Problems related to hardy weinberg law

1. If 98 out of 200 individuals in a population express the recessive phenotype, what percent of the population would you predict would be heterozygotes?Figure out the % of homozygous dominant?

q2 = (98/200) = 0.49 (or 49%)
q =  0.7 (70%)
 As p + q = 1.
p = 1-q
p = 1 - 0.7 = 0.3 (30%)

% of homozygous dominant-

p2 = (0.3)(0.3) = 0.09 (or 9%)

What is the frequency of heterozygotes?


Frequency of heterozygotes = 2pq
                                          = 2X0.3X0.7
                                          =0.42
p2 + 2pq + q2 = 1
0.09 + 0.42 + 0.49 = 1.0(PROVED)

2. Red hair is a recessive trait in humans. In a 

randomly mating population in Hardy-
Weinberg equilibrium, approximately 9% of 
individuals are red-haired. What is the 
frequency of heterozygotes? 
1. 81% 2. 49%

3. 42% 4. 18%
Ans.Given - recessive trait (9%)
q2=9/100=0.09
q=0.3
We know that p+q=1
So,p= 1-q=0.7
Now, the question is to find out frequency of heterozygous i,e 2pq
=2*0.3*0.7
=0.42
So, the correct option is 3.

3.In a population which is in Hardy-Weinberg equilibrium, the frequency of a
recessive allele for a certain genetic trait is 0.40. What percentage of individuals
would be expected to show the dominant trait in the next generation?
(A) 16%
(B) 32%
(C) 84%
(D) 96%

Ans.c(0.4=q recessive gene, so p=0.6 dominant. As p+q=1formula is: p2+q2+2pq=1. (0.4)2+ (0.6)2 + 2*0.4*0.6=1. dominant % is 0.36+0.48=0.84 that is  84%

4.In a population at Hardy-Weinberg equilibrium ,the genotype frequencies are 
f(A1A1 )=0.59 ,f(A1A2)= 0.16 ,f(A2A2)=0.25.What are the frequencies of the two alleles at this locus?
1.A1=0.59  A2= 0.41
2.A1=0.75  A2=0.25
3.A1=0.67 A2= 0.33
4.A1=O.55 A2=0.44

Answer-
A1A1=p2= 0.59 
A1A2=2pq=0.16
A2A2=q2=0.25
A1=p=p2+1/2 2pq=0.59+0.08=0.67(Ans)
q=1-p
  =1-0.67
  =0.33 (Ans)

5.In a random sample of 400 individuals from a population with allele of trait in Hardy weinberg equilibrium 36 individuals are homozygous for allele a.How many individuals in the sample are expected to carry atleast one allele A?
A.36                                                 C.364
B.168                                               D.196                  

Ans.364 ( Let us say q is the frequency of allele a and p is the allele frequency of allele A.p+q=1
36 in 400 individuals are aa=q^2=36/400=0.09 q=0.3
p= 0.7 
Individuals carrying at least one AA and Aa.
Their frequency will be p^2+2pq=0.49+ 2.0.7.0.3=0.91
Therefore 400x 0.91 =364  individuals with either AA or Aa.

6.The following genotypes are found in a population AA:Aa:aa::60:20:20. What are the the allele frequencies of A and a?

a.A=0.7 and a=0.3

b.A=0.6 and a=0.2

c.A=0.8 and a=0.2

d.A=0.3 and a=0.7

Ans.The number of individuals = 100 
The number of alleles = 200 (because they are diploid) 
Calculate p(A) from homozygotes AA and heterozygotes Aa, and do for q(a) as well. 
p(A) = ((2x60) + 20)/200 = 0.7 
q(a) = ((2x20) + 20)/200 = 0.3 
so the answer is
 
a) A=0.7 ; a=0.3

7. If 16% of newborns in a population have sickle cell anemia what percentage of the population will be more resistant to malaria-

a.84%       b.36%                  c.24%  d.     48%

Ans. D (48%) 

Homozygous Hb^s Hb^s= sickle cell anemia (die)
HeterozygousHb^N Hb^s=Normal(survive)
if 16%=Hb^s Hb^s q=16%=.4  p+q=1 So, p=.6 2pq=2*.6*.4=.48= 48%

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