Question 7. A scientist is studying a population of butterfly fish, a species co

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Question 7. A scientist is studying a population of butterfly fish, a species commonly observed on coral reefs. She has discovered a set of alleles: the dominant form gives the fish a white color, while the recessive allele causes a slight green color. She samples the reef population and observes the following: out of 2500 fish, she counted 225 that were green. Estimate the allele frequencies for these two alleles. List the information given: AA and Aa = white, aa = green; 225 green fish out of 2500 List the term(s) to find: p, q Question 8. Five years later she returns to the same reef and performs another count. This time she counts 2000 fish, and 720 are green. Estimate the allele frequencies for the population five years after the initial count. Has evolution occurred? If so, suggest a possible reason. List the information given: AA and Aa 2000 white, aa-green; 720 green fish out of List the term(s) to find: p, q Hardy-Weinberg Problems and Post Lab Questions Question 9. If the frequency of a dominant allele in a population in equilibrium is 0.35, then the frequency of the corresponding recessive allele in the same population is: List the information given: List the term(s) to find:

Explanation / Answer

Ans. 7 As per Hardy - Weinberg equation

P2 + 2pq + q2 = 1, p + q = 1

where p = frequency of dominant allele, q = frequency of recessive allele

q2 = aa = 225/2500*100 = 9% = 0.09

q = 0.3

As p+q = 1 therefore p = 0.7

Hence p2 = 0.7 * 0.7 = 0.49, hence AA = p2 = 49%

So 2pq = 2 * 0.7*0.3 = 0.42 = 42%

Hence aa = green fish = p2 = 9%

AA = white fish = q2 = 49%

Aa = heterozygous = 2pq = 42%

Ans 8. q2 = aa = 720/2000*100 = 0.36 = 36%

q = 0.6

p = AA = 1-q = 1 - 0.6 = 0.4

p2 = 0.4*0.4 = 0.16 = 16%

Aa = 2pq = 2*0.4*0.6 = 0.48 = 48%

The most possible reason for those rise in green coloured fish may be due to evolution as a result of which the recessive may be expressed in umber of offsprings

Ans 9. As per the equation p + q = 1

hence recessive allele = q = 1-p = 1-0.35 = 0.65

therefore the frequency of the corresponding recessive allele would be 0.65.

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