3) A mountain has a population of 100 wild goats, with the following number of i
ID: 64683 • Letter: 3
Question
3) A mountain has a population of 100 wild goats, with the following number of individuals of each genotype: AA: 49, Aa: 42, and aa: 9.
a) Calculate the allele frequencies and the genotypic frequencies expected under Hardy-Weinberg. Show your work and circle your final answers.
b) Is the population in H-W equilibrium? Why or why not? (You do not need to analyze this statistically).
4) In a population of 1,000 fern plants, there are 480, 200, and 320 individuals with genotypes QQ, Qq, and qq, respectively.
a) Calculate the allele frequencies and the expected Hardy-Weinberg genotypic frequencies. Show your work and circle your final answers .
b) Is this population in Hardy-Weinberg equilibrium? Why or why not? (You do not need to analyze this statistically).
5) In a breed of dogs 1 pup in 900 is born albino. Albinism is due to a recessive gene at a coat pigment locus. Assume that the population is in Hardy-Weinberg equilibrium.
a) What is the frequency of albino alleles in the population? Show your work and circle your final answer.
b) What is the expected frequency of individuals with two normal alleles? Show your work and circle your final answer..
c) What is the expected frequency of individuals that are heterozygotes at this locus? Show your work and circle your final answer..
Explanation / Answer
a) calculation of gene frequency for S is based on the gene counting method, every SS individuals have 2 S genes and a SF individual has 1.
This is relative to all genes in the population, which are 2*106.
The frequency of S is calculated as p = (2*36 + 47)/(2*106) = 0.56
do F do q = (2*23 + 47)/(2*106) = 0.44
1.00
The combination of two gene pairs, A and B, each with
two alleles are shown below in the classical two-by-
two table, where r, s, t, and u ar
e the observed gamete frequencies of
the gametes AB, Ab, aB and ab, see
Figure 2.5.
gene A gene B B b frequency
------------------------------------------------
A | r=p(A)*p(B)+D s=p(A)*q(b)-D | p(A)
a | t=q(a)*p(B)-D u=q(a)*q(b)+D | q(a)
-----------------------------------------------
Frequency | p(B) q(b) | 1
Gene frequencies can be calculated by means of the
gene counting method and they correspond to the border
distributions. The expected frequency of
a gamete is the product of the bord
er distributions which is equal to
p(A) = r + s = p(A)*p(B) + D + p(
A)*q(b) - D = p(A)[p(B) + q(b)] =
p(A). The deviation which occurs
between observed and expected numbers is assigned
the symbol D (Disequilibrium). Note that the
deviations are the same size for all
cells (D), but with a negative sign
for the repulsion phase gametes, Ab and aB.
18
A calculation example: The observed number
of gametes is AB=21, Ab=49, aB=19, ab=11.
gene A/gene B B b Sum (Freq)
A | 21 (r=0.21) 49 (s=0.49) | 70 (p(A)=0.7)
a | 19 (t=0.19) 11 (u=0.11) | 30 (q(a)=0.30)
Sum (Freq) | 40 (p(B)=0.4) 60 (q(b)=0.6) | 100 1
Frequencies are given within brackets: Chi-squared = 9.7
Df=1
D= u-q(a)*q(b)=0.11 - 0.3*0.6= -0.07
Statistically significant deviation occurs from random combination between the genes A and B
corresponding to a disequilibrium equal to -0.07.
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