Is the slug population in HardyWeinberg equilibrium? the slug population has 2 a

Question

Is the slug population in HardyWeinberg equilibrium? the slug population has 2 alleles: slime 1(very slimy) and slime 2 ( barely slimy). They arecodominant. The total population consists of 200 very slimyslugs along with 400 barely slimy; there are also 400 slugs withmedium sliminess. 1. Calculate the actual population'sgenotype frequencies. 2. Calculate q and p for thepopulation. 3. Calculate expected frequencies ofeach genotype if the population is in HWequilibrium. Show all calculations! 4. Is the population in Hardy-Weinbergequilibrium? BONUS: What University's mascot is thefighting Banana Slugs? Is the slug population in HardyWeinberg equilibrium? the slug population has 2 alleles: slime 1(very slimy) and slime 2 ( barely slimy). They arecodominant. The total population consists of 200 very slimyslugs along with 400 barely slimy; there are also 400 slugs withmedium sliminess. 1. Calculate the actual population'sgenotype frequencies. 2. Calculate q and p for thepopulation. 3. Calculate expected frequencies ofeach genotype if the population is in HWequilibrium. Show all calculations! 4. Is the population in Hardy-Weinbergequilibrium? BONUS: What University's mascot is thefighting Banana Slugs?

Explanation / Answer

the slug population has 2 alleles:slime 1 (very slimy) and slime 2 ( barely slimy). They arecodominant. The total population consists of 200 very slimy slugs alongwith 400 barely slimy; there are also 400 slugs with mediumsliminess. Therefore, we have 3 genotypes: AA---200 Aa----400 aa----400 The genotype frequencies of the actual population is: AA = 200/1000       = 0.2 Aa = 400/1000      = 0.4 aa = 400/1000     = 0.4 p = frequency of the dominant allele in thepopulation
q = frequency of the recessive allele in the population Therefore, p[AA+Aa] = 200 x 2 + 400/1000 x 2                            = 800/2000                           = 0.4 p = 0.4 q[aa+Aa] = 400 x 2 + 400/1000 x 2                =1200/2000               = 0.6 q = 0.6 we see that p+q = 1 as 0.4+0.6 = 1 p2+2pq+q2 = 1 p2 = 0.16 q2 = 0.36 2pq = 0.48 Therefore the expected frequencies of each genotype is: AA = 0.16 x 1000 = 160 Aa = 0.48 x 1000 = 480 aa = 0.36 x 1000 = 360 WE can compare the actual and the HWE frequencies to knowwhether the population is in Hardy-weinberg Equilibrium ornot: Genotype Actual HWE AA 200 160 Aa 400 480 aa 400 360 No, the population is not in HW equilibrium. The fighting Banana Slugs is the mascot of the Santa cruz ofuniversity of California Genotype Actual HWE AA 200 160 Aa 400 480 aa 400 360

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