Is the slug population in Hardy Weinbergequilibrium? the slug population has 2 a
ID: 4488 • Letter: I
Question
Is the slug population in Hardy Weinbergequilibrium?
the slug population has 2 alleles: slime 1 (very slimy) andslime 2 ( barely slimy). They are codominant.
The total population consists of 200 very slimy slugs along with400 barely slimy; there are also 400 slugs with mediumsliminess.
1. Calculate the actual population's genotypefrequencies.
2. Calculate q and p for the population.
3. Calculate expected frequencies of each genotype ifthe population is in HW equilibrium.
Show all calculations !
4. Is the population in Hardy-Weinbergequilibrium?
BONUS: What University's mascot is the fighting BananaSlugs?
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 thepopulationq = 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 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
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