HARDY WEINBURG PRACTICE

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HARDY WEINBERG PRACTICE
1. If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what
percentage of the population will be more resistant to malaria because they are
heterozygous (Ss) for the sickle-cell gene?
Ss or 2pq = ?
ss or q2 = 9% or 0.09
q
= 0.09
= 0.3
p
=1–q
= 1 – 0.3
= 0.7
2pq
= 2 (0.7) (0.3)
= 0.42
= 42%
2. This is a classic data set on wing coloration in the scarlet tiger moth (Panaxia dominula).
Coloration in this species had been previously shown to behave as a single-locus, two-allele
system with incomplete dominance. Data for 1612 individuals are given below:
White-spotted (AA) =1469 Intermediate (Aa) = 138 Little spotting (aa) =5
Calculate the allele and genotypic frequencies:
p
=?
q
=?
AA = p2 = ?
Aa= 2pq = ?
aa = q2 = ?
aa = q2 = 5/1612 = 0.003
q
= 0.003
= 0.05
p
=1-q
= 1 – 0.05
= 0.95
AA = p2 = (0.95)2
= 0.90
Aa = 2pq = 2 (0.95) (0.05)
= 0.095
3.Seventy percent of a population has the dominant trait while 30% are recessive. What
percentage of the population are carriers?
A_ = 70%
aa = q2 = 30%
Aa = 2pq = ?
aa = q2 = 0.30
q = 0.30
= 0.548
p=1–q
= 1- 0.548
= 0.452
Aa = 2pq = 2 (0.452) (0.548)
= 0.495
4. It is determined in a certain population that 1% of the people do not have freckles. What
percentage of the population are carriers for this trait? Note: Freckles are caused by a dominant
allele!
ff = q2 = 1% = 0.01
Ff = 2pq = ?
q2
= 0.01
q
= 0.01
= 0.1
p
=1–q
= 1 – 0.1
= 0.9
Ff = 2pq = 2 (0.9) (0.1)
= 0.18
= 18%
5. For foxes there exists a single gene that controls coat thickness. Allele C confers a thick coat
while allele c a thin coat. In a certain population of 540 foxes, 49 have a thin coat. What are the
allelic and individual frequencies?
C – thick coat
c – thin coat
Population size = 540 foxes
cc - 49 foxes
Find p, q, p2, 2pq, and q2.
cc = q2 = 49/540 = 0.09
q = 0.09
= 0.3
p=1–q
= 1 – 0.3
= 0.7
CC = p2 = (0.7)2 = 0.49
Cc = 2pq = 2 (0.7) (0.3) = 0.42
6. In a certain population of flowers, when red and white flowers are crossed they produce
pink flowers. A certain field has 1190 red flowers, 1070 pink flowers and 240 white
flowers. What is the frequency of individuals and alleles?
CRCR = 1190 flowers
CRCW = 1070 flowers
CWCW = 240 flowers
Total flowers = 2500
Find p, q, p2, 2pq, and q2.
CWCW =
q2
q
= 0.096
= 0.31
p
=1–q
= 1 – 0.31
= 0.69
= 240/2500 = 0.096
CRCR
= p2 = (0.69)2
= 0.48
CRCW = 2pq
= 2 (0.69) (0.31)
= 0.43
7. Pea seeds can be round (R: dominant) or wrinkled (r: recessive). In a field of wild peas
ninety-six percent (96%) of the peas are round. What are the allelic and genotypic
frequencies?
R – round
r – wrinkled
R_ = 96%
Find p, q, p2, 2pq, and q2.
If R_ = 96%, then rr = q2 = 4% = 0.04
q
= 0.04
= 0.2
p
=1–q
= 1- 0.2
= 0.8
RR = p2
= (0.8)2 = 0.64
Rr = 2pq
= 2 (0.8) (0.2)
= 0.32
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