3. 1. In a population of 1000 individuals, 360 belong to genotype PP, 480 to
Pq and the remaining 160 to qq. Based on this data, the frequency of allele
P in the population is:
1) 0.4 2) 0.5
3) 0.6 4) 0.7
4. 2. At a particular locus, frequency of A allele is 0.6 and that of a is 0.4. what
would be the frequency of heterozygotes in a random mating population at
equilibrium?
1) 0.36 2) 0.16
3) 0.24 4) 0.48
5. 3. Consider a population of sheep to be in Hardy-Weinberg equilibrium. The allele
for black wool(w) has an allele frequency of 0.81 while the allele for white
wool(W) has an allele frequency of 0.19. Then the percentage of heterozygous
individuals in the population is:
1) 4% 2) 15% 3) 31% 4) 66%
6. 4. A gene locus has two alleles A and a. If the frequency of dominant allele A is
0.4, then the frequency of homozygous dominant, heterozygous and homozygous
recessive individuals in the population is
1) 0.16 (AA); 0.48 (Aa); 0.36 (aa)
2) 0.16 (AA); 0.24 (Aa); 0.36 (aa)
3) 0.16 (AA); 0.36 (Aa); 0.48 (aa)
4) 0.36 (AA); 0.48 (Aa); 0.16 (aa)
7. 5. A sampled population has 36% of homozygous recessive genotype (aa). Then
the frequency of allele “a” is:
1) 0% 2) 20% 3) 60% 4) 70%
8. 6. In a population at Hardy Weinberg equilibrium, the allele frequency of ‘A’ is 0.3,
the expected frequency of ‘Aa’ individuals is
1) 0.21 2) 0.42 3) 0.63 4) 0.18
9. 7. In the Caucasian population of the US, 1 in 2500 babies is affected by a
recessive condition – cystic fibrosis. In this population, the frequency of the
dominant allele is:
1) 0.02 2) 0.36 3) 0.56 4) 0.98
10. 8. In a random mating population of 28,800 individuals percentage of dominant
homozygous individuals is 49%. Find out the percentage of the heterozygous
individuals?
1) 21% 2) 42% 3) 32% 4) 9%
11. 9. There is an irregular mating population. If the frequency of an autosomal
recessive lethal gene is 0.4, then the frequency of the carriers in a population of
200 individuals is:
1) 30 2) 80 3) 96 4) 104
12. 10. In a random mating population frequency of recessive allele is 0.5. What is the
frequency of dominant phenotypes in population?
0.25 2)0.50 3)0.75 4) 0.20
13. Answer Key:
1. 3) 2. 4)
3. 3)
The number of heterozygous individuals (Aa) = 2pq = 2 x 0.19 x 0.81 = 0.3078
Therefore in a population of 100, the number of heterozygotes are = 30.78 = 31
4. 1)
5. 3)
The frequency of aa is 36%, i.e., q2 = 0.36
If q2 = 0.36, then q = 0.6 = 60%.
6. 2)
14. 7. 4)
In this question, q2 is 1/2,500 = 0.0004.
q = 𝑥 = 0.0004 = 0.02.
P= 1 - q = 1 – 0.02 = 0.98
8. 2)
In the question, the frequency of dominant homozygous individuals p2= 49% or 0.49. Thus, the
frequency of a dominant allele p = 0.7.
The frequency of recessive allele (q) = 1 - p = 1−0.7=0.3.
Thus, the frequency of heterozygotes = 2pq = 2 X 0.7 X 0.3 = 0.42 or 42%.
15. 9. 3) According to the question, the frequency of a recessive lethal allele is (q = 0.4).
Frequency of wild-type allele : p = 1- q, i.e., 1- 0.4=0.6
The frequency of carrier genotype in total population (2pq) = 2 X 0.6 X 0.4 = 0.48.
And the number of carriers in the population of 200 is = 0.48 X 200= 96.
10. 3)
The frequency of dominant allele p = 1 – q = 1 – 0.5 = 0.5
The frequency of homozygous dominant allele p2 = (0.5)2 = 0.25
The frequency of heterozygous dominant allele 2pq = 2 * 0.5 * 0.5 = 0.5
Frequency of dominant phenotypes = frequency of homozygous dominant + frequency
of heterozygous dominant
Frequency of dominant phenotypes = 0.25 + 0.5 = 0.75
17. 1. There are 100 students in a class. 96 did well in the course whereas 4 blew it
totally and received a grade of F. In the highly unlucky event, that these traits are
genetic rather than environmental, if these traits involve dominant and recessive
alleles, and if the 4% represent the frequency of the homozygous recessive
condition, then calculate the following:
A. The frequency of the recessive allele
B. The frequency of the dominant allele
C. The frequency of the heterozygous individuals
18. 2. The next generation of lizards has 1092 individuals with green scales and 108
individuals with blue scales. Is the population in Hardy-Weinberg Equilibrium?
Solve for p and q.
19. 3. A very large population of randomly mating laboratory mice contains 35 % white
mice. White colouring is caused by the double recessive genotype aa. Calculate
allelic and genotypic frequencies for this population.
20. 4. If 98 out of 200 individuals in a population express the recessive phenotype,
what percent of the population would you predict would be heterozygotes?
21. 5. In a population that is in Hardy Weinberg equilibrium, the frequency of a
recessive allele for a certain hereditary trait is 0.20. What percentage of the
individual in the next generation would be expected to show the dominant trait?
22. 6. In a random mating population, frequency of disease causing recessive allele is
80%. What would be the frequency of carrier individual in population?
23. 7. The frequency of two alleles in a gene pool is 0.19 ( A) and 0.81 ( a) . Assume
that the population is in Hardy-Weinberg equilibrium and
(a) Calculate the percentage of heterozygous individuals in the population.
(b) Calculate the percentage of homozygous recessive in the population .
24. 8. A population of cats can be either black or white; the black allele (B) has
complete dominance over the white allele (b). Given a population of 1,000 cats,
840 black and 160 white, determine the allele frequency, the frequency of
individuals per genotype, and number of individuals per genotype.
25. 9. Scale coloration of lizards has a complete dominance relationship where green
scales are dominant over blue scales. There are 1,024 individuals with the
genotype GG, 512 individuals with the genotype Gg, and 64 individuals with the
genotype gg. Find: the frequency of the dominant and recessive alleles and the
frequency of individuals with dominant, heterozygous, and recessive genotype.
26. 10. Within a population of butterflies, the colour brown (B) is dominant over the
colour white (b). And, 40% of all butterflies are white. Given this sample
information, calculate the following:
A. The percentage of butterflies in the population that are heterozygous
B. The frequency of homozygous dominant individuals
28. Ans: 1
A. The frequency of the homozygous recessive genotype = 4%
i.e., q2 = 4% (i.e., 0.04)
q = √.04 = 0.2
B. The frequency of the dominant allele
If q = 0.2 and p+q = 1
The P= 1- 0.2 = 0.8
C. The frequency of the heterozygous individuals = 2 pq
= 2 x 0.8 x 0.2 = 0.32 = 32%
29. Ans: 2. Total lizards in this generation = 1092 +108 = 1200
q² = 108 /1200 = 0.09
q = 0.3
p= 1 – 0.3 = 0.7
No, the population is not in a state of Hardy-Weinberg Equilibrium because the
allele frequencies are not the same as the preceding generation.
30. Ans: 3
White mice (aa / q2) = 35% = 0.35
q = √0.35 = 0.59
P= 1-q = 1-0.59 = 0.41
Frequency of remaining genotypes (AA and Aa)
AA = p2 = 0.41 x 0.41 = 0.17 = 17%
Aa = 2pq = 2 x 0.41 x 0.59 = 0.48 = 48%
31. Ans : 4. To determine what the actual frequency is, simply divide 98/200=0.49.
q2=0.49.
q=0.7
p+0.7=1
p=0.3
Frequency of heterozygous genotypes=2pq
=2(0.3)(0.7)=0.42
Converting this into a percent, we see that 42% of the population is heterozygous.
32. Ans :5 According to the question, the frequency of recessive allele for a certain
hereditary trait = q = 0.20
The frequency of dominant allele : p+q=1, Thus, p=1 -q
P= 1 - 0.2 = 0.8
The frequency of homozygous dominant genotype in total population : p2 = (0.82)
= 0.64 or 64%".
The frequency of heterozygous dominant genotype in total population (2pq) = 2 X
0.8 X 0.2 = 0.32 or 32%".
Hence, the percentage of dominant individuals in next population = (64+32)% =
96%.
33. Ans: 6. The frequency of disease causing recessive allele is given i.e q=80% or o.8
If p+q = 1 then, p = q - 1 = 1 - 0.8 = 0.2 or 20%
Thus, the frequency of carrier will be = 2pq = 2 x 0.8 x 0.2 = 0.32 or 32%
34. Ans: 7. Frequency of allele (A) in a gene pool = 0.19
Frequency of allele (a) in a gene pool = 0.81
a) The number of heterozygous individuals
(Aa) is equal to 2pq which equals to
2 x 0.19 x 0.81 = 0.3078
Therefore in a population of 100, the number of heterozygotes are = 30.78 = 31
b) b) The homozygous recessive individuals (aa) are represented by the q² term in the
Hardy-Weinberg equilibrium equation which equals to
0.81 x 0.81 = 0.6561
Therefore in a population of 100, the number of homozygous recessive are = 65.61 =
66
35. Ans: 8. Frequency of white cats = 𝑞² = 160 / 1,000 = 0.16
𝑞 = 0.4
𝑝 + 𝑞 = 1
𝑝 = 1 − 𝑞
𝑝 = 1 − (0.4) = 0.6
The frequency of individuals with the dominant genotype: 𝑝² = 0.36
The frequency of individuals with the heterozygous genotype: 2𝑝𝑞 = 0.48
The frequency of individuals with the recessive genotype: 𝑞² = 0.16
𝑝² × 𝑡𝑜𝑡𝑎𝑙 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 0.36 × 1,000 = 360 black cats, BB genotype.
2𝑝𝑞 × 𝑡𝑜𝑡𝑎𝑙 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 0.48 × 1,000 = 480 black cats, Bb genotype
𝑞² × 𝑡𝑜𝑡𝑎𝑙 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 0.16 × 1,000 = 160 white cats, bb genotype.
36. Ans: 9. Total number of individuals in that population = 1024+ 512 +64 = 1600
GG (p²) = 1024 /1600 = 0.64
𝑝 = 0.8
𝑞 = 0.2
𝑝² = 0.64
2𝑝𝑞 = 0.32
𝑞² = 0.04
41. 1. If finches with large beaks and finches with small beaks increase in a
population, while finches with average sized beaks decreases, which pattern of
selection has occurred?
1) Directional selection
2) Disruptive selection
3) Stabilizing selection
4) Artificial selection
42. 2. Robins usually lay 4 eggs. If more than 4 eggs are produced, then the baby
chicks are malnourished, while smaller numbers of eggs result in no viable baby
chicks. This is an example of:
1) Directional selection
2) Disruptive selection
3) Stabilizing selection
4) Artificial selection
43. 3. What is the correct explanation for what is happening in this graph?
1) There are a lot of birds having average sized beaks
2) There are no birds having large beaks
3) There are a lot of birds having small beaks only
4) There are a lot of birds having small and large beaks, but no bird with average
sized beaks
44. 4. Scientists found that most feral cats in a neighbourhood have whiskers about 12
cm long. After about 15 generations of stabilizing selection, what would you
expect to see?
1) even more cats with 12 cm long whiskers
2) Most cats with less than 12 cm long whiskers
3) Most cats with more than 12 cm long whiskers
4) Most cats with whiskers either less than or more than 12 cm long, very
few with whiskers 12 cm.
45. 5. Human babies don’t do well if they are born too small with low birth weight (<2
kg weight) . If they are too big (>4.5 kg weight), complications can occur during
birth. By which method, baby size should then be selected for?
1) Directional selection
2) Disruptive selection
3) Stabilizing selection
4) Artificial selection
46. 6. When a group of deer migrated to a cold mountain. The thin furred deer and
average furred deer died. The thick furred deer survived and eventually the
population became thick furred. What type natural selection is this?
1) Directional selection
2) Disruptive selection
3) Stabilizing selection
4) None of these
47. 7. Small turtles are too weak to protect themselves from predators and large turtles
are too slow to hide from predators. Average sized turtles have high fitness. What
type natural selection is this?
1) Directional selection
2) Disruptive selection
3) Stabilizing selection
4) None of these
48. 8. An extreme phenotype is favoured causing the alleles to shift towards that
phenotype. This is:
1) Directional selection
2) Disruptive selection
3) Stabilizing selection
4) Artificial selection
49. 9. Genetic diversity decreases and individuals at the centre of the curve have
higher fitness. This is:
1) Directional selection
2) Disruptive selection
3) Stabilizing selection
4) Artificial selection
