The HW model (and its assumption of random mating) is often used for genetic cou
ID: 276400 • Letter: T
Question
The HW model (and its assumption of random mating) is often used for genetic counseling. For example, one birth in a population of 1800 individuals is homozygous recessive (q2) for cystic fibrosis (genotype frequencies are at HW equilibrium at birth). Answer the following questions: 5) (4 pts.) What is the frequency of the q allele? 6) (4 pts.) How many individuals are predicted to be heterozygotes (2pg)? 7) bonus question (2 pts) What is the probability of marriage between heterozygotes? 8) (6 pts) 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(p) (AA) blood, 75 have type AB (AB) blood, and 25 have type B(q) (BB) blood, what are the allele frequencies of this population (i.e., what are p and q)? Do not assume HW equilibrium.Explanation / Answer
Answer:
5). Frequency of q^2 = 1/1800 = 0.00056
Frequency of allele, q= SQRT of q^2 = 0.024
6). As the q = 0.0024
The frequency of allele, p = 1-0.024 = 0.976
The frequency of heterozygotes = 2pq = 2 * 0.976 * 0.024 = 0.047
Number of heterozygous individuals = 2pq * total number of individuals = 0.047 * 1800 = 85
7). The probability of marriage between heterozygotes = 0.047 * 0.047 = 0.002 = 0.2%
8).
AA = 200
AB = 75
BB = 25
Total “A” alleles = 200+200+75= 475
Total “B” alleles = 25+25+75=125
Total alleles = 475+125=600
The frequency of allele “A” = 475/600 = 0.79
The frequency of allele “B” = 125/600 = 0.21
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