Hardy-Weinberg Practice Problems

Hardy-Weinberg Practice Problems

Sickle Cell Anemia, Cystic Fibrosis, Tay-Sachs, and PKU are recessive disorders.

1. 16% of a population has sickle cell anemia. 84% are normal.

a. What is the frequency of Ss?

b. What is the frequency of S?

c. What is the frequency of ss?

d. What is the frequency of SS?

e. What is the frequency of s?

2. 81% of a population has Tay-Sachs. 19% are normal.

a. What is the frequency of TT?

b. What is the frequency of Tt?

c. What is the frequency of T?

d. What is the frequency of tt?

e. What is the frequency of t?

3. 49% of a population has cystic fibrosis. 51% are normal.

a. What is the frequency of C?

b. What is the frequency of c?

c. What is the frequency of CC?

d. What is the frequency of Cc?

e. What is the frequency of cc?

4. 36% of the population has PKU. 64% are normal.

a. What is the frequency of Pp?

b. What is the frequency of p?

c. What is the frequency of P?

d. What is the frequency of PP?

e. What is the frequency of pp?

5. 96% of a population is normal. 4% has cystic fibrosis.

a. What is the frequency of c?

b. What is the frequency of CC?

c. What is the frequency of C?

d. What is the frequency of cc?

e. What is the frequency of Cc?

Answers are on the back.

1. 16% have SCA so

0.16 = ss = q2

q = √q2 = √0.16 = 0.4 = s

p = 1 - 0.4 = 0.6 = S

p2 = (0.6)2 = 0.36 = SS
2pq = 2 (0.6) (0.4) = 0.48 = Ss

a. 0.48
b. 0.6
c. 0.16
d. 0.36
e. 0.4

2. 81% have Tay-Sachs so

0.81 = tt = q2

q = √q2 = √0.81 = 0.9 = t

p = 1 - 0.9 = 0.1 = T

p2 = (0.1)2 = 0.01 = TT

2pq = 2 (0.1) (0.9) = 0.18 = Tt

a. 0.01
b. 0.18
c. 0.1
d. 0.81
e. 0.9

3. 49% have cystic fibrosis so

0.49 = cc = q2

q = √q2 = √0.49 = 0.7 = c

p = 1 - 0.7 = 0.3 = C

p2 = (0.3)2 = 0.09 = CC

2pq = 2 (0.3) (0.7) = 0.42 = Cc

a. 0.3
b. 0.7
c. 0.09
d. 0.42
e. 0.49

4. 36% have PKU so

0.36 = pp = q2

q = √q2 = √0.36 = 0.6= p

p = 1 - 0.6 = 0.4 = P

p2 = (0.4)2 = 0.16 = PP

2pq = 2 (0.4) (0.6) = 0.48 = Pp

a. 0.48
b. 0.6
c. 0.4
d. 0.16
e. 0.36

5. 96% are normal so 4% have cystic fibrosis so

0.04 = cc = q2

q = √q2 = √0.04 = 0.2= c

p = 1 - 0.2 = 0.8 =C

p2 = (0.8)2 = 0.64 = CC

2pq = 2 (0.8) (0.2) = 0.32 = Cc

a. 0.2
b. 0.64
c. 0.8
d. 0.04
e. 0.32