This document discusses key concepts in population genetics, including defining a population as a group of the same species living in a specific area. It provides an example of studying the genetics of a population of racers lizards, looking at the frequency of dominant and recessive alleles for stripe color. The values for p (frequency of dominant allele) and q (frequency of recessive allele) are calculated based on observing 110 lizards with the dominant white stripe allele out of 200 total alleles observed. This yields values of p=0.55 and q=0.45 for this population.

1. Mrs. Nelson

2.  Thestudy of how genetic principles apply to
an entire population of organisms.
 Population:
A group of organisms of the
same species that live in a specific area.
 Species:A group of organisms that can
interbreed and produce fertile offspring.
 To
study the genetics of a
population, scientists must collect a
population sample.

3. A population sample is random collection of
organisms of same species used to represent
the entire population.
 Population
genetics involves studying the
frequency at which certain alleles occur in a
population.

4.  Undercertain conditions, the frequency of
the dominant and recessive alleles will stay
the same from one generation to the next.
 Undercertain conditions, p and q will stay
the same.

5.  The term “allele frequency” refers to how
common an allele is in a population.
 Many traits in a population are controlled by
two alleles – dominant and recessive allele.
 Scientists assign: (variables to represent alleles)
p= dominant allele
q= recessive allele
 For any population, p + q = 1
 Allele frequencies can be written as a
decimal or percentage.

6. •Lizard found commonly in mid-west.
Population Indiana Dunes State Park.
•Six stripes along its back and head.
•White stripes (T) dominant to yellow
stripes (t).

7.  In this population of racerunners, 110/200
alleles are dominant.
 110/200 = 0.55 or 55%
 The frequency of the dominant allele is
assigned a variable (p).
 Therefore in this population, p = 0.55 or 55%

8.  In this population of racerunners, 90/200
alleles are recessive.
 90/200 = 0.45 or 45%
 The frequency of the recessive allele is
assigned a variable (q).
 Therefore, in this population q = 0.45 or 45%

9.  Scientists monitor frequency of the dominant and
recessive allele in a population year after year.
 In other words, scientists may monitor the values of p
and q for a population over several generations.
 If the values of p and q…
-remain the same the population is not
evolving according to scientists.
-change then scientist say the group is
evolving.
 Evolution- “genetic change” amongst population
 Scientist argue all populations will evolve unless
“certain conditions” are upheld within the group

10. 1. No mutations
2. No individual can be more adapted to
survive than any other (no “survival
fittest”).
3. The population must be large and stay the
same size (no immigration or emigration).
4. Mating must be completely random.

11.  How do scientists determine values of p and
q in a population?
 First, they collect a population sample.
 For example, 100 racerunners of which 85
have white stripes and 15 yellow stripes.
 What is the genotype 85 with white stripes?
-TT or Tt
 What is the genotype of 15 with yellow
stripes?
-tt
 Using this information, scientists can find
values for p and q of the population.

12.  Forany trait, there will always be a certain
percentage of individuals who are…
 Homozygous dominant (p2)
 Heterozygous (2pq)
 Homozygous recessive (q2)
 P2 + 2pq + q2 = 1 or 100%

13.  Thefollowing data was taken from a
population sample of racerunners in Dunes:
 85 white racerunners – either homozygous
dominant or heterozygous
 15 yellow racerunners - are homozygous
recessive
 Determine p, q, p2, 2pq, and q2

14.  The overall goal of this problem is really to get to
the values for p and q.
 The key is to focus on the yellow racerunners
because they are homozygous recessive = q2
 Mathematically to solve the problem start with q2
and solve for q.
 Solve for q by taking the √of q2.
 q2= 15/100 or .15
To go from q2  q, take the square root each side
q = √.15 = 0.4
 Remember, p + q = 1
 p + 0.4 = 1
 p = 0.6

15.  Use the decimal forms of the p and q values to
find p2, 2pq, and q2
 p = 0.6
 q = 0.4
 p2 = (0.6 x 0.6) = 0.36 or 36%
 q2 = (0.4 x 0.4) = 0.16 or 16%
 2pq = (2 x 0.6 x 0.4) = 0.48 or 48%
 Check your work…
 .36 + .16 + .48 = 1