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)

 
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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?

 
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Lab: RNA

Lab: RNA

In this lab, you will learn about the types of RNA molecules and you will study the details of transcription and translation.

Objectives:

Learn about the types of RNA molecules produced by cells.
Study the details of transcription and translation.

Time Requirements:

This lab should take two hours to complete.

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RNA Lab Report

If a codon is mutated, say from GGU to CGU, is the same amino acid specified? _______

Using information from the course and online research, answer the questions.

Searching Websites:

Conduct a keyword search to find websites. To begin a keyword search, start by searching broad terms such as RNA, transcription, translation, or the scientist’s name. The resulting list of websites can give you a lot of information, but how do you know if it is reliable? Some basic guidelines can help you when you open a website and try to determine its reliability.

Always consider a site’s:

· Objectivity – Excessive expressions of emotion, opinions, and stereotyping are tip-offs that the information on a site may be biased.

· Ownership and contributors – Go to the Home or About page of the website and find out who sponsors and writes for the site. Look for contributors who have reliable credentials, such as “Harvey Jones, Professor, University of Wisconsin—Madison.”

· Writing style and mechanics – Check the grammar, spelling, and writing style on the site. Errors and awkwardness are signs of a nonprofessional website.

· Currency – Look for publication or copyright dates associated with the site; the more current the better.

· Links – What links does the site contain? A reliable website will offer links to other reliable websites, not to “junk” sites.

Keyword Search: RNA, transcription, translation, or the scientist’s name.

Scientific Analysis:

1. Scientists use observations and inferences for many discoveries and conclusions in genetics. What is the difference between an observation and an inference? Give an example of each.

2. If DNA from a cell is placed in a test tube containing the enzyme DNA polymerase and nucleotides, the result is new copies of the original DNA. This evidence was used to understand DNA replication (DNA passing information from one generation to the next). Write an explanation detailing why this clue represents solid data and is not an opinion or an inference.

3. In your own words, describe Crick’s Central Dogma. Explain why this is an inference.

Scientific Contributions

There were many scientists that contributed to decoding the genetic code. Explore the contributions of some scientists.

4. Describe Marshall Nirenberg’s contribution to determining the role of RNA in genetics.

5. Describe how Seymour Benzer’s phage experiments contributed to deciphering the genetic code.

6. Describe Seymour Brenner’s contribution to cracking the genetic code with respect to translation.

Replication, Transcription, and Translation

Place an x in the box for which process each item is a characteristic.

Characteristic Replication Transcription Translation
1. RNA polymerase unwinds the DNA double helix.      
2.  This is the first part of protein synthesis.      
3. After the stop codon is reached, the protein is separated from the ribosome.      
4. Amino acids sequences are made from RNA sequences.      
5. A copy of RNA is made from DNA.      
6. Peptide bonds form between amino acids.      
7. Uracil hydrogen bonds with adenine.      
8. tRNA brings amino acids to the ribosome.      
9. Copies of the original DNA are made.      
10.  Thymine hydrogen bonds with adenine.      

 

7. Use the table and your research to compare and contrast transcription and translation.

8. In your own words, explain the function of mRNA, tRNA, and rRNA.

9. Use the codon chart to convert this sequence into an amino acid:

UCU-CGA-GCC-GUU-GGG-UGA

10. Explain why an mRNA strand composed of only uracil nucleotides (UUUUUUUUUU) did conclusively prove that the DNA code was a triplet code.

11. How could this mRNA sequence support a triplet code? ACACACACACAC

12. List the steps involved in protein synthesis, beginning with the DNA code in the nucleus and ending with a polypeptide. Try to explain it in 10 steps.

13. A mutation in a codon may not result in a change of an amino acid in the protein sequence. Explain why, using the genetic code chart to give an example.

© KC Distance Learning

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The Lower Limit Of The Intertidal Zone Is The

Question

1 of 25

The lower limit of the intertidal zone is the

average high tide.

average middle tide.

lowest tide.

lowest edge of the rocky substrate.

None of the above choices are correct: it varies according to type of substrate.

Question

2 of 25

The particular characteristic most widely used in classifying intertidal communities is which of the following?

Type of tides

Relative exposure to air

Type of substrate

Type of seaweeds

Relative immersion by water

Question

3 of 25

Which of the following strategies is not used to avoid desiccation in the intertidal?

Closing shells

Crowding in areas that are always moist

Burrowing into substrate

Mucus

Moving out of tide pools

Question

4 of 25

Most sessile animals living on rocky shores are which of the following?

Deposit feeders

Carnivores

Detritus feeders

Filter feeders

Grazers

Question

5 of 25

If a rocky shore highly exposed to wave action is compared to a similar rocky shore that is a lot less exposed to wave action, one should expect that barnacles living on the more exposed shore show a

wider vertical distribution.

narrower tolerance to salinity.

narrower tolerance to temperature.

narrower vertical distribution.

higher resistance to predators.

Question

6 of 25

Byssal threads are used by mussels to cope with.

desiccation.

wave shock.

restricted feeding.

salinity changes.

temperature changes.

Question

7 of 25

Vertical zonation on rocky shores is mostly the result of differences in tolerance to

 

wave action.

exposure.

predation.

light.

salinity.

Question

8 of 25

This organism is typical of the upper intertidal on rocky shores.

Mussels

Barnacles

Sponges

Sea anemones

Periwinkles

Question

9 of 25

The middle intertidal is characterized by which of the following?

Constant wetting by splash and spray

Long exposure to air

Steady immersion

Exposure and immersion on a regular basis

Splashing during high tide and complete exposure at low tide

Question

10 of 25

Intertidal organisms from exposed areas sometimes have thicker shells than their counterparts from less exposed areas. This is thought to be an adaptation to

desiccation.

restricted feeding.

wave shock.

salinity changes.

temperature changes.

Question

11 of 25

Predation by sea stars on rocky shores ultimately results in

fewer species.

less wave action.

a decrease in the number of seaweeds.

an increase in the number of mussels.

more species.

Question

12 of 25

Ecological succession ultimately results in

a climax community.

an upper-limit stage.

competitive exclusion.

keystone predation.

vertical zonation.

Question

13 of 25

Which of the following is generally considered to be the greatest limiting resource in the rocky intertidal?

Salinity

Space

Prey

Availability of mates

Light availability

Question

14 of 25

Which of these organisms is typically a very rare component of soft-bottom intertidal communities?

Burrowing organisms

Detritus feeders

Seaweeds

Infauna

Deposit feeders

Question

15 of 25

Most animals living on sandy beaches are included among which of the following?

Infauna

Deposit feeders

Macrofauna

Producers

Grazers

Question

16 of 25

Fine sediments are characteristic of

shores exposed to wave action.

rocky shores.

areas with wide temperature fluctuations.

calm, less exposed shores.

areas that experience considerable water flow.

Question

17 of 25

The interstitial water in muddy bottoms

has plenty of oxygen since temperature is much higher than in the water column.

accumulates oxygen as a result of photosynthesis by inhabitants of the sediment.

recirculates oxygen very frequently so it is high.

is deficient in oxygen.

is deficient in oxygen but only during the day.

Question

18 of 25

The main source of food in muddy-bottom intertidal communities is

seaweeds

detritus

plankton

large prey

epifauna

Question

19 of 25

On Atlantic shores, the dog whelk (Nucella lapillus) has two color forms: white-shelled and brown-shelled. Survival of one form over the other form in a region appears to be related to

temperature.

salinity.

substrate type.

prey availability.

predators.

Question

20 of 25

What organism would be most likely to be found in the upper intertidal zone of a rocky shoreline?

Mussels

Seaweed

Encrusting algae

Barnacles

Irish moss

Question

21 of 25

In soft-bottomed intertidal communities, locations with strong waves and currents are most likely to have which type of bottom?

Gravel

Sand

Silt

Clay

Mud

Question

22 of 25

Meiofauna live

on rocky shores.

on sandy bottoms.

between grains of sediment.

in burrows.

in the water column.

Question

23 of 25

A species that significantly influences the structure and diversity of its community is a ______________.

predator species

keystone species

prey animal

primary producer

apex predator

Question

24 of 25

The best reason for putting a cage over certain organisms in an intertidal study area is so that __________________.

other organisms can be kept out of the study area

biologists can control when certain animals mate

their respiration rates can be measured

they can be removed from the area

body measurements can be taken

Question

25 of 25

Two sediments that are very rich in detritus are ______________.

sand and gravel

sand and clay

silt and gravel

clay and sand

silt and clay

 

 
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