Answers to Hardy Weinburg and to Chapter Review 23.1 - 23.8

Hardy Weinburg Problems:

1. "cc" is most significant in that it is the only genotype that is solely connected to the phenotype. In other words, the the dominant phenotype is exhibited by two genotypes: "CC" and "Cc" and you cannot distinguish them.

2. 1/1700 = .059%

3. a) we calculate "q" 1st because this is the only variable we CAN calculate initially by using the information that we have.

b) c = q = the square root of q squared = .024 or 2.4%

c) It is easy to now find "p" since p= 1- q:

d) 1 -(2.4%) = 97.6%

4. a) CC = p squared = (.976)(.976) = .953 or 95.3%

b) Cc = 2pq = 2(.024)(.976) = 4.7%

5. (.953)(1700) = approx 1620 have the CC genotype

6. (.0468)(1700) = approx 79 have the Cc genotype

7. The "c" allele would increase in the population due to heterozygous advantage. Those with a heterozygous genotype would better survive the disease and pass this allele to more offspring.

II. 9% = ss, therefore q suared = .09. q = the square root of this which would be .3 or 30 %. p = 1 - q = .7.
Therefore, the heterozygous condition, which = 2pq = 2(.7)(.3) = .42 or 42% are heterozygous.

III. In this problem, you are given all the information needed to calculate the allele frequency WITHOUT using the Hardy Weinburg equations. For example:
A= (2 x 1469) + 138 / 2(1469 + 138 +5) - in other words the total number of "A" alleles divided by the total number of alleles. This equals .954 or 95.4%
a = 1 - .954 = .046 or 4.6%
AA = 1469/1612 = .911 or 91.1%
Aa = 138/1612 =.086 or 8.6%
aa = 5/1612 =.003 or.3%
If you used the hardy Weinburg, you would find similar, but not exact answers, perhaps due to rounding errors.

Answers to interactive questions 23.1 - 23.8

23.1 a. freq of "B" allele: (2 x 98) + 84 / 400 =.7 freq of "b" allele: (2 x 18) + 84/400 = .3

b. Genotype frequency for BB = 98/200= .49 or 49%; for Bb = 84/200 = .42 or 42% ; bb= 18/200 = .09 or 9%

23.2 The answers to this question are the same as 23.1 as we are assuming this population is in Hardy Weinburg equilibrium - therefore the allele frequencies remain constant from one generation to the next.

23.3 a. AA = (.6) squared = .36 = 36%; aa = (.4) squared = .16 or 16%; Aa = 2(.6)(.4) = .48 or 48%
b. A = .6; a = .4 - there would be no change in allele frequency when in Hardy Weinburg equilibrium

23.4 a. mutation (with some recombination)
b. sexual recombination
c. Bacteria and viruses have very short generation times and a new beneficial mutation can increase in frequency rapidly
in an asexual reproducing population. Although mutations do introduce new alleles in a large, diploid population, they
are so infrequent that their contribution to genetic variation is minimal. However, sexual recombination in the
production and union of gametes produces zygotes with fresh combinations of alleles each generation.

23.5 a. natural selection b. genetic drift c. gene flow d. better reproductive success e. small population
f. bottleneck effect g. founder effect h. genetic variation between populations

23.6 a. less variation; we have about a tenth of the variability found in fruit fly populations
b. human have about 0.1% nucleotide diversity.

23.7 a. The two given fitness values are less than one, so Bb must produce the most offspring and have a relative fitness of 1
b. 0

23.8 a. Diploidy -The sickle cell allele is hidden from selection in heterozygotes
b. Heterozygote advantage - Heterozygotes are resistant to malaria and have a selective advantage in areas where malaria is a major cause of death