POPULATION GENETICS PROBLEM SET
These genetic problems can be use AS ASSESSMENT and OF ASSESSMENT of
learning. There are too many concepts that must be learned before students can
successful investigate and solve these problems on their own.
Name:____________________________________________
1.
Mark: /16
The allele y occurs with a frequency of 0.8 in a population of clams. Give the frequencies of
the genotypes YY, Yy, and yy. (3 marks)
Y is dominant
y is recessive
Y+y=1
Y2 + 2Yy + y2 = 1
If y = 0.8 then Y= 0.2
The frequency of YY is equal to Y2 = (0.2 x 0.2 = 0.04) = 4%
The frequency of Yy is equal to 2Yy = (2 x 0.2 x 0.8 = 0.32) = 32%
The frequency of yy is equal to y2 = (0.8 x0.8 = 0.64) = 64%
2.
In a population of rabbits, brown colour is dominant over white colour. 25% of the population of rabbits are
white. Calculate the allelic frequencies. Would you expect this frequency to stay constant for the next few
generations? Explain. (4 marks)
B = brown, b = white
B2 + 2Bb + b2 = 1
b2 = bb = 25% Therefore, by definition b = 0.5
If B + b = 1 Then B = 0.5
The frequency of BB is equal to B2 = (0.5 x 0.5 = 0.25) = 25%
The frequency of Bb is equal to 2Bb = (2 x 0.5 x 0.5 = 0.50) = 50%
The frequency of bb is equal to b2 = (0.5 x0.5 = 0.25) = 25%
B
b
B
BB
Bb
b
Bb
bb
I would not expect this frequency to stay constant for the next few generations because the second generation has
rabbits that are homozygous dominant and homozygous recessive. If these rabbits were to mate, you would get a
100% brown rabbits that are heterozygous for the trait.
3.
Cystic fibrosis is a recessive disease that has a high frequency in Caucasians (1 in 2500 newborns). Until
recently, it was lethal in childhood. Calculate the frequency of the recessive allele. In a population of 2
million Caucasian how many would be carriers? (3 marks)
q2 is 1/2,500 or 0.0004
Therefore, q is the square root, or 0.02.
The frequency of the cystic fibrosis (recessive) allele in the population is 0.02 (or 2%).
The frequency of the dominant (normal) allele in the population (p) is simply 1 - 0.02 = 0.98 (or 98%).
The percentage of heterozygous individuals (carriers) in the population equals the frequency of 2pq = (2)(.98)(.02)
= 0.04 or 1 in 25 are carriers.
4.
In a given population of 610 individuals, the gene frequencies of the LM and LN alleles were found to be
0.62 and 0.38, respectively. Calculate the number of individuals with M, MN and N blood types.
(3 marks)
M + N = 1, 0.62 + 0.38 = 1
The frequency of MM is equal to M2 = (0.62 x 0.62 = 0.385) = 39%
The frequency of MN is equal to 2MN = (2 x 0.62 x 0.38 = 0.4712) = 47%
The frequency of NN is equal to N2 = (0.38 x 0.38 = 0.1444) = 14%
Therefore,
Number of M individuals is 39% of 610 = 238
Number of MN individuals is 47% of 610 = 287
Number of N individuals is 14% of 610 = 85
5.
The S-s antigen system in humans is controlled by two codominant alleles S and s. In a group of 3146
individuals the following genotypic frequencies were found: 188 SS, 717 Ss, and 2241ss. Calculate the
frequencies of each allele. (3 marks)
S+s=1
S2 + 2Ss + s2 = 1
The frequency of S equals the following:
2 x (number of SS) + (number of Ss) divided by 2 x (total number of individuals)
2 x (188) + 717 / 2 (3146) = 1093 / 6292 = 0.174 = S
Since s = 1 – S, then s = 1 – 0.174 = 0.826
S = 0.174
s = 0.826
Evaluation Rubric
Knowledge &
Understanding
Application
Communication
Level 1
Demonstrates little
knowledge of population
genetics
Uses formulas inaccuracy
to solve population
genetics problems
Incompletely explains the
concepts of genotype,
phenotype, alleles,
Level 2
Demonstrates some
knowledge of population
genetics
Uses formulas with some
accuracy to solve
population genetics
problems
Somewhat explains the
concepts of genotype,
phenotype, alleles,
Level 3
Demonstrates knowledge
of population genetics
Uses formulas to solve
population genetics
problems
Level 4
Demonstrates thorough
knowledge of population
genetics
Accuratley uses formulas
to solve population
genetics problems
Completely explains the
concepts of genotype,
phenotype, alleles,
Thoroughly explains the
concepts of genotype,
phenotype, alleles,
Thinking & Investigation
dominance, incomplete
dominance,
codominance,
recessiveness, and sex
linkage according
to Mendelian laws of
inheritance
Investigates and predicts
allelic frequencies with
limited effectiveness
dominance, incomplete
dominance,
codominance,
recessiveness, and sex
linkage according
to Mendelian laws of
inheritance
Investigates and predicts
allelic frequencies with
some effectiveness
dominance, incomplete
dominance,
codominance,
recessiveness, and sex
linkage according
to Mendelian laws of
inheritance
Investigates and predicts
allelic frequencies with
considerable
effectiveness
dominance, incomplete
dominance,
codominance,
recessiveness, and sex
linkage according
to Mendelian laws of
inheritance
Thoroughly investigates
and predicts allelic
frequencies