AP Biology Lab 8 Hardy-Weinberg problems? (Includes the questions)? - ap bio lab 8 answers hardy-weinberg problems 5 and 6
I need help with # 3, 4 a.m. to 5 p.m. in the problems of the Hardy-Weingberg lab AP Bio 8th Please give us the answer and explain each step, how to get it.
3) The allele for the hair as a "widow's peak" is dominant over the allele for non-peak "widows." In a population of 1,000 people, 510 show the dominant phenotype. How many people are expected, all three possible genotypes for this character?
4) In the United States and about 16 of the population is Rh-negative. The Rh-negative allele is recessive to the allele of the Rh-positive. If the population of secondary students in the United States is 2000, how many students can choose from any of the three expected genotypes are possible?
5) In some African countries than 4% of newbornsSickle cell anemia, a recessive trait. In a random population of 1000 newborns, how many can be used for each of the three genotypes expected are possible?
Thank you very much!
Friday, February 12, 2010
Ap Bio Lab 8 Answers Hardy-weinberg Problems 5 And 6 AP Biology Lab 8 Hardy-Weinberg Problems? (Includes The Questions)?
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First, some background information. If we have the dominant allele frequency peak (widow) of P represented and represented the frequency of the recessive allele (No Widow's Peak) from q, then the sum of the frequencies of the alleles is should be:
p + q = 1
If you square both sides of the equation, we get:
(p + q) ^ 2 = 1 ^ 2
(p + q) ^ 2 = 1
(p + q) x (P + Q) = 1
p ^ 2 + 2PQ + Q ^ 2 = 1
where p ^ 2 is the dominant frequency of homozygotes is the frequency of heterozygotes is 2pq, and q ^ 2 is the frequency of recessive homozygotes in the population. If you look at a different angle, is the probability that a person has (two recessive alleles of one homozygous recessive) is the product of the probability thatAllele is this person:
DXD = q ^ 2
The same reasoning in determining the frequency of homozygous dominant (p ^ 2) and heterozygous (2pq).
But their problems ...
3) If more than 1000 people and 510, the dominant phenotype, which means that have:
1000 - 510 = 490 have the recessive phenotype, so
q ^ 2 = 490/1000 = 0.49
q = 0.7
and
p + q = 1
p = 1 - q
p = 1 to 0.7
p = 0.3
Thus, the frequency of each genotype:
p ^ 2 = 0.3 x 0.3 = 0.09 (frequency of homozygous dominant)
2 pq = 2) x 0.7 x 0.3 = 0.42 (frequency of heterozygotes
q ^ 2 = 0.49 (frequency of homozygous recessive)
Finallydetermine "the number of people ..." You multiply the frequencies by the total number of persons in the community:
Homozygous dominant: 1000 x 0.09 = 90
Heterozygous: 1000 x 0.42 = 420
Homozygous recessive: 1000 x 0.49 = 490
Problems with 4 a.m. to 5 p.m. are calculated the same way.
4) Q ^ 2 = 16% = 0.16, then q = 0.4, so that p = 0.6, thus:
p ^ 2 = 0.36, ie 0.36 * 2000 = 720 homozygous dominant (Rh + / Rh +)
2pq = 0.48, ie 0.48 * 2000 = heterozygous 960 (Rh + / Rh)
q ^ 2 = 0.16, ie 0.16 * 2000 = 320 homozygous recessive (Rh-/Rh-)
Was 5) I no longer do it on your own, good luck!
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