The step-by-step procedure should be documented in the lab notebook. This includes recording the calculations of concentrations, etc. as well as the actual weights and volumes used.
8. The results should be recorded (including drawings, photos, data print outs).
9. The analysis of results should be recorded.
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All Rights Reserved, Printed in the U.S.A.
(continued on next page)
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CELL PROCESSES: CELLULAR RESPIRATION – TEAChER’S GUIdE notes
Kit #36-7406
(ContInUed)
10. Draw conclusions based on how the results compared to the predictions.
11. Limitations of the conclusions should be discussed, including thoughts about improving the experimental design, statistical significance and uncontrolled variables.
12. Further study direction should be considered.
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Ecology: BEhavior – TEachEr’s guidE
Kit # 36-7412
Organisms orient to stimuli that are important to their survival. Movement toward or away from important stimuli (taxis) depends upon both the sensory and motor abilities of the organism. This lab explores the chemotactic behaviors that fruit flies and/or pill bugs exhibit when exposed to the controlled environment of a choice chamber. Students identify patterns in the behaviors and make inferences based on the composition of the tested materials and the organisms’ responses. Students then determine what materials and experimental paradigms will be tested further.
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Ecology: BEhavior – TEachEr’s guidE
Kit # 36-7412
1
1 pkg.
1
1
1
8
1 pkg./300
1 pkg./100
1
Units per kit
1
MAterIAls ProvIded In kIt
Description pH Paper, 1-14 range, Vial/100
Filter Paper, Medium Grade,
Ward’s Dual Magnifier, 3X & 6X
Disposable Petri Dish, Pkg/20
Vinegar, 473 mL, White
Animal Behavior Trays
Cotton Balls
Pipets
Instructions (this booklet)
Redemption coupon for pill bugs and Drosophila *
Order organisms to be delivered a week in advance of lab
MAterIAls needed bUt not ProvIded
Clear plastic bottles
(e.g., soda bottles) with caps
Household substances, condiments, foods with heavy odors
Clear plastic packing tape
Water
Masking tape
Funnel
Morgue (beaker filled with salad oil or alcohol)
Light
Alka Seltzer tablets oPtIonAl MAterIAls ( not ProvIded)
Fine paintbrushes
Cold packs or crushed ice
Aluminum foil
Other materials as determined by students’ experimental design
* - It is recommended that you redeem your coupon for live/ perishable materials as soon as possible and specify your preferred delivery date. Generally, for timely delivery, at least a week’s advance notice is preferred.
Call “Us” at
1.800.962.2660 for
Technical Assistance or
Visit “Us” on-line at www.wardsci.com
for U.S. Customers www.wardsci.ca
for Canadian Customers
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Ecology: BEhavior – TEachEr’s guidE
Pre-Lab Prep:
Redeem Live Materials Coupon
Part 1: Structured Inquiry –
Chemotaxis
Part 2: Guided Inquiry –
Test Variables of
Student’s Choice
Part 3: Open Inquiry
At least 1 week prior to lab
Total of 30 minutes*:
10 minutes set up
10 minutes observation and recording
10 minutes analysis
*Optional: Teacher may decide to break this into two observation periods – 30 minutes for control (no stimulus in tray) and
30 minutes for experimental)
Total of 20 minutes:
5 minutes set up
10 minutes observation and recording
5 minutes analysis
Total depends on student/teacher scheduling and parameters of experiment
Kit # 36-7412
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Ecology: BEhavior – TEachEr’s guidE
Kit # 36-7412 objeCtIves
‹
‹
‹
‹
‹
‹
‹
Refine scientific models and questions about the effect of complex biotic and abiotic interactions on all biological systems from cells and organisms to populations, communities, and ecosystems.
Design a plan for collecting data to show that all biological systems (cells, organisms, populations, communities, and ecosystems) are affected by complex biotic and abiotic interactions.
Analyze data to identify possible patterns and relationships between a biotic or an abiotic factor and a biological system (cells, organisms, populations, communities, or ecosystems).
Analyze data to support the claim that response to information and communication of information affect natural selection.
Justify claims, using evidence, to describe how timing and coordination of behavioral events in organisms are regulated by several mechanisms.
Connect concepts in and across domain(s) to predict how environmental factors affect response to information and change behavior.
Apply mathematical routines to quantities that describe interactions among living systems and their environment that result in the movement of matter and energy.
(continued on next page)
Organisms orient to stimuli that are important to their survival.
Movement toward or away from important stimuli depends upon both the sensory and motor abilities of the organism. A stimulus might involve anything that can be sensed, like light, sound, touch, heat, or chemicals. For example, humans do not sense a magnetic field and cannot orient towards it (without instruments like a compass).
Therefore, we may infer that magnetic fields have not been very important for human survival as a species. Once an organism with sensory/motor abilities perceives a stimulus, it can orient or move either towards or away from that stimulus depending upon the nature of the stimulus (opportunity or threat). Movement in response to a stimulus is classified as taxis, whereas random movement or movement irrespective of stimulus is classified as kinesis. Generally, the more critical a stimulus is to an organism’s survival, the stronger the response to that stimulus. Therefore, an organism that senses an optimal food source will usually orient strongly toward it. In an animal that senses primarily through smell, movement towards an appropriate food sourcewould be called positive chemotaxis. In the same vein, orienting with reference to light is called phototaxis, and orienting in response to gravity is called geotaxis, etc.
Behavior can be classified as innate or learned. Innate behavior is inherited and instinctive, and develops independently of the experience of an organism in its environment over time. On the other hand, learned behaviors are not inherited and can be changed as a result of the animal’s experience with its environment and other organisms.
In this laboratory, you will be investigating and observing the taxis and kinesis of model organisms- either fruit flies or pill bugs. As you make behavioral observations, think about how those behaviors contribute to making the species an evolutionary success in its natural environment.
Since differential reproduction is a strong driver of evolution, you may want to take special note of any taxis related to reproductive behaviors.
Drosophila melanogaster represents a model organism with well analyzed genetics and many mutant strains available (see www.fruitfly.
org or www.flybase.or
g as well as the care sheet in this booklet).
Further, this organism has several distinct developmental stages that can be investigated separately (larval, pupae, and adult). Larval stages are advantageous to study since the organisms are slow moving and do not fly, however, they may be too slow to make relevant behavioral investigations in the supplied behavioral tray (students may want to
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Ecology: BEhavior – TEachEr’s guidE
Kit # 36-7412 objeCtIves
(ContInUed)
‹
‹
‹
Use visual representations to analyze situations or solve problems qualitatively to illustrate how interactions among living systems and with their environments result in the movement of matter and energy.
Predict the effects of a change of matter or energy availability on communities.
Use data analysis to refine observations and measurements regarding the effect of population interactions on patterns of species distribution and development.
(ContInUed) consider making modifications). Pupae do not display behavior, and adults are very fast moving, and since they fly it is challenging to get the specimens into and out of the supplied behavioral tray. Cooling the vial of flies for 10-20 minutes in a refrigerator or on ice or a cool pack will slow the adults down enough to be handled more easily.
Fine paintbrushes are an excellent tool for handling adults that have been slowed. Adult male Drosophila are distinguishable from females in that they are smaller than females, and they have dark sexcombs on their first (most anterior) pair of legs. Drosophila will eat many different fruits and vegetables.
Isopods (pill bugs or sow bugs) do not have a bank of mutants available for behavioral experiments but they do provide the advantage of being easy to handle in the context of these behavioral experiments.
They also represent organisms with a different evolutionary history, and they are adapted to environments that overlap with fruit fly environments. Isopods, however, are distinct and have evolved different developmental patterns as well as different sensory and motor responses to stimuli. To determine the sex of an isopod use a stereomicroscope or magnifying glass to observe the underside of the specimen near the posterior end. Males have two white, elongated appendages that serve as copulatory organs. These are modified
“pleopods” and are absent in the females. Pill bugs and sowbugs eat decaying plant material.
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Ecology: BEhavior – TEachEr’s guidE
Kit # 36-7412
ProCedUre tIPs
‹
‹
When performing this lab, all data should be recorded in a lab notebook. You will need to construct your own data tables, where appropriate, in order to accurately capture the data from the investigation.
Record all data
ImmEDIAtEly in your laboratory notebook.
Your teacher will assign you an organism, either fruit flies or pill bugs, to observe and investigate its behaviors.
1. To become familiar with the organisms, sketch an illustration of your organism in your laboratory notebook. Label all anatomical structures that you recognize. Can you differentiate male from female?
2. Place one piece of masking tape on the outside of each chamber of the behavior tray; label them A, B, C, D, and E (central chamber).
3. Place a drop or two of vinegar on a cotton ball or on a small section of filter paper and place this in chamber A (not central chamber). Plain water on a similar paper in the opposite chamber can provide one control condition.
4. Place several (2 or fewer in each chamber for a total of 10. organisms in the behavior tray, cover the tray with clear cover, and carefully observe the specimens for at least 10 minutes.
(NOTE: If you are using fruit flies, place them in the refrigerator for five minutes before they are needed. They will be easier to deposit into the behavior trays as the cooler temperature slows them down. Use funnel if necessary.) Document any behaviors you see in a list. Remember to record even seemingly unimportant behaviors.
5. Every minute for 10 minutes, count the number of organisms in each chamber and record observations in Table 1 (on next page).
‹ Do not disturb the pill bugs or fruit flies; shaking or tipping the tray will introduce additional stimuli.
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Ecology: BEhavior – TEachEr’s guidE
Kit # 36-7412
(ContInUed) table 1: Organism taxis
Time (min)
#
Organisms in Chamber
A vinegar
#
Organisms in Chamber
B
2 2
#
Organisms in Chamber
C
Water
(opposite
A)
2
#
Organisms in Chamber
D
#
Organisms in Chamber
E central
2 2
7
8
5
6
2
3
0
1
4
9
10
Ave.
Class avg.
Average
6. Graph your results in the space below.
Class
Average
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Ecology: BEhavior – TEachEr’s guidE notes
Kit # 36-7412
(ContInUed)
7. Calculate the average number of organism in each chamber in the 10-minute period of time. Add this to Table 1.
8. Using the data from every group in the class, calculate the class average for number of organisms in each chamber in a
10-minute time period. Enter this data in Table 1. Calculate standard error and deviation. Draw a graph to discuss whether the results are significantly different between empty chambers and vinegar chambers. You may want to use data points only after 5 minutes for comparisons to allow for taxis to occur. If two types of organisms were used in class – compare whether there were significant differences between organisms.
At end of lab:
• Fruit flies and pill bugs are living organisms that should not be released to the environment. After all the investigations are complete, organisms should be tapped into a “morgue” through a funnel. The morgue typically is a 150-mL beaker that contains about 50 mL of salad oil or 70% alcohol.
• All laboratory bench tops should be wiped down with a 20% bleach solution or disinfectant to ensure cleanliness.
• Wash your hands thoroughly with soap and water before leaving the laboratory.
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Ecology: BEhavior – TEachEr’s guidE
Kit # 36-7412
ProCedUre tIPs
‹ When performing this lab, all data should be recorded in a lab notebook. You will need to construct your own data tables, where appropriate, in order to accurately capture the data from the investigation.
1. Identify a stimulus you would like to test for chemotaxis on your organism. Household items or foods that have a noticeable odor, and/or volatile chemicals associated with decay or fermentation might be good choices.
2. Put this stimulus in the chamber opposite the vinegar stimulus.
3. Introduce all 10 organisms into the central chamber and cover.
4. Observe every minute for 10 minutes and record observations and analyze as described in Part 1.
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Page 20

EnzymE Activity: tEAchEr’s GuidE
Kit # 36-7413
Chemical reactions underlie metabolism. Organisms have evolved catalytic proteins, called enzymes, that can make the reactions more efficient by lowering the activation energy of a chemical reaction.
Catalytic efficiency of enzymes is dependent upon the precise shape of the active site in the protein that interacts with substrates and products. The abiotic conditions of the reaction affect the rate of enzyme-mediated conversion of substrate to product by affecting the conformation of this active site.
In this set of investigations, students will use an extract of turnips containing the enzyme peroxidase to react with its substrate (hydrogen peroxide) and a color indicator (guaiacol) for formation of product (O2), to estimate the rate of peroxidase activity under a variety of experimental conditions.
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Page 1

EnzymE Activity: tEAchEr’s GuidE
Kit # 36-7413
ProCedUre tIPs
‹
‹
When performing this lab activity, all data should be recorded in a lab notebook. You will need to construct your own data tables, where appropriate, in order to accurately capture the data from the investigation.
If directed to do so by your teacher, this part of the lab may be done at the same time as Part
2 of the lab.
q q q q q q q q q q
2
1
1
5
1 mL syringes
2.5 mL syringe
10 mL syringe
15 x 125 mm tubes
1 Disposable 9” transfer pipet
1 mL Guaiacol
1 mL Turnip peroxidase solution
2 mL Dilute hydrogen peroxide
4 mL pH 7 buffer solution
Timer
Shared Materials q 1 box of Parafilm
1. Label the test tubes and syringes , as follows:
2.5 mL syringe labeled ‘E’ for enzyme – turnip peroxidase solution.
2.5 mL syringe labeled ‘P’ for product as represented by indicator . Guaiacol reacts with free O
2 color.
(product) to form brown
10 mL syringe labeled ‘NB’ for neutral buffer – pH 7.
2.5 mL syringe labeled ‘S’ for substrate – 0.1% H
2
O
2
Test tube labeled ‘SPNB’ for mixture A
Test tube labeled ‘ENB’ for mixture B
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EnzymE Activity: tEAchEr’s GuidE notes
©2012, Ward’s Natural Science
All Rights Reserved, Printed in the U.S.A.
Kit # 36-7413
(ContInUed)
2. Fill and prepare the labeled syringes and tubes with the appropriate solutions (provided).
a. Dispense the following reagents in “ Tube SPNB ”:
2 mL of ‘S’ substrate
1 mL of ‘P’ indicator for product
1 mL of ‘NB’ neutral buffer b. Cap tube ‘SPNB’ with parafilm and gently invert two times to mix.
c. Dispense the following reagents in “ Tube ENB ”:
1 mL ‘E’ enzyme (turnip peroxidase)
3 mL‘NB’ neutral buffer d. Cap tube ‘ENB’ with parafilm and gently invert two times to mix.
e. Using a disposable transfer pipet, transfer the mixture from
Tube ‘SPNB’ into Tube ‘ENB’.
f. Cap with parafilm and invert two times to mix.
3. Make observations and record data.
a. Using the color palette provided by your instructor, immediately observe and compare the color of your reaction to the color palette and record the tube # (color) of the mixture over time. This data should be recorded in 1-minute intervals for a total incubation period of five minutes.
‹ OPTIONAL: If the class has access to a probe to measure the transmittance of light or to a spectrophotometer, more quantitative results with better resolving power can be generated by measuring the changes in the transmittance or absorbance of light.
b. Plot the increase in color intensity (product formation) relative to your color palette over the 5-minute interval and calculate the rate of enzymatic reaction under the baseline conditions of this experiment.
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EnzymE Activity: tEAchEr’s GuidE notes
Kit # 36-7413
(ContInUed)
3. Make observations and record data (continued)
‹ Note: The color palette represents a range of indicator color that increases 10% between tubes 1-10. Tube 0 represents no indicator, no color.
‹ NOTE: The color intensity is used as way to quantify the amount of oxygen that is produced in the enzymatic reaction. The brown color is produced when the guaiacol reacts with oxygen (product of the enzyme substrate reaction). Therefore, the more intense the color, the more oxygen is produced in the reaction.
4. Set your labeled syringes aside for use in the following parts of this investigation.
‹ NOTE: Do not throw away your labeled syringes.
You will need them for the remaining parts of this investigation.
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EnzymE Activity: tEAchEr’s GuidE
Kit # 36-7413
ProCedUre tIPs
‹ When performing this lab activity, all data should be recorded in a lab notebook. You will need to construct your own data tables, where appropriate, in order to accurately capture the data from the investigation.
‹ If directed to do so by your teacher, this part of the lab may be done at the same time as Part
1 of the lab.
q q q q q q
12
3
15 x 125 mm tubes
2.5 mL syringes (from Part 1)
1 10 mL syringe (from Part 1)
6 mL ‘P’ product guaiacol
Jana, please note...
add p, S, etc.
6 mL ‘E’ enzyme turnip peroxidase solution
12 mL ‘S’ substrate dilute hydrogen peroxide (0.1% H
2
O
2
) q q q q q q q
Shared Materials
1 ‘NB’ 10 mL Syringe to dispense all buffer solutions
500 mL pH 3 Buffer Solution
500 mL pH 5 Buffer Solution
500 mL pH 6 Buffer Solution
500 mL pH 7 Buffer Solution
500 mL pH 8 Buffer Solution
500 mL pH 10 Buffer Solution
1. In your laboratory notebook or sheet, record the baseline rate
(from Part 1???) .
2. Label twelve 13 x 100 mm tubes 1 through 12, respectively.
To simplify the process, pair the tubes according to the chart below:
‹
Tube
Tube/pH
1
3
2
5
4
6
9
7
11
8
12
10
NOTE: All of the pH tubes will contain the respective pH buffer solution, in case the tubes get mixed up.
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EnzymE Activity: tEAchEr’s GuidE notes
Kit # 36-7413
(ContInUed)
3. In each of tubes 1, 2, 4, 9, 11, and 12, dispense:
2 mL ‘S’ substrate dilute hydrogen peroxide using ‘S’ syringe
1 mL‘P’ product indicator guaiacol using ‘P’ syringe
1 mL ‘NB’ neutral buffer pH 7
‹ NOTE: All of these tubes contain the substrate.
4. In tube 3 dispense the following volumes of reagents:
1 mL of turnip peroxidase solution, using the syringe labeled ‘E’
3 mL of pH 3 solution, using the rinsed ‘NB’ syringe
5. Repeat step 4 for tubes 5, 6, 7, 8, and 10 with their respective pH buffer s .
6. Using a disposable 9” transfer pipet, combine the reagents from tube 1 with the reagents in tube/pH 3.
7. Observe the enzyme reaction mixtures every minute for
5 minutes by comparing to the color palette (or optional measurement of absorbance/transmittance). Record your observations in your laboratory notebook.
8. Refer back to the tube pairing chart and mix the remaining pairs of tubes. Repeat Steps 6 and 7 for the remaining pairs of tubes.
9. Calculate the rate of reaction for each tube as described in Part
1.
In your laboratory notebook, graph your rate results relative to pH.
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Why Teach the Polymerase
Chain Reaction (PCR)?
•   Hands-on biotechnology lab
•   Real-world applications to forensics (unlike CSI), medicine, research.
•   Reinforces Laboratories 7 & 8 and human evolution
What can you do with the Chromosome
16 PV92 PCR kit ?
5/20/09
•   Besides the kit itself, you will need:
–   Microcentrifuge
–   Water bath (56C)
–   Micropipets
–   Salt (pure NaCl, Kosher or sea salt)
–   Thermocycler
–   Students
PCR Procedures
Add Master Mix containing:
• Nucleotides
• Primers
• Reaction buffer
• Electrophoresis dyes
• Taq polymerase
Day 1
Day 2
Day 3
Mg ++
Mg ++
Mg ++
Mg ++
Mg ++
Mg ++
1