ERIE COMMUNITY COLLEGE
TITLE III
Algebra Project
Interdisciplinary Course Materials
Biology
Course: MT 007 Elementary Algebra I/ MT013 Elementary Algebra I and II
Course Outline Topic:
 Demonstrate an understanding of additive and multiplicative inverse.
 Perform fundamental operations with algebraic fractions, simplify complex fractions and solve rational
equations.
Project Title: Effective Population Size
Project description: In this project, students will determine if the census population size differs from the
effective population size.
Author: Dianna Cichocki
Curriculum Expert: Rosanne Redlinski
Semester Created: Fall 2008
A.
Essential Question
Is the census population size a good measure of the effective population size?
B.
Introduction
Ecologists and evolutionary biologists often measure population size (a census) to asses population
health and to identify processes that shape evolution. Population size, however, may not always be a
good measure to assess the health and changes in a population. For example, a population may consist of
95% sexually immature and 5% sexually mature individuals. In such cases, effective population size
may be computed based on the number of individuals that actually take part in reproduction.
C.
Basic Directions
Students will perform fundamental operations with algebraic fractions.
D.
Things to Learn Before Starting the Project
Students should how to substitute for a variable. They should also know how to solve an algebraic
equation for a variable.
E.
The Project Assignment
In this project we use the formula for effective population size to answer three questions.
F.
Student Resources
Formula for Effective Population size (given).
G.
Grading Rubric
30 points total
Erie Community College
Title III Grant
Name _____________________________
Interdisciplinary Project (Math/Biology)
EFFECTIVE POPULATION SIZE
Effective population size and census population size may be drastically different when the number of breeding
males and breeding females in a population are drastically different (e.g. polygamous species and social
insects). In this case, we can express effective population size using a rational function as,
where Nf and Nm are the number of breeding females and breeding males, respectively. We will also assume
that, Nf + Nm = N.
Problem 1 -Find the effective population size given the number of breeding males and females.
If the number of breeding males and females are equal, what is the effective population size?
A. 4N
B.
N
C.
N/4
D.
2N
Problem 2 - Find the number of breeding males in a population.
If the effective size of a given population is 1000 and the number of breeding females is a constant C,
where 500 <C< 1000 is positive integer, what is the number of breeding males?
A. Nm = 4C
B. Nm = 996 C
C.
D. Nm = 1000 - 4C
Problem 3 - Determine the effective population when there are twice as many breeding males.
What is the effective population size when there are two breeding males for every one breeding female?
A. Ne= 8N/3
B. Ne= 8N/9
Erie Community College
Title III Grant
C.
Ne= 2N
D. Ne= 8N
Answer key:
Problem 1 -Find the effective population size given the number of breeding males and females.
If the number of breeding males and females are equal we have Nm = Nf and,
Substituting the above equation of Ne gives,
Therefore, if the number of breeding males and females are equal, the effective population size is equal to the
census population size.
Problem 2 - Find the number of breeding males in a population.
We are given that the effective population, Ne, is 1000 and that the number of breeding female, Nf, is a constant
represented by C, where 500 < C < 1000, and we are asked to find the number of breeding males, Nm. We begin
by substituting the given information into the equation for Ne as,
We must now solve the above equation for Nm by bringing all terms involving Nm to one side as,
Erie Community College
Title III Grant
Notice that if Ne = 1000, C must be greater that 250 to guarantee Nm > 0.
Problem 3 - Determine the effective population when there are twice as many breeding males.
We are given that there are twice as many breeding males as breeding female, which we represent as Nf = 2 Nm
and we have,
Therefore, we have Nm = N/3 and Nf = 2 Nm = 2N/3, and we substitute these values into the equation for Ne as,
Erie Community College
Title III Grant