Question

1a. A human protein has a run of 14 glutamine residues in a row, leading to a 42 bp repeat sequence in an exon that is prone to the type of rapid mutation described above. Long insertions in this region lead to a breakdown of the protein’s tertiary structure resulting in a loss of enzyme function and a childhood condition for those homozygous for these alleles that sometimes leads to major complications and even death. A pharmaceutical cure is developed that completely alleviates the symptoms of this condition. The presence of this new drug means that selection is no longer going to operate against this allele. If the mutation rate for the recessive deleterious alleles is 5 x 10^-4, and the starting frequency of the allele in the population is q= 0.025, what would be the expected frequency of this allele (q) after ten generations of no selection.

1b. Assuming that the population is otherwise in HWE for this locus, what is the change in the proportion of people that need the medication for this condition over these ten generations (remember that it’s recessive)?

1c. If the starting point of q= 0.025 above was the result of long-term mutation/selection balance at this locus, what was the selection coefficient against the disease state (homozygous recessive genotype) prior to the pharmaceutical treatment?

Answer 1

The system involved here is a single-gene model with the only driving force being mutation. The mutation is a very weak force in evolutionary time scales albeit a major force since it produces almost all the variations observed in the population.

Here we are considering a gene with two alleles. The normal allele produces poly-glutamine stretch. Let this allele be represented by G. The mutant allele has insertions and fails to produce a poly-glutamine stretch in the homozygous scenario. Let this allele be represented as g. There is a high mutation rate from G to g and this mutation rate is u = 5*10^-4 bases per generation. The reverse mutation from g to G is less frequent and can be neglected for our purpose.

The frequency of the recessive allele g is q = 0.025. Since the locus has only one allele, the frequency of allele G will be p = 1-q = 1-0.025 = 0.975.

a) After N generations, the frequency of the allele g can be calculated as:

b = Nb

010 = 0.025(1-5x 10-40 = 0.0250.99570 = 0.0250.9511 = 0.0299

Now the frequency of the dominant allele p after 10 generations would be

P10 = 1-0.0299 = 0.9701

Now, we can look at the genotypes of the individuals in this 10th generation.

b) The frequency of genotypes at 0th generation can be obtained by assuming the population to be in HWE equilibrium

The frequency of GG will be

p = 0.9752 = 0.9506

The frequency of Gg will be

Pogo = 0.975 x 0.025 = 0.0244

The frequency of gg will be

gå = 0.0252 = 0.0006

The treatment is aimed at homozygous recessive individuals. Hence at this stage, the treatment is required by 0.06% of individuals.

Since the population is in Hardy Weinberg Equilibrium, the frequency of the genotypes can be calculated as the product of the frequency of each allele that makes the genotype.

Thus, we have

Frequency of GG will be

pio = 0.97012 = 0.9411

Frequency of Gg will be

P10910 = 0.9701 x 0.0299 = 0.0290

Frequency of gg will be

qn = 0.02992 = 0.0000

At this stage, the treatment will be required by 0.009% of individuals. Hence 0.009-0.006 = 0.003% more people would need the treatment after 10 generations.

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c) The reproductive fitness of the individual depends on the survival rate and the reproductive rate. Here, the population is assumed to be ideal and hence the reproductive rate of each of the genotypes is equal. Thus the fitness depends only on the survival rate of the individual genotypes.

The survival rate is calculated as the proportion of a given genotype compared to the genotype that is most abundant.

Since the reproductive fitness in our scenario is a measure of the selective fitness, we have:

The reproductive fitness of GG is 0.9506/0.9506 = 1.

Reproductive fitness of Gg is 0.0244/0.9506 = 0.0257

Reproductive fitness of gg is 0.0006/0.9506 = 0.0006

Now, the selection coefficient is given by 1-reproductive fitness. Hence the selection coefficient of homozygous recessive phenotype or gg is 1-0.0006 = 0.9994.