Hardy-Weinberg Assignment I
1. The frequency of the A allele is .65 in the Mad Hatter population of squirrels. What is the frequency of the different genotypes in the Mad Hatter population?
The frequency of a is .35, so the frequency of AA is (.65)2 and Aa is 2(.65)(.35) and aa is (.35)2 Making AA 42%; Aa 46%; aa 12%
2. There are 1260 squirrels in the Mad Hatter population. How many are heterozygous for the A allele?
Multiply the frequency of Aa and the population (.46)(1260) = 580 heterozygous squirrels
3. A population of finches lives on one of the Galapagos Islands. 76 have the CC genotype, 82 have the Cc genotype and 22 have the cc genotype. What is the frequency of the C allele in this population?
Determine the total population (76+82+22 = 180) Double the population number to get the total number of alleles at the c location 2*180 = 360). Now determine the number of C alleles in the population. CC has 152 C alleles and Cc has 82 C alleles. Total number of C alleles is 234. Now divide the number of C alleles by the total number of alleles. 234/360 = .65
4. A population of 210 snow leopards in Afghanistan have an unusual light gray coat color that is due to the homozygous recessive allele "g". The frequency of this allele is .1. How many snow leopards have this allele?
The frequency of the G allele is 1 - .1 = .9 Now find the frequency of the different genotypes. GG is (.9)2 = .81 Gg is 2(.9)(.1) = .18 and gg is (.1)2 = .01 Now determine the number of individuals that have these genotypes by multiplying the frequency by the population. GG is (.81)(210) = 170 Gg is (.18)(210) = 38 and gg is (.01)(210) = 2. Add the two genotypes up that have the g allele 38 + 2 = 40 snow leopards have the rare g allele.
5. In a study of 4550 fruit flies, scientists found that the frequency of the F allele was .8. What was the number of fruitflies that made up the different genotypes?
If F is .8, then f is .2 Use the HW equation to determine the frequency of fruit flies of each genotype and then multiply the frequencies by the population size.
FF is (.8)2 = .64 Ff is 2(.8)(.2) = .32 and ff is (.2)2 = .04 Multiplying the total population by the frequencies will give the number of individuals with that particular genotype. FF = (.64)(4550) = 2912 fruit lies Ff = (.32)(4550) = 1456 ff = (.04)(4550) = 182