EVOLUTION Name: _____________________________ BIOL 1001 Fall 2014 Hardy Weinberg
ID: 67466 • Letter: E
Question
EVOLUTION Name: _____________________________ BIOL 1001 Fall 2014
Hardy Weinberg problems
PROBLEM #1.
You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:
The frequency of the "aa" genotype.
The frequency of the "a" allele.
The frequency of the "A" allele.
The frequencies of the genotypes "AA" and "Aa."
The frequencies of the two possible phenotypes if "A" is completely dominant over "a."
PROBLEM #2.
Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the following.
The frequency of the recessive allele in the population.
The frequency of the dominant allele in the population.
The percentage of heterozygous individuals (carriers) in the population.
PROBLEM #3.
In a given population, only the "A" and "B" alleles are present in the ABO system; there are no individuals with type "O" blood or with O alleles in this particular population. If 200 people have type A blood, 75 have type AB blood, and 25 have type B blood, what are the alleleic frequencies of this population (i.e., what are p and q)?
Your Answers:
1A
1B
1C
1D
1E
2A
2B
2C
3A
p=
3B
q=
1A
1B
1C
1D
1E
2A
2B
2C
3A
p=
3B
q=
Explanation / Answer
Solution 1
The frequency of (aa) is 36/100 = 0.36
Th frequency of an allele is the square root of .36 is equal to 0.6.
The frequency of genotype aa = 0.36 and AA = 1-0.36 = 0.64
Th frequency of A allele is the square root of .64 is equal to 0.8.
Solution 2
The frequency of recessive gene = 1/2500 = 0.0004
The frequency of recessive allele = under root 0.0004 = 0.02
The frequency of dominant allele = 0.98
The percentage of carrier is = 2pq = 2 x 0.98 x 0.02 *100 = 0.392*100 = 39.2%.
Solution 3
The frequency of Total number of individuals in the population = (200+75+25) = 300
the frequency with gene A blood group (p) = 200/300 = 0.6, so allele frequency = under root 0.6 = 0.77
the frequency with allele showing B blood group (q) = 1- 0.77 = 0.23
the frequency with allele showing AB blood group (2pq) = (2 x p x q) 2 * 0.77 * 0.23 = 0.35.
p & q are the allele frequency representing the concerned gene.
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