Title: Answers - HW problems
Course: BIO-182 RTZ

Answers to Hardy-Weinberg Problems (dragons)

1. a) 7/21 = 0.33 = q^2

b) √.33 = 0.57 = q

c) p = ?   p+q =1

p + 0.57 = 1

p = 1 - 0.57

p = 0.43

d) p^2 = (0.43)^2

p^2 = 0.18

e) 2pq = 2 (0.43)(0.57) = 0.49

frequency (f) of heterozygotes = 0.49

2.  %aa = 0.36

a)      “aa” = 0.36                  (q^2) = 36%

b)     “a” = 0.60                    (q) = 60%

c)      “A” = 0.40                   (p) = 40%

d)     “AA” = 0.16                                (p^2) = 16%

Aa = 2pq = 2(0.40)(0.60) = 0.48 or 48%

e)      Because A is dominant over a, dominant phenotypes will appear if genotypes are AA or Aa;

recessive phenotype = aa. Therefore f of a dominant phenotype = Aa + aa = 0.16 + 0.48 = 0.64 or 64%. Frequency of homozygous recessive = 0.36 or 36%.

3.  If 0.9% of population is ss, what % of population is Ss

a.  s = 0.09 = q^2  &  √0.09 = 0.3 = q

b.  p+q=1   à  p + 0.3 = 1  à p = 0.7

c.  2pq = 2(0.7)(0.3) = 0.42

d.  0.42 or 42% of population = Ss

4. 4% is frequency of homozygous recessive genotyp

a.  f of r = 0.20 = 20%

b. f of R = 0.80 = 80%

c.  2pq = 2 (0.20)(0.80) = 0.32 = 32%

5. 40% are white therefore 0.40 is frequency of bb

a. ? % of Bb

0.40 = bb so b = 0.63 = q

if b = 0.63, then B = 0.37   and B = p

so 2(0.63)(0.37) = 0.47 or 47% (frequency of Bb)

b. ? % of BB (homozygous dominant)

From above: B = 0.37

So B^2  = p^2 = (0.37)^2 = 0.14 or 14% of population is BB

6.  Twenty (20) people, therefore 40 alleles & 2 of these alleles are c

So,  2/40 = 0.05 of all alleles = c or q; therefore 0.05 = q

q^2 = (0.05)^2 = 0.0025 = 0.25% of population may potentially have cystic fibrosis.

7.  1/2500 babies have cystic fibrosis; therefore they are homozygous recessive.

1/2500 = qq = 0.0004

Frequency of recessive allele = √q^2  = q

therefore √0.0004 = 0.02

Frequency of dominant allele

p + q = 1

p + 0.02 = 1

p = 1 - 0.02

p = 0.98

2(pq) = 2(0.02)(0.98) = 0.04

8.  Will vary