A large population of fish live in a lake. The color of their scales is determined by two different alleles of the autosomal S gene, S1 and S2. Homozygous S1S1 fish have yellow scales, S2S2 homozygotes have blue scales, and S1S2 heterozygotes are green. Scientists caught 100 fish at random and recorded their color. Among those 100 fish were 30 yellow, 50 blue, and 20 green ones. After the original sample was analyzed, an asteroid landed in the lake and killed all the fish that happened to be in one particular location. When 100 fish were sampled again, 15 were yellow, 75 were blue, and 10 were green. Was the fish population in Hardy-Weinberg equilibirim (HWE) before the asteroid hit? In an ideal population, how many generations would it take to establish HWE after the asteroid hit?

Question

Rowan Davies

Biology

11 February, 07:56

A large population of fish live in a lake. The color of their scales is determined by two different alleles of the autosomal S gene, S1 and S2. Homozygous S1S1 fish have yellow scales, S2S2 homozygotes have blue scales, and S1S2 heterozygotes are green. Scientists caught 100 fish at random and recorded their color. Among those 100 fish were 30 yellow, 50 blue, and 20 green ones. After the original sample was analyzed, an asteroid landed in the lake and killed all the fish that happened to be in one particular location. When 100 fish were sampled again, 15 were yellow, 75 were blue, and 10 were green. Was the fish population in Hardy-Weinberg equilibirim (HWE) before the asteroid hit? In an ideal population, how many generations would it take to establish HWE after the asteroid hit?

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Answers (1)

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Rory Salinas · Answer 1

Yes it was

two generations

Explanation:

H0: population is in Hardy-Weinberg equilibrium

Ha: population is not in Hardy-Weinberg equilibrium

P (S1) = (2*obs (S1S1) + obs (S1S2)) / (2*fish (S1S1) + 2*fish (S1S2) + 2*fish (S2S2))

P (S1) = (60+50) / (60+100+40)

P (S1) = 0.55

P (S2) = 1-P (S1)

P (S2) = 0.45

Exp (S1S1) = 0.55*0.55*100=30.25

Exp (S1S2) = 2*0.45*0.55*100=49.5

Exp (S2S2) = 0.45*0.45*100=20.25

x²=∑ (Observed-Expected) ² / Expected

= (30-30.25) ²/30.25 + (50-49.5) ²/49.5 + (20-20.25) ²/20.25

= 0.010203

The value at 5% significance level and degree of freedom 1 from table is 3.84. Since x² value is less than value we accept null hypothesis

It takes two generations to achieve Hardy-Weinberg equilibrium if conditions of Hardy-Weinberg are met.