8. [From Lecture 2 practice problems] Consider the following pair of subpopulati

Question

8. [From Lecture 2 practice problems] Consider the following pair of subpopulations of a
tropical butterfly, with genotype frequencies as shown:
Genotype Lowland Mountain Top
AA                   0.64                  0.16
Aa                   0.32                   0.48
aa                    0.04                   0.36
Three biologists head to the field to collect this species, and each ends of collecting 500
individuals.
The first biologist gets altitude sickness easily, so he samples only individuals from the
lowlands.
(a) What allele frequencies does the first biologist find (assuming he has a random
sample of the population)?
(b) What are the expected genotype frequencies assuming Hardy-Weinberg equilibrium?
(c) How do the observed genotype frequencies compare to the expected frequencies?
The second biologist hates the heat, so she drives up the mountain and samples only
individuals from the top.
(d) What allele frequencies does the second biologist find?
(e) What are the expected genotype frequencies assuming Hardy-Weinberg equilibrium?
(f) How do the observed genotype frequencies compare to the expected frequencies?
The third biologist is a bit more adventurous, and samples individuals continuously as
she hikes up the mountain. Assume that she ends up sampling an equal number from
the lowland and mountain top populations.
(g) What are the observed genotype frequencies in the sample obtained by the third
biologist?
(h) What are the allele frequencies in this sample?
(i) What are the expected genotype frequencies assuming Hardy-Weinberg equilibrium?
(j) Which genotype(s) is overrepresented in the observed sample compared to the
expected sample? Which is underrepresented?
(k)What would the third biologist conclude about Hardy-Weinberg equilibrium?
(l) What assumption that we used in deriving the Hardy-Weinberg equilibrium is violated
in this case?

Explanation / Answer

Answer.

1st Biologist,

Low Land, Genotype Frequency given
AA = 0.64
Aa = 0.32
aa = 0.04

No. of Individuals = Genotype Frequency * no. of Individuals
AA individuals = 0.64 * 500 = 320
Aa individuals = 0.32 * 500 = 160
aa individuals = 0.04 * 500 = 20

Ans. a
Now,
Allele Frequency = no. of alleles / total no. of alleles
for A = 2 * 320 + 160 / 2 * 500
   = 640 + 160 / 1000
= 800 / 1000
   = 0.8
for a = 2 * 20 + 160 / 2 * 500
   = 40 + 160 / 1000
= 200 / 1000
   = 0.2

Ans. b

Expected Genotype Frequency will be,
AA = 0.8 * 0.8 = 0.64
aa = 0.2 * 0.2 = 0.04
Aa = 1 - 0.68 = 0.32

Ans. c

No Change in Observed Genotype frequency to Expected Genotype Frequency.

2nd Biologist

Mountain Top, Genotype Frequency given
AA = 0.16
Aa = 0.48
aa = 0.36

No. of Individuals = Genotype Frequency * no. of Individuals
AA individuals = 0.16 * 500 = 80
Aa individuals = 0.48 * 500 = 240
aa individuals = 0.36 * 500 = 180

Ans. d
Now,
Allele Frequency = no. of alleles / total no. of alleles
for A = 2 * 80 + 240 / 2 * 500
   = 160 + 240 / 1000
= 400 / 1000
   = 0.4
for a = 2 * 180 + 240 / 2 * 500
   = 360 + 240 / 1000
= 600 / 1000
   = 0.6

Ans. e

Expected Genotype Frequency will be,
AA = 0.4 * 0.4 = 0.16
aa = 0.6 * 0.6 = 0.36
Aa = 1 - 0.52 = 0.48

Ans. f

No Change in Observed Genotype frequency to Expected Genotype Frequency.

3rd Biologist

Ans. g

Now,
No. of Individuals = Genotype Frequency * no. of Individuals
Lowland
AA individuals = 0.16 * 500 = 80
Aa individuals = 0.48 * 500 = 240
aa individuals = 0.36 * 500 = 180
Mountain Top
No. of Individuals = Genotype Frequency * no. of Individuals
AA individuals = 0.64 * 500 = 320
Aa individuals = 0.32 * 500 = 160
aa individuals = 0.04 * 500 = 20
On Average,
AA = 320 + 80 = 400/ 2 = 200
Aa = 160 + 240 = 400/ 2 = 200
aa = 20 + 180 = 200/ 2 = 100

Ans. h
Now Allele frequency,

Allele Frequency = no. of alleles / total no. of alleles
for A = 2 * 200 + 200 / 2 * 500
   = 400 + 200 / 1000
= 600 / 1000
   = 0.6
for a = 2 * 100 + 200 / 2 * 500
   = 200 + 200 / 1000
= 400 / 1000
   = 0.4

Ans. i

Expected Genotype Frequency will be,

AA = 0.6 * 0.6 = 0.36
aa = 0.4 * 0.4 = 0.16
Aa = 1 - 0.52 = 0.48

Ans. j

AA and aa are Over Represented
Aa is Under Represented

Ans. k

She need to Conduct a Chi Square Test to Test the Hypothesis whether her Population obtained is in Hardy-Weinberg equilibrium or Not.

Ans. l

Population Size.
Allele Frequencies are equal in the sexes

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