sickel cell anemia is caused by recessive allele. frequency of sicke.pdf

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sickel cell anemia is caused by recessive allele. frequency of sickel cell allele is 0.1. according to Hardy-Weinberg law frequency of sickel cell allele (Recessive allele) (ss) = 0.1 = q2 q = square root of 0.1 = 0.316 then p = 1 - q = 1 - 0.316 = 0.684 p = frequency of healthy allele = 0.684 Frequency of healthy allele = 0.684 2.Ans. how many people in a population of 10,000 will have sickel cell anemia. sickel cell anemia (ss) = (0.316)(0.316)(10,000) = 999 people have sickel cell anemia. 3.Ans. how many people in a population of 10,000 will be sickel cell carriers. sickel cell carriers (Ss) = 2pq = 2 (0.316)(0.684)(10,000) = 4323 people are sickel cell carriers 4.Ans. mutations occurs and new allele introduced into the population. small population which causes genetic drift. the above TWO conditions we would NOT expect Hardy-Weinberg equlibrium to be maintained. 3. Alkaptonuria caused by recessive allele. survey revels that 5 in 50 people have Alkaptonuria a. if a population of 50 people if is caused by Recessive allele and 5 in 50 people have Alkaptonuria = q2 is 5/50 or 0.1 q2 = 0.1 then q = square root of 0.1 = 0.316 q = 0.316 p = 1 - q =0.684 p = 0.684 b. Heterozygous to Alkaptonuria = 2pq 2pq = 2(0.684) (0.316) (50) = 22 people are Hetrozygous for Alkaptonuria Allele C. Homozygous Dominant = AA =p2 AA = (0.684)(0.684)(50) = 23 people have Heterozygous Dominant Solution sickel cell anemia is caused by recessive allele. frequency of sickel cell allele is 0.1. according to Hardy-Weinberg law frequency of sickel cell allele (Recessive allele) (ss) = 0.1 = q2 q = square root of 0.1 = 0.316 then p = 1 - q = 1 - 0.316 = 0.684 p = frequency of healthy allele = 0.684 Frequency of healthy allele = 0.684 2.Ans. how many people in a population of 10,000 will have sickel cell anemia. sickel cell anemia (ss) = (0.316)(0.316)(10,000) = 999 people have sickel cell anemia. 3.Ans. how many people in a population of 10,000 will be sickel cell carriers. sickel cell carriers (Ss) = 2pq = 2 (0.316)(0.684)(10,000) = 4323 people are sickel cell carriers 4.Ans. mutations occurs and new allele introduced into the population. small population which causes genetic drift. the above TWO conditions we would NOT expect Hardy-Weinberg equlibrium to be maintained. 3. Alkaptonuria caused by recessive allele. survey revels that 5 in 50 people have Alkaptonuria a. if a population of 50 people if is caused by Recessive allele and 5 in 50 people have Alkaptonuria = q2 is 5/50 or 0.1 q2 = 0.1 then q = square root of 0.1 = 0.316 q = 0.316 p = 1 - q =0.684 p = 0.684 b. Heterozygous to Alkaptonuria = 2pq 2pq = 2(0.684) (0.316) (50) = 22 people are Hetrozygous for Alkaptonuria Allele C. Homozygous Dominant = AA =p2 AA = (0.684)(0.684)(50) = 23 people have Heterozygous Dominant.

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q = square root of 0.1 = 0.316
then p = 1 - q = 1 - 0.316 = 0.684
p = frequency of healthy allele = 0.684
Frequency of healthy allele = 0.684
2.Ans.
how many people in a population of 10,000 will have sickel cell anemia.
sickel cell anemia (ss) = (0.316)(0.316)(10,000) = 999 people have sickel cell anemia.
3.Ans.
how many people in a population of 10,000 will be sickel cell carriers.
sickel cell carriers (Ss) = 2pq = 2 (0.316)(0.684)(10,000) = 4323 people are sickel cell carriers
4.Ans.
mutations occurs and new allele introduced into the population.
small population which causes genetic drift.
the above TWO conditions we would NOT expect Hardy-Weinberg equlibrium to be maintained.
3. Alkaptonuria caused by recessive allele.
survey revels that 5 in 50 people have Alkaptonuria
a. if a population of 50 people
if is caused by Recessive allele and 5 in 50 people have Alkaptonuria = q2 is 5/50 or 0.1
q2 = 0.1 then q = square root of 0.1 = 0.316
q = 0.316
p = 1 - q =0.684
p = 0.684
b. Heterozygous to Alkaptonuria = 2pq
2pq = 2(0.684) (0.316) (50) = 22 people are Hetrozygous for Alkaptonuria Allele
C. Homozygous Dominant = AA =p2
AA = (0.684)(0.684)(50) = 23 people have Heterozygous Dominant

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