Ecology
Revised December 2018
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Click & Learn
Student Supplement
CSI Wildlife: Frequency Primer
INTRODUCTION
This worksheet complements the Frequency Primersection in Case One of the CSI Wildlife
Click & Learn. In
this section, you will use allele frequencies to determine the chances of two elephants sharing the same genetic
profile.
PROCEDURE
Open the Click & Learn and select Case One at the top. Click on the last section of Case One, Frequency Primer.
As you go through each part of the Frequency Primer, follow the instructions below and answer the questions in
the spaces provided.
1. Explain what is meant by an alleles relative frequency.
2. The gel on the screen shows a genetic profile of the elephant identified in Case One. Which allele has
the lowest relative frequency, and which has the highest? (Remember that an individual can have one or
two alleles for any marker.)
3. Imagine an STR marker that has four known alleles in a population: A, B, C, and D. Below are the alleles’
relative frequencies in one population of elephants. Fill in the missing frequency for allele D.
Allele
Relative frequency
A
0.25
B
0.15
C
0.45
D
CSI Wildlife: Frequency Primer
Ecology
Revised December 2018
www.BioInteractive.org
Page 2 of 4
Click & Learn
Student Supplement
4. Let’s take a closer look at how the relative frequencies in
Question 3 were calculated. Imagine that the population has 10
elephants. Each elephant’s alleles for this STR are shown in the
table to the right.
Using the information above, complete the table below. The first row has been filled in for you.
Number of the allele
in the population
Total number of alleles
in the population
Relative frequency of the allele
(Number of that allele/Total)
5
20
5/20 = 0.25
20
20
20
Do the last columns of the tables in Question 3 and Question 4 match up? If not, check your work.
5. In the probability formulas shown, what do p and q represent?
6. In terms of p and q, what is the formula for the probability of having a homozygous genotype?
7. In terms of p and q, what is the formula for the probability of having a heterozygous genotype?
8. Using the table of allele frequencies in Question 3, calculate the probability that an elephant born in this
population will:
a. have a copy of allele B
b. have the homozygous genotype BB (Hint: Use the formula from Question 6.)
c. have the heterozygous genotype BD (Hint: Use the formula from Question 7.)
Elephant
Allele 1
Allele 2
1
A
A
2
C
C
3
B
C
4
B
D
5
A
A
6
C
C
7
B
C
8
A
D
9
C
C
10
C
D
CSI Wildlife: Frequency Primer
Ecology
Revised December 2018
www.BioInteractive.org
Page 3 of 4
Click & Learn
Student Supplement
Let’s now apply these formulas to some real data. Look at the gel shown
on the screen. It represents a four-marker genetic profile for an elephant
in a different population. The relative frequencies for the alleles shown
are summarized in the table to the right.
9. Why is there only one frequency for the alleles for the FH19 and FH71
STR markers, but two frequencies for the alleles for the FH127 and
FH67 markers?
10. The elephant profiled here is homozygous for an allele of the FH71 marker. As shown in the table, this allele
has a relative frequency of 0.09. Based on the formula in Question 6, the probability of an elephant from this
population having the profiled elephants genotype for the FH71 marker (in other words, for being
homozygous for the same FH71 allele) is (0.09)
2
= 0.008.
This elephant is also homozygous for an allele of the FH19 marker. What is the probability of an elephant
from this population being homozygous for the same FH19 allele? Show your work.
11. The elephant profiled here is heterozygous for the FH67 marker; it has two different alleles for this marker
with relative frequencies of 0.1 and 0.24. Based on the formula in Question 7, the probability of an elephant
from this population having the profiled elephants genotype for the FH67 marker is 2(0.1)(0.24) = 0.048.
What is the probability of an elephant from this population having the profiled elephant’s genotype for the
12. Using the probabilities from Questions 10 and 11, calculate the probability of an individual from this
population having this exact four-marker genetic profile. Show your work. (Note that the answer in the Click
& Learn was calculated without rounding the probabilities from Questions 10 and 11. You may get a slightly
different answer if you round these probabilities.)
Marker
Relative frequency
of allele(s) shown
FH19
0.06
FH127
0.08, 0.05
FH67
0.10, 0.24
FH71
0.09
CSI Wildlife: Frequency Primer
Ecology
Revised December 2018
www.BioInteractive.org
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Click & Learn
Student Supplement
Go to the Review section. The gel shown on the screen represents a more detailed, 16-marker genetic profile for
the same elephant. Follow the instructions below to help answer the review question.
13. The relative frequencies for the alleles shown in the gel are summarized in the table below. Fill in the last
column, showing your work for each row. Some of the rows have been filled in for you.
Marker
Relative frequency
of allele(s) shown
Probability of an individual from this
population having this genotype
FH39
0.16, 0.26
2(0.16)(0.26) = 0.0832
FH153
0.05, 0.06
2(0.05)(0.06) = 0.006
FH40
0.49
(0.49)
2
= 0.2401
FH94
0.20, 0.44
2(0.20)(0.44) = 0.176
FH102
0.13, 0.12
2(0.13)(0.12) = 0.0312
FH19
0.06
Hint: This is the answer to Question 10.
FH48
0.19, 0.14
FH127
0.08, 0.05
Hint: This is the answer to Question 11.
FH129
0.16, 0.08
FH60
0.26, 0.06
LAFMS04
0.78
FH103
0.15
LAFMS03
0.10
(0.10)
2
= 0.0100
FH126
0.14
(0.14)
2
= 0.0196
FH67
0.10, 0.24
2(0.1)(0.24) = 0.048
FH71
0.09
(0.09)
2
= 0.0081
14. Using the probabilities from Question 13, calculate the probability of an individual from this population
having this exact 16-marker genetic profile. (Note that the answer in the Click & Learn was calculated
without rounding the probabilities from Question 13. You may get a slightly different answer if you round
these probabilities.)
15. Using the probability you calculated in Question 14, fill in the following blank:
The chance that another elephant from this population would have the same 16-marker genetic profile is 1
in ______________.
16. There are many more than 16 STRs in an elephant’s genome. Why do you think that Sam Wasser and his
colleagues reasoned they did not need to use more than 16 markers for their genetic profiles?