Question

Cystic fibrosis is a genetic disorder in homozygous recessives that causes death during the teenage years. If 9 in 10,000 newborn babies have the disease, what are the expected frequencies of the dominant (A1) and recessive (A2) alleles according to the Hardy—Weinberg model?

Answer 1

Explanation:

Sequences of DNA make up genes which can have different forms called alleles. DNA, which makes up the genotype, is transcribed into mRNA and later translated into amino acids which are linked together by rRNA to form proteins which make up the phenotype of an organism. Mutations in DNA sequences form new alleles, and affect the corresponding mRNA and thus the protein encoded.  Along with genetic drift, selective mating and natural selection, evolution may occur within populations.

Some alleles become fixed within a population over time. In Hardy-Weinberg equilibrium for a specific gene is not undergoing evolution- allelic frequencies are stable over several generations. During this period, mutations do not occur within the population, this is shown as

$p^{2} + 2pq +q^{2} = 1$

where...

• p represents the dominant allele frequency
• q represents the frequency of the recessive allele
• q^2 = the genotypic frequency of the homozygous recessive
• p^2= the genotypic frequency of the homozygous dominant

for the homozygous recessive disease CF, 9/10000 = 0.0009 = q^2

∴ $q= \sqrt{0.0009} = 0.03$

Thus, since p + q = 1

                   p + 0.03 = 1

                   p = 1 - 0.03= 0.97

f(A1) the dominant allele = 0.9700

f(A2) the homozygous recessive allele = 0.0300

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