Fatty Acids, Nutrition and Health Exercise

Fatty Acids, Nutrition and Health Exercise. Scenario

 

Fatty Acids, Nutrition and Health

 

Students please note : It is assumed that you have an understanding of organic molecules as presented in your text book. If not, you should read chapter 3 in your text book dealing with organic molecules. At a minimum, you need to look over the sections associated with lipids. At the end of the scenario, there are some links listed to web sites that you might find useful.

 

 

Its time for your annual physical examination…

 

Suppose that you were visiting your doctor for an annual physical examination. While you were in the waiting room, you picked up and began to read a pamphlet about dietary fats (lipids) and their impacts on health. In order to remember them later, you typed some of the important points into a file on your laptop. To refresh your memory, and for your convenience (and the rest of the class), these notes can be seen below.

 

Notes on Lipids and Health

 

True lipids (triglycerides)

· composed of a glycerol molecule covalently bound to three fatty acid side chains

· hydrophobic molecules, meaning they are non-polar and don’t mix with water

· a group of triglycerides is called a fat

 

Saturated fats

· solid at room temperature (think of fat on an uncooked steak)

· hydrocarbon chains in fatty acids have no carbon-carbon double covalent bonds

· maximum number of hydrogens are covalently bound to the carbons, thereby “saturating” them

· ( Saturated Fatty Acid )animal fats such as lard and butter are usually high in saturated fatty acids

 

 

 

 

Unsaturated fats

· liquid (oils) at room temperature

· hydrocarbon chains of the fatty acids have at least one (monounsaturated) or more (polyunsaturated) carbon-carbon double covalent bonds

· double bonds cause bends in the molecules and also leave them “unsaturated” with hydrogens

· unsaturated fats are found in plant oils such as olive or canola oil

( Monounsaturated Fatty Acid )

 

Dietary fatty acids

· fatty acids are a necessary component of a complete diet

· fatty acids are found in foods such as fatty meats, plant oils and dairy products

· certain polyunsaturated fatty acids (called essential fatty acids) cannot be synthesized by the human body and must come from the diet

· one such essential fatty acid is linoleic acid which can be found in foods such as sunflower oil and almonds

· both monounsaturated and polyunsaturated fatty acids can be found in plant oils

 

( Polyunsaturated Fatty Acid )

 

 

Trans fats

· are triglycerides that contain trans fatty acids

· Trans means “across” so…

· a trans fatty acid is an unsaturated one in which the hydrogens attached to adjacent carbons in a carbon-carbon double covalent bond are on opposite sides of the molecule

· produced by the process of adding hydrogens to unsaturated vegetable oils

· hydrogenation decreases the number of carbon-carbon double covalent bonds in the molecules and creates what are known as “hydrogenated” or “partially hydrogenated” vegetable oils

· trans configuration gives the fatty acids chemical properties more similar to saturated fatty acids (such as lack of bends in the molecules) and can also lead to some accumulation of “bad” types of cholesterol when consumed

· trans fats are found in any hydrogenated oils, so margarines and shortening made from vegetable oils are major sources

· since they are made from plant oils, these were once touted as being much healthier than saturated animal fats

 

 

· in cis fatty acids, the hydrogens attached to adjacent carbons in a carbon-carbon double covalent bond are on the same side of the molecule.

· Almost all naturally occurring unsaturated fatty acids are cis isomers

 

 

 

 

So, the doctor says your cholesterol is too high…

 

When you were finally called back to a room, your doctor said that a standard analysis of your blood showed that your total cholesterol level was 250mg/dL (a dL=deciliter, or 1/10 of a liter, 100mL). He/she explained that a total cholesterol level of 240mg/dL or above is considered high and is a risk factor for coronary artery disease (CAD).

 

In continuing conversations with your physician, you learned that as a general rule cholesterol is not always “bad.” In fact, cholesterol is a necessary component of cell membranes and is a vital part of normal metabolic processes, including formation of other steroids. In addition to looking at total cholesterol concentration in your blood, your doctor emphasized that it is important to take notice of high density lipoprotein (HDL) cholesterol and low density lipoprotein (LDL) cholesterol, which make up major fractions of the total cholesterol and may be more accurate indicators of CAD risk.

 

Your LDL fraction was found to be 195mg/dL which is considered very high.

 

After seeing your very high LDL number, your physician went on to say that the role of LDLs is to carry cholesterol around in the bloodstream and deposit it where it is needed. Unfortunately, if the level of LDLs becomes too high they can also deposit the cholesterol in arteries, forming “plaques” and clogging the arteries. On the other hand, HDLs are responsible for carrying excess cholesterol away from the arteries to the liver where it can be metabolized or “reprocessed” for other purposes. So, with this in mind, cholesterol bound in LDLs is often called “bad cholesterol” and cholesterol bound in HDLs is called “good cholesterol”. In order to avoid CAD, the average person should attempt to raise the level of HDLs in their blood and lower the level of LDLs, although both are needed at some level in the blood.

 

At this point, your doctor strongly urged you to exercise regularly and watch your diet because consumption of fats can have a profound effect on cholesterol levels in the body. However, it turns out that it is probably the types of fat, not the total amount of fat that you eat that may determine incidence of CAD. In fact, a 2001 study by Frank Hu and colleagues at the Harvard School of Public Health showed that different parts of the world with similar total fat intake had very different amounts of death from CAD. Individuals in countries where the fats eaten were mainly in the form of saturated fats and trans fats had much higher incidences of death from CAD than individuals in countries where the fats were mainly consumed in the form of polyunsaturated fats containing omega-3 fatty acids. In general, the study showed that consumption of saturated fats and trans fats tended to increase LDL cholesterol while consumption of polyunsaturated fats tended to both decrease LDL cholesterol and increase HDL cholesterol.

 

References

Hu, FB, Manson, JE, and Willett, WC. 2001. Types of Dietary Fat and Risk of Coronary Heart Disease: A Critical Review. Journal of the American College of Nutrition 20:5-19.

 

Useful Links:

More explanation about trans fats from the FDA

http://www.fda.gov/Food/LabelingNutrition/ConsumerInformation/ucm109832.htm#unhide

 

Understanding cholesterol numbers website

http://www.webmd.com/cholesterol-management/guide/understanding-numbers

 

Government information about reading nutrition labels

http://www.fda.gov/Food/LabelingNutrition/ConsumerInformation/ucm078889.htm

 

 

Pre-Activity Assignment: Fatty Acids, Nutrition and Health ________________________ Name

Read the Fatty Acids, Nutrition and Health reading and Chapter 3 in the text book and then answer the following questions before class.

 

1. A trans fatty acid is one

a. that has no carbon-carbon double bonds directly adjacent to each other.

b. that is a major component of phospholipids in cell membranes.

c. in which the hydrogens attached to adjacent carbons in a carbon-carbon double covalent bond are on opposite sides of the molecule.

d. in which the hydrogens attached to adjacent carbons in a carbon-carbon double covalent bond are on the same side of the molecule.

e. that is saturated with hydrogens.

 

2. Types of polyunsaturated fatty acids that are necessary in the human diet because they cannot be synthesized by the body are called _____________ fatty acids.

a. essential

b. important

c. trans

d. omega-3

e. hydrophobic

 

3. Generations of Americans were introduced to trans fats in their diet in the form of ____________ which was hailed as a healthy alternative to the saturated fats found in butter and lard.

a. Coconut oil

b. Olive oil

c. Margarine

d. Canola oil

e. Beef tallow

 

4. HDL stands for

a. Highly dense lipid.

b. Hydrogenated dark lipid.

c. High density lipid.

d. Hydrogenated dense lipoprotein.

e. High density lipoprotein.

 

5. In the next class meeting you will work in small collaborative groups to answer four question sets. Each person in the group will act as the ‘facilitator’ for one question set, leading the group discussion, promoting input from each of the other students (who will be acting as ‘discussants’) and formalizing the group response. In the role of a discussant, students provide their knowledge, experience and perspectives, compare and contrast the inputs of other members of the group and collaborate in the formulation of the group response. At the end of the activity, you may be called on to present your group’s answers to one of the question sets (not necessarily the one you were the facilitator for). You will act as both a facilitator and a discussant in the activity.

Make sure you have critically analyzed the Fatty Acids, Nutrition and Health scenario and chapter 3 in the text book so that you are prepared to participate in all aspects of the group activity. Bring the scenario with you to the next class meeting.

 

Activity—Fatty Acids, Nutrition and Health (Critical Question Sets)

1. Plaque Formation in Arteries: High levels of LDL cholesterol in the bloodstream can lead to formation of “plaques” in the arteries.

 

_________________________________

Facilitator

 

 

· What are the potential health consequences of coronary arteries being clogged by “plaques”?

 

 

 

 

 

· Name some specific foods that might lead to increased LDL cholesterol in the bloodstream.

 

 

 

 

 

· What do these foods contain that could cause this?

 

 

 

 

2. Nutrition Labels: Analyze the nutrition label below and answer the following questions.

 

______ ___________________________

Facilitator

 

· How much total fat, saturated fat and trans fat does this product contain per serving?

 

 

 

 

· How many grams of total fat would a person consume if he/she ate the whole container?

 

 

· How many calories are from fat per serving?

 

 

 

· Calculate the percentage of the total calories per serving that come from total fat?

 

 

 

 

· Based on a 2,000 calorie diet, what percent of the USDA’s percent daily value for total fat would be consumed per serving? Calculate the percent if the whole container was consumed?

 

 

 

· Based on what you know about trans fats, do you think there are hydrogenated oils in this product? Explain.

 

 

 

3. Hydrogenation of Oils: Answer the following questions about hydrogenation of vegetable oils.

_________ _____________ ___________

Facilitator

 

· What is the reason behind hydrogenating vegetable oils? What effect does the hydrogenation process have on their chemical and physical properties?

 

 

 

 

 

· Draw both a cis isomer and a trans isomer of a polyunsaturated fatty acid. Which would be produced by the hydrogenation process?

 

 

 

 

 

 

 

 

 

4. Cholesterol Profiles: Answer the following question using what you know about cholesterol and the tables to the right that show health categories based on blood levels of different fractions of cholesterol.

 

___________ _____________________

Facilitator

 

 

· Circle the blood cholesterol profile below that you think would be most desirable? What about this profile made you choose it over the others? Explain why you didn’t choose each of the other two.

 

 

 

 

(a) (b) (c)

 

 

 

Post-Activity Assignment: Fatty Acids, Nutrition and Health ________________________

Name

 

6. Which one of the following diagrams represents a trans fatty acid?

 

 

a.

 

 

 

 

 

 

 

 

b.

 

 

 

 

 

 

 

c.

 

 

 

d.

 

 

 

 

 

 

 

7. Which one of the following would be solid at room temperature?

a. Cis fatty acids

b. Corn oil

c. Peanut oil

d. Saturated fats such as lard (pig fat)

e. Unsaturated fats

 

 

 

8. A true lipid is composed of glycerol and three fatty acids. What type of reaction is used to link each of the fatty acids to a glycerol molecule?

a. Dehydration

b. Hydrolysis

c. Dehydrohalogenation

d. Hydrogenation

e. Hydroxylation

 

9. Food companies can tag their products on the nutrition label as having 0g of trans fats if they have <0.5g of trans fat per serving. What could be found in the ingredients list that is probably a better indicator of the presence of trans fats in foods than the trans fat line on the nutrition label?

a. Lard

b. Hydrogenated oils

c. Palm oil

d. Olive oil

e. Almonds

 

10. Consumption of which of the following is most likely to raise your HDL and also lower your LDL levels?

a. Trans fats

b. Saturated fats

c. Lard

d. Polyunsaturated fats

e. Margarine

 

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Fatty Acids, Nutrition and Health Exercise

 
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Lab 5: Meiosis INSTRUCTIONS: And 3425

Lab 5: Meiosis INSTRUCTIONS: And 3425. lab biology questions
BiolLab_5.docx
BiolLab_5.docx

Your Full Name:

102/103
Lab 5: Meiosis
INSTRUCTIONS:

and submit it via the Assignments Folder by the date listed in the Course
Schedule (under Syllabus).

To conduct your laboratory exercises, use the Laboratory Manual located under
Course Content. Read the introduction and the directions for each exercise/experiment
carefully before completing the exercises/experiments and answering the questions.

Save your Lab 5 Answer Sheet in the following format: LastName_Lab5 (e.g.,
Smith_Lab5).

You should submit your document as a Word (.doc or .docx) or Rich Text Format
(.rtf) file for best compatibility.

© eScience Labs, LLC 2014

Pre-Lab Questions
1. Compare and contrast mitosis and meiosis.

2. What major event occurs during interphase?

Experiment 1: Following Chromosomal DNA Movement
through Meiosis
Data Tables and Post-Lab Assessment
Trial 1 – Meiotic Division Beads Diagram:
Prophase I

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Metaphase I

Anaphase I

Telophase I

Prophase II

Metaphase II

Anaphase II

Telophase I

Cytokinesis

© eScience Labs, LLC 2014

Trial 2 – Meiotic Division Beads Diagram:
Prophase I

Metaphase I

Anaphase I

Telophase I

Prophase II

Metaphase II

Anaphase II

Telophase I

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Cytokinesis

Post-Lab Questions
1. What is the ploidy of the DNA at the end of meiosis I? What about at the end of meiosis

II?

2. How are meiosis I and meiosis II different?

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3. Why do you use non-sister chromatids to demonstrate crossing over?

4. What combinations of alleles could result from a crossover between BD and bd

chromosomes?

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5. How many chromosomes were present when meiosis I started?

6. How many nuclei are present at the end of meiosis II? How many chromosomes are in

each?

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7. Identify two ways that meiosis contributes to genetic recombination.

8. Why is it necessary to reduce the number of chromosomes in gametes, but not in other

cells?

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9. Blue whales have 44 chromosomes in every cell. Determine how many chromosomes

you would expect to find in the following:

a.i.

Sperm Cell:

a.ii.

Egg Cell:

a.iii.

Daughter Cell from Mitosis:

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a.iv.

Daughter Cell from Meiosis II:

10. Research and find a disease that is caused by chromosomal mutations. When does the

mutation occur? What chromosomes are affected? What are the consequences?

11. Diagram what would happen if sexual reproduction took place for four generations using

diploid (2n) cells.

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Experiment 2: The Importance of Cell Cycle Control
Data Tables and Post-Lab Assessment
1.

2.

3.

4.

5.

Post-Lab Questions
1. Record your hypothesis from Step 1 in the Procedure section here.

2. What do your results indicate about cell cycle control?

3. Suppose a person developed a mutation in a somatic cell which diminishes the

performance of the body’s natural cell cycle control proteins. This mutation resulted in
© eScience Labs, LLC 2014

cancer, but was effectively treated with a cocktail of cancer-fighting techniques. Is it possible
for this person’s future children to inherit this cancer-causing mutation? Be specific when
you explain why or why not.

4. Why do cells which lack cell cycle control exhibit karyotypes which look physically

different than cells with normal cell cycle.

5. What are HeLa cells? Why are HeLa cells appropriate for this experiment?

Experiment 1:
Following
Chromosomal DNA
Movement through
Meiosis
In this experiment, you
will model the movement
of the chromosomes
through meiosis I and II to
create gametes.

© eScience Labs, LLC 2014

Materials

2 Sets of Different Colored Pop-it® Beads (32 of e
may be any color)
8 5-Holed Pop-it® Beads (used as centromeres)
Procedure:
Part 1: Modeling Meiosis
without Crossing Over
As prophase I begins, the
replicated chromosomes
coil and condense…
1.

Build a pair of
replicated,
homologous
chromosomes. 10
beads should be
used to create each
individual sister
chromatid (20
beads per
chromosome pair).
Two five-holed
beads represent
each centromere.
To do this…

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Figure 3: Bead set-up. The blue beads
represent one pair of sister chromatids
and the black beads represent a second
pair of sister chromatids. The black and
blue pair are homologous.
a.

Start with
20 beads of
the same
color to
create your
first sister
chromatid
pair. Five
beads must
be snapped
together for
each of the
four
different
strands.
Two
strands
create the
first
chromatid,
and two
strands
create the
second
chromatid
with a 5holed bead
at the
center of
each
chromatid.
This creates
an “I”
© eScience Labs, LLC 2014

shape.
b.

Connect the
“I” shaped
sister
chromatids
by the 5holed beads
to create
an “X”
shape.

c.

Repeat this
process
using 20
new beads
(of a
different
color) to
create the
second
sister
chromatid
pair.

2.

Assemble a second
pair of replicated
sister chromatids;
this time using 12
beads, instead of
20, per pair (six
beads per each
complete sister
chromatid strand).

3.

Pair up the
homologous
chromosome pairs
created in Step 1
and 2. DO NOT
SIMULATE
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CROSSING
OVER IN THIS
TRIAL. You will
simulate crossing
over in Part 2.
4.

Configure the
chromosomes as
they would appear
in each of the
stages of meiotic
division (prophase
I and II, metaphase
I and II, anaphase I
and II, telophase I
and II, and
cytokinesis).

5.

Diagram the
corresponding
images for each
stage in the
sections titled
“Trial 1 – Meiotic
Division Beads
Diagram”. Be sure
to indicate the
number of
chromosomes
present in each cell
for each phase.

© eScience Labs, LLC 2014

Figure 4: Second set of replicated
chromosomes.
6.

Disassemble the
beads used in Part
1. You will need to
recycle these beads
for a second
meiosis trial in
Steps 8 – 13.

Part 1 – Meiotic Division
Beads Diagram
Prophase I
Metaphase I
Anaphase I
Telophase I
Prophase II
Metaphase II
Anaphase II

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Telophase II
Cytokinesis
Part 2: Modeling Meiosis
with Crossing Over
7.

Build a pair of
replicated,
homologous
chromosomes. 10
beads should be
used to create each
individual sister
chromatid (20
beads per
chromosome pair).
Two five-holed
beads represent
each centromere.
To do this…
a.

Start with
20 beads of
the same
color to
create your
first sister
chromatid
pair. Five
beads must
be snapped
together for
each of the
four
different
strands.
Two

© eScience Labs, LLC 2014

strands
create the
first
chromatid,
and two
strands
create the
second
chromatid
with a 5holed bead
at the
center of
each
chromatid.
This creates
an “I”
shape.
b.

Connect the
“I” shaped
sister
chromatids
by the 5holed beads
to create
an “X”
shape.

c.

Repeat this
process
using 20
new beads
(of a
different
color) to
create the
second
sister
chromatid
© eScience Labs, LLC 2014

pair.
8.

Assemble a second
pair of replicated
sister chromatids;
this time using 12
beads, instead of
20, per pair (six
beads per each
complete sister
chromatid strand).
Snap each of the
four pieces into a
new five-holed
bead to complete
the set up.

9.

Pair up the
homologous
chromosomes
created in Step 8
and 9.

10.

SIMULATE
CROSSING
OVER. To do this,
bring the two
homologous pairs
of sister
chromatids
together (creating
the chiasma) and
exchange an equal
number of beads
between the two.
This will result in
chromatids of the
same original
length, there will
now be new

© eScience Labs, LLC 2014

combinations of
chromatid colors.
11.

Configure the
chromosomes as
they would appear
in each of the
stages of meiotic
division (prophase
I and II, metaphase
I and II, anaphase I
and II, telophase I
and II, and
cytokinesis).

12.

Diagram the
corresponding
images for each
stage in the section
titled “Trial 2 Meiotic Division
Beads Diagram”.
Be sure to indicate
the number of
chromosomes
present in each cell
for each phase.
Also, indicate how
the crossing over
affected the genetic
content in the
gametes from Part1
versus Part 2.

Part 2 – Meiotic Division
Beads Diagram:
Prophase I
Metaphase I

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Anaphase I
Telophase I
Prophase II
Metaphase II
Anaphase II
Telophase II
Cytokinesi

Experiment 2: The Importance of Cell Cycle Control
Some environmental factors can cause genetic mutations which result in a
lack of proper cell cycle control (mitosis). When this happens, the possibility
for uncontrolled cell growth occurs. In some instances, uncontrolled growth
can lead to tumors, which are often associated with cancer, or other biological
diseases.
In this experiment, you will review some of the karyotypic differences which
can be observed when comparing normal, controlled cell growth and
abnormal, uncontrolled cell growth. A karyotype is an image of the complete
set of diploid chromosomes in a single cell.
Procedure
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Materials
*Computer Access
*Internet Access

1.

*You Must Provide

Begin by constructing a hypothesis to explain what differences you
might observe when comparing the karyotypes of human cells which
experience normal cell cycle control versus cancerous cells (which
experience abnormal, or a lack of, cell cycle control). Record your
hypothesis in Post-Lab Question 1.
Note: Be sure to include what you expect to observe, and why you think
you will observe these features. Think about what you know about
cancerous cell growth to help construct this information

2.

Go online to find some images of abnormal karyotypes, and normal
karyotypes. The best results will come from search terms such as
“abnormal karyotype”, “HeLa cells”, “normal karyotype”, “abnormal
chromosomes”, etc. Be sure to use dependable resources which have
been peer-reviewed

3.

Identify at least five abnormalities in the abnormal images. Then, list and
draw each image in the Data section at the end of this experiment. Do
these abnormalities agree with your original hypothesis?
Hint: It may be helpful to count the number of chromosomes, count the
number of pairs, compare the sizes of homologous chromosomes, look
for any missing or additional genetic markers/flags, etc.

Data
1.
2.
3.

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4.
5.
Click here to download and solve a few questions.

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Bio lab 5 answer.docx
Bio lab 5 answer.docx

Your Full Name:

102/103
Lab 5: Meiosis
INSTRUCTIONS:

and submit it via the Assignments Folder by the date listed intheCourse
Schedule (under Syllabus).

To conduct your laboratory exercises,…

Lab 5: Meiosis INSTRUCTIONS: And 3425

 
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Life Table Analysis For Black-Striped Wallabies

Life Table Analysis For Black-Striped Wallabies.

ECOL203/403 Assignment 1: Age Structure of a Population Using Life Tables Introduction to Life Tables

Before you begin this exercise (or read any further) you should:

1. Read Chapter 13 of Attiwill and Wilson (2006), particularly the section on life tables on page 220 – 223.

2. Make sure you have the life_table.xls file from the Assignment 1 folder)

3. Do the Molar Index and Skull aging Tutorial (Assignment 1 folder)

4. Download the Box of Skulls (Assignment 1 folder)

5. It is also advisable to read through this exercise completely before starting on the spreadsheet in excel.

Background to the Data

The Black-striped wallaby, Macropus dorsalis

The black-striped wallaby is a medium-sized macropod (females 7kg; males 16kg) that occurs from northern Queensland to northern NSW. The species is listed as ‘Endangered’ in NSW, but can become overabundant in some parts of Queensland – so wildlife ecologists need to manage their numbers in some regions so that they do not cause over-grazing of livestock pastures, while in other place, the population needs to be stimulated to increase in numbers to prevent them from becoming locally extinct. The wallabies shelter in dense scrub thickets (e.g. Brigalow) by day and graze adjacent pasture or natural grasslands by night. Debra White did her UNE Master of Natural Resource Science on black- striped wallabies at the Brigalow Research Station near Theodore in central Queensland (White 2004). She found that there was a high density of wallabies sheltering in the patches of brigalow by day, and that at night, these animals moved onto pasture, which they grazed heavily. White (2004) also looked at age structure of wallabies at the site by aging skulls she collected, and using these in a life table analysis. We will do a similar exercise in this assignment using skulls collected at the same site used by White (2004), and we will compare our results from the results from Debra White’s much larger dataset.

 

 

Molar Progression in Macropods

Molar progression occurs only in the marsupial genera Macropus, Petrogale and Peradorcas (Jackson 2003). These marsupials are among only a relatively few mammals worldwide whose teeth erupt at the posterior end of the jaw, and migrate forward along the jaw during life (the others are the elephants). As the teeth wear down and become less useful for grazing, they have moved sufficiently anterior in the jaw that they can fall out, ‘pushed’ from behind by newly erupted teeth. In this way, macropods can maintain good functioning teeth with high cusps for grazing on tough fibrous grasses throughout life. This ‘molar progression’ is a handy way to age kangaroo and wallaby skulls, and was used to generate the dataset you will use in this assignment to examine the life history parameters of a ‘population’ of black- striped wallabies (Macropus dorsalis) from the Brigalow Research Station in southern Queensland.

Aging of Macropods Using Molar Index (MI)

Molar Index (MI) is calculated by measuring the position of molariform teeth on the upper tooth row relative to a reference line drawn across the skull in line with the anterior limits of the orbits (see Figure 1 below). For convenience, ten stages of molar progression are recognized per tooth, and given decimal notation in tenths. In the example below (Fig 1a), molar M1 has progressed beyond the orbit, while molar M2 is given a score of about 0.7, according to Figure 1b. Therefore, the skull has a molar index (MI) of 1.7. Consulting Table 1 below of published estimates from red-necked wallabies (Macropus rufogriseus), a similar-sized macropod to the black- striped wallaby, we see that this animal was around age 2 (in years) when it died.

Figure 1. Skull showing reference line for age determination, and one-tenth division in length for molar teeth of kangaroos and wallabies. Figures from Kirkpatrick (1964), and Jackson (2003).

 

 

In this assignment, you will generate age data from the skulls of black-striped wallabies collected in the field, and use them to generate a life table for the population. You will measure these skulls and calculated their molar index, and then convert these into ages for each animal (using Table 1). You can then use those data to generate a life table for the population. Incidentally – the skulls come from animals that died ‘naturally’ – that is, they weren’t culled or harvested. Some might have died of old age; others may have been caught up on a barbwire fence, while a dingo perhaps ate others. In any event, from the age structure of the population, we can determine a profile for the population in terms of average life expectancy, mortality rates, probability of surviving to the next age bracket and so forth. Such tools are useful for the ecologists, because they provide rich demographic data that allows one to determine which individuals should be culled in order to contain or reduce a population that is overabundant, or conversely, whether certain cohorts (age groups) are particularly susceptible to some kind of mortality agent (such as predation) and need to be protected so that the population has the best chance or recovering to more sustainable numbers.

Black-striped wallaby painting by John Gould

Data for the life table analysis Download and open the excel spreadsheet exercise called ‘life_table.xls’ from the Assignment 1 folder on the unit website. Spend a minute looking around this spreadsheet. There are three ‘worksheets’ (clickable tabs at the bottom left of the opened workbook) within the spreadsheet: 1. Molar Index 2. Life Table 3. Data from White (2004)

MI Age (years) <0.8 0

0.9 – 1.6 1 1.7 – 2.1 2 2.2 – 2.5 3 2.6 – 2.7 4 2.8 – 2.9 5 3.0 – 3.1 6 3.2 – 3.3 7

3.4 8 3.5 9 3.6 10 3.7 11 3.8 12 3.9 13 4.0 14 4.1 15 4.2 16 4.3 17 4.4 18+

Table 1: Relationship between molar index (MI) and age (in years) of red- necked wallabies. Equation modified from Kirkpatrick (1965).

 

 

The first, Molar Index’, is where you will enter the age data of the population. The second, ‘Life Table’ is where you will calculate the life history parameters of the population. The third worksheet is data from Debra White’s thesis. White examined the demography of wallabies at the Brigalow Research Station in the exact way that we are doing – but because she worked with 667 skulls, where you only have 49 – we might expect her results to be a robust estimate against which you can compare your own. Graphs appearing on the ‘Life Table’ worksheet template already show White’s results for some key life table parameters (take a look at these now – Whites results are in light grey on Graphs A and B). As you complete this exercise, your results will be graphed in red, so you can compare your results to White’s results. STEP 1 – Click the tab to bring up the worksheet labelled ‘Molar Index’. Enter the molar index data you have gathered in the columns labelled ‘MI’ against each of the 49 skulls. While in this worksheet, use Table 1 (above) to determine the age at death for each wallaby, and put that number alongside the molar index value (i.e. Column C) labelled ‘Age’ (Given age as the nearest whole number in years). [HINT: use the sort function to sort Cells 2 to 50 in Column A and B (as a block, sorted by MI) before you begin assigning ages – this will make it much easier to generate ages, because wallabies will be arranged from youngest to oldest – don’t worry that the skull numbers will be jumbled – we won’t be using these in the analysis]. STEP 2 – Once all the MI and age data are entered alongside the corresponding skull ID, click on the worksheet ‘Life table’. You will see here that there is a template for a life table that has the following columns: Column A – Actual numbers of skulls at each age Column B – Age in years (x) Column C – Age interval (yrs) Column D – number surviving (nx) Column E – Proportion surviving (lx) Column F – Deaths at each age interval (dx) Column G – mortality rate (qx) Column H – Number surviving at agex at last birthday (bx) Column I – Expectation of further life (ex) in years If you had a look at this spreadsheet before you entered the age data in the ‘Molar Index’, you will now see that the sheet has changed – the ‘Actual Numbers of skulls at each age’ column (Column A) is now filled with the data you entered, neatly compiled by age class – you can click on a cell to see the underlying formula for doing this – but don’t change the formula! Definitions for columns requiring calculation of life table parameters are as follows: nx – The number of animals from the original cohort, that are still alive at each age interval

 

 

lx – Proportion of animals surviving from birth to age x. Because all individuals born are alive, l0 is proportional to the total number of animals sampled (i.e., is ‘1’), and successive values of lx get smaller, as fewer of those l0 animals live to older and older ages. dx – Deaths at each age interval (dx) is the number of individuals alive at age x that will die before age x+1. qx – the mortality rate (qx) is the per capita mortality rate during an age interval. bx – Number surviving at agex at last birthday – this is a prediction of the average number of individuals alive at the midpoint of age interval x, based upon number alive at one interval (x), and the next interval (x+1). ex – the expectation of further life (in years) for an individual that makes it to age x. This is calculated by summing all the values of bx from that age interval to the bottom of the table, divided by lx So, to recap, you should now have a spreadsheet that has the first three columns complete – is this correct? If so, is now time to calculate the parameters that we can use to describe the population. We will do these calculations, and compile the life table, using some simple Microsoft Excel equations. STEP 3 – Number Surviving (nx). Recall that nx is the number of wallabies surviving from birth to age x. So, we need to write a formula to go in Column D that describes survivorship at each age class. Starting at Cell D2, write the following formula:

=SUM(A5:A$23)

Then hit the return key. What does this formula do? Well, simply put, it sums all of the individuals in Column A, to generate a value that describes the number of individuals surviving to Age 0. Because all individuals survive to this age (i.e., if a wallaby was born, it must have survived to Age 0), this value should be equal to the total animals in our sample – i.e., 49. OK – next step is to copy this cell down to fill the column as far as Cell D23. How do we do this? Hold the mouse over the bottom right corner of the cell – the cursor should change to a square with arrows at top left and bottom right… click and hold the mouse button down while dragging the cursor to Cell D22, then release the mouse. What happened? Hopefully, you generated a series of numbers that describe survivorship at each age class. When you highlight Cell D6, you should see the following formula:

=SUM(A6:A$23) Note that the formula now sums from Cell A6 to A23 (rather than from A5), similarly, in Cell D7, the formula will sum from A7, and so on (the ‘$’ preceding

 

 

the ‘23’ keeps Cell A23 constant in each subsequent equation, as the formula changes). When you give this some thought, this makes sense – because we only want survivorship from any particular age class, to the last (oldest) age class. If you have completed this step correctly, you should see fewer and fewer surviving wallabies with increasing age, until at age 18, only 1 wallaby survives (and for our purposes, we will assume that this animal will die before reaching the next age class). STEP 4 – Proportion Surviving If this is to be a representation of the whole population, we need to start converting out numbers to something more universal – the previous column simply told us how many of the 49 animals survived, but by converting this to a proportion, we have a number at each interval that we could apply to a population of any size, to predict how it might behave. So, the next step involves a simple calculation to determine the proportion of wallabies in the original cohort that survive to each age class. Since Column D represents the number surviving to the next age class, Column E is simply the number surviving, divided by our entire sample (N = 49). So, in Cell E2, type the following formula:

=D5/49 – and hit return. Then, as before, click the mouse in the bottom right hand corner of Cell E5, hold the mouse button down, drag the cursor down to Cell E23, and then release the mouse. You should now have filled Column E, which is a calculation of the proportion of animals born into the population that survive to each age class. You will also see that “Figure A, the Proportion of M. dorsalis Surviving (lx)” also was created, so you can, for the first time, see graphically how you data compares to that of White’s (2004) data. STEP 5 – Deaths at each age interval (dx), and mortality rate (qx) Survivorship and mortality are clearly inter-related; mortality at any one age interval is simply the difference between the numbers surviving from that age interval to the next. So, the calculation of dx is simply nx – nx+1, so for our spreadsheet, type the following in Cell F5:

=D5-D6

Copy this formulae down to Cell F23, as previously described at STEP 4. STEP 6 – Mortality Rate Now we are in a position to calculate an important life history parameter for black-striped wallabies that is important in understanding any population – the age-specific mortality rate, or put another way, the rate at which animals in any particular age class would be expected to die before reaching the next age class. Incidentally, this is one of the calculations insurance companies use when determining what to charge you on a life insurance policy (see Table 13.6 on page 221 of Attiwill and Wilson).

 

 

Look at Column D of the life table in the spreadsheet. For age interval 0 – 1 we should have 49 animals, but at other age intervals (at least, after about age 3) we see less that this. What happened to those other animals? For whatever reason, they didn’t make it to that particular age class – they died before the next census period. So, the mortality rate at each age class is clearly equal to dx/nx, and in the spreadsheet, we can calculate this by typing the following equation in Cell G5:

=F5/D5 and then copying this down all the way to Cell G23 in the usual manner. You now have an estimate of the rate at which animals at different age classes in the population are dying – or put another way, “what is the likelihood of death for a wallaby at age x before the next census date”. And, as for proportion surviving in STEP 4, you now also have a graphical representation of this in “Figure B: Mortality rate (qx) for M. dorsalis”, and can compare your data for mortality directly with White’s (2004) data. STEP 7 – Number surviving at agex at last birthday (bx) and expectation of further life (ex). Simply put, bx is the average number of individuals alive at the midpoint of age interval x. So, to calculate this for each age class, you need to calculate the average of one age class (x) and the next age class (x+1). So, at Cell H5, enter the formula that describes this parameter:

=(D5+D6)/2 then, copy this down to H23. You now have a calculation of the ‘age structure’ of the population. You may be asking at this point, so what? Well, this is an intermediate calculation that will allow us to calculate a very useful parameter for the population – expectation of further life. For any age class, our calculations so far allow us to answer questions like: “What is the mortality or survival rate of animals in that age class?” But importantly, as an individual animal survives each age class, its expectation of further life should increase, because it managed to survive (through finding food, evading predation and disease etc) where others died. Life tables allow us to ask the question “For animals at a given age, how much longer should we expect those animals to live?” To calculate this, we sum the age structure of the remainder of the population older than any particular age class, and divide this value by the number of animals surviving at that age class. So, type the following formula into Cell I5:

=SUM(H5:H$23)/D5 And copy this down to Cell I23, as previously described. Have a look at the values you generated in this column. Starting at Age 0, we can see that all animals, at birth, have an expectation of further life of about 7.4 years (or something similar). But, if we look at Age 8, we see that animals that survived to 8 years of age have an expectation of further life of about 5

 

 

years. At age 11, these animals can be expected to live another 3.5 years, and so on. Once again, insurance companies use calculations of this nature on human census data to determine life insurance premiums. STEP 8 – Generating a Survivorship Curve The final step involves generating a survivorship curve for black-striped wallabies that we can compare to theoretical survivorship curves commonly used to summarise a population’s demography. These curves (e.g. Figure 13.12 on page 222 in Attiwill and Wilson) have a logarithmic scale on the y- axis, so to compare to our survivorship for wallabies to these theoretical curves, we must convert our data for nx to log values. To do this, type:

=LOG(D5) in Cell D26, and copy this down to Cell D44. Presto! – you have a survivorship curve for the population. References Attiwill, P. and Wilson, B. (2006). Ecology: An Australian Perspective. 2nd Edition. Oxford

University Press, Melbourne. Jackson, S. (2003). Australian Mammals: Biology and Captive Management. CSIRO

Publishing, Melbourne. Kirkpatrick, T.H. (1964). Molar progression and macropod age. Qld J. Ag. Anim. Sci. 21:

163–165. Kirkpatrick, T.H. (1965). Studies of Macropodidae in Queensland. 2. Age estimation in the

grey kangaroo, the eastern wallaroo, and the red-necked wallaby, with notes on dental abnormalities. Qld J. Ag. Anim. Sci. 22: 301–317.

White, D. (2004). Utilisation of remnant Brigalow communities and adjacent pasture by the

black-striped wallaby (Macropus dorsalis). Master of Resource Science, University of New England, Armidale NSW.

 

 

Assessment Questions As part of your assessment (the other part being the completed spreadsheet), you are required to write a paragraph on each of the following questions (use Microsoft Word for this), and submit your responses as part of your assignment. The total length of this section should be no more than 600 words.

1. How do the graphs you made compare to White’s data? Was your sample a reasonable representation of the black-striped wallaby population?

2. Given your results, how would you respond if someone asked the

question: “To what age does a black-striped wallaby live at the Brigalow Research Station?”

3. In reference to hypothetical survivorship curves (see pages 221-222 in

Attiwill and Wilson), what type of survivorship do black-striped wallabies most likely exhibit? Does this fit the typical curve for mammals?

4. Does the mortality rate fit the prediction for mammals? In very brief

terms, explain the pattern of mortality in the black-striped wallaby population.

 

TO COMPLETE THIS ASSIGNMENT, DON’T FORGET TO UPLOAD BOTH:

1. THE COMPLETED SPREADSHEET (results page

only, as a pdf file)

*AND*

2. THE ANSWERS TO ASSESSMENT QUESTIONS (as a pdf file)

Life Table Analysis For Black-Striped Wallabies

 
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UMUC Biology 102/103 Lab 6: Taxonomy

UMUC Biology 102/103 Lab 6: Taxonomy.

 

Full Name:

 

 

UMUC Biology 102/103

Lab 6: Taxonomy

INSTRUCTIONS:

 

·         On your own and without assistance, complete this Lab 6 Answer Sheet electronically and submit it via the Assignments Folder by the date listed in the Course Schedule (under Syllabus).

·         To conduct your laboratory exercises, use the Laboratory Manual located under Course Content. Read the introduction and the directions for each exercise/experiment carefully before completing the exercises/experiments and answering the questions.

·         Save your Lab 6 Answer Sheet in the following format:  LastName_Lab6 (e.g., Smith_Lab6).

·         You should submit your document as a Word (.doc or .docx) or Rich Text Format (.rtf) file for best compatibility.

 

Pre-Lab Questions

 

  1. Use the following classifications to determine which organism is least related out of the three. Explain your rationale.
Table 2: Classifications
  Classification Level American Green Tree Frog European Fire- Bellied Toad Eastern Newt
Domain Eukarya Eukarya Eukarya
Kingdom Animalia Animalia Animalia
Phylum Chordata Chordata Chordata
Class Amphibia Amphibia Amphibia
Order Anura Anura Caudata
Family Hylidae Bombinatoridae Salamandridae
Genus Tursipops Bombina Notophthalmus
Species cinerea bombina viridescens

 

 

  1. How has DNA sequencing affected the science of classifying organisms?

 

 

 

 

  1. You are on vacation and see an organism that you do not recognize. Discuss what possible steps you can take to classify it.

 

 

Exercise 1: Dichotomous Key Practice

 

Table 3: Dichotomous Key Results
Organism Binomial Name
i  
ii  
iii  
iv  
v  
vi  
vii  
viii  
ix  
x  
xi  
xii  
xiii  

 

 

Post-Lab Questions

1.    What do you notice about the options of each step as they go from number one up?

 

 

2.    How does your answer from Question 1 relate to the Linnaean classification system?

 

 

 

Exercise 2: Classification of Organisms

Data Tables and Post-Lab Assessment

Table 2: Key Characteristics of Some Organisms

Organism Kingdom  

Domain

Defined Nucleus Motile Cell Wall Photosynthesis Unicellular
E. coli         Yes   Yes
Protozoa     Yes Yes     Yes
Mushroom     Yes   Yes    
Sunflower     Yes   Yes Yes  
Bear     Yes Yes      

 

lab006fig004
Figure 4:Exercise 2 – Classification of Organisms Flow Chart

 

 

Post-Lab Questions

 

1.    Did this series of questions correctly organize each organism? Why or why not?

 

2.    What additional questions would you ask to further categorize the items within the kingdoms (Hint: think about other organisms in each of the kingdoms and what makes them different than the examples used here)?

 

 

3.    What questions would you have asked instead of the ones that you answered above when classifying the organisms?

UMUC Biology 102/103 Lab 6: Taxonomy

 
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