u observed an initial population 1000 individuals and obtained the following res
ID: 136013 • Letter: U
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u observed an initial population 1000 individuals and obtained the following results wn haired rabbits and 250 white rabbits apply the hardy weinberg equation and determine the 750 bro number of individuals mozyeous dominant individuals, heterorygous individusls and homorvsous recessie After resurveying the population after 25 generations, you obtain the results of 850 brown haired ndividuals and 150 white individuals apply the hardy Weinberg equation and determine the number of homozygous dominant individuals, heterozygous individuals and homozygous recessive individuals. What type of natural selection is occurring? 9. What type of natural selection is occurring if a population of rabbits have initial gene frequencies of AA 25 (Brown), Aa(Brown)-5 and aalWhite]-.25 and gene frequencies of AA.2(Brown), Aa-.6(Brown) and aa-.2(White) after 25 generations? 10. What type of natural selection is occurring if a population of rabbits have initial gene frequencies of AA-.25 (Brown), Aa(Brown)- S(tan) and aalWhite).25 and gene frequencies of A-3(Brown), Aa 4(Tan) and aa 3(White) after 25 generationsExplanation / Answer
The Hardy-Weinberg equation is expressed as: p2 + 2pq + q2 = 1;
where ‘p’ is the frequency of the "A" allele and q is the frequency of the "a" allele in the population. In the equation, p2 represents the frequency of the homozygous genotype AA, q2 represents the frequency of the homozygous genotype aa, and 2pq represents the frequency of the heterozygous genotype Aa. In addition, the sum of the allele frequencies for all the alleles at the locus must be 1, so p + q = 1.
Ans 8: lets consider 750 brown are both ‘AA’ and ‘Aa’ and 250 are ‘aa’. If we consider the above formula ‘q’ = 250 (aa), hence ‘p’=250 (A) then ‘pq’ is 500 (Aa) and ‘p’ is 250 (AA)
Homozygoes brown = 250, Heterozygoes brown = 500, Homozygoes white = 250
Frequency 0.25+0.5+0.25 = 1; Type of natural selection is stabilising selection
After 25 generations 150 white that means ‘q’ = 150 (aa), hence ‘p’=150 (A) then ‘pq’ is 300 (Aa) and ‘p’ is 550 (AA)
Homozygoes brown = 550, Heterozygoes brown = 300, Homozygoes white = 150
Frequency 0.55+0.3 +0.15 = 1; Type of natural selection is directional selection
Q9) In the first case Type of natural selection is stabilising selection
In the second case Type of natural selection is directional selection
Q10) In the first case Type of natural selection is disruptive selection (because tan frequency is 0.5)
In the second case Type of natural selection is disruptive selection (because tan frequency is 0.4)
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