POB II Practicing Hardy-Weinberg 1. Assuming that there are only two alleles at
ID: 146063 • Letter: P
Question
POB II Practicing Hardy-Weinberg 1. Assuming that there are only two alleles at a given locus. what is the frequency of the other allele? the frequency of one allele is 0.6, 2. In a hypothetical population of sawilies, 20 percent of the population is o are the only gous for tof the Po that allele A and 45 percent is homozygous for allele a. Assuming thata at this locus, what percent of the population is heterozygous A and a are the only alleles 3. In a hypothetical population of tree swallows, 18 individuals are homozygous for the c4 allele, is the frequency of the c4 allele? 22 individuals are heterozygous for the allele, and 10 individuals lack the allele. What lation of 200 individuals, 72 are homozygous recessive for the character of eye color One hundred individuals from this population die from a fatal disease. Thirty-six of the (cc). survivors are homozygous recessive What is the frequency of the dominant allele in the original population? a.Explanation / Answer
Answer:
1. Assuming Hardy Weinberg equilibrium, p and q are the allele frequencies, if one of them, p=0.6 (given), the other allele frequency, q=0.4. since p+q=1 and q=1-p; q=1-0.6; Thus q=0.4.
2. AA=p2=20% or 0.2
aa=q2=45% or 0.45
p2=0.2 ; q2=0.45
According to Hardy Weinberg euilibrium, p2+q2+2pq=1
p2+q2+X=1; 0.2+0.45+X=1; X=1-0.65; X=0.35; i.e Heterozygote genotypic frequency is 0.35 or 35%.
3. Individuals homozygous for c4 allele = 18
Individuals heterozygous for c4 allele = 22
Individuals lacking the c4 allele =10
So 18+22+10=50
Therefore, 18/50=0.36
22/50 =0.22; but since in heterozygote condition, only half of them are carrying the c4 allele we divide, 0.22/2=0.11
Therefore c4 allele frequency = 0.36+0.11=0.47.
4.
(a) Frequency of dominant allele 'p' in original population is:-
cc = 72/200 = 0.36 ; homozygous recessive q2= 0.36 or q= 0.6.
Since p+q=1; p=1-q or 1-0.6; p=0.4
(b) In the new population, 100/200 are dead, that is 100 more survivors are remaining.
Therefore, homozygous recessive cc= 36/100 (given).
q2=0.36 or q= 0.6 and sice p+q=1; p =1-q or 1-0.6 or 0.4
Thus even in new population, frequency of dominant allele remains 0.4.
(c) Heterozygous individuals = 2pq= 2*0.4*0.6= 0.48 or 48% individuala are heterozygous in new population.
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