A rare human disease is due to a genetic defect in enzymecalled AP. In a certain

Question

1. A rare human disease is due to a genetic defect in enzymecalled AP. In a certain, isolated population, a heterozygous childcarrying one mutant (defective) allele and one wild-type allele ofAP gene has 2% less chance of surviving into an reproductive adultcompared to a homozygous child with two wild-type alleles. Ahomozygous child with two mutant alleles suffers 25% decline insurvival relative to a wild-type homozygote. (Once becoming adults,individuals with three genotypes all leave the same number ofoffspring, on average.) It was found that one out of a millionchildren suffers from the severe form of this disease becausehe/she is homozygous for the defective AP allele. Assuming that thefrequency of this disease remains constant through time, what isthe probability that a wild-type AP allele will mutate to adefective allele during meiosis?

1. A rare human disease is due to a genetic defect in enzymecalled AP. In a certain, isolated population, a heterozygous childcarrying one mutant (defective) allele and one wild-type allele ofAP gene has 2% less chance of surviving into an reproductive adultcompared to a homozygous child with two wild-type alleles. Ahomozygous child with two mutant alleles suffers 25% decline insurvival relative to a wild-type homozygote. (Once becoming adults,individuals with three genotypes all leave the same number ofoffspring, on average.) It was found that one out of a millionchildren suffers from the severe form of this disease becausehe/she is homozygous for the defective AP allele. Assuming that thefrequency of this disease remains constant through time, what isthe probability that a wild-type AP allele will mutate to adefective allele during meiosis?

Explanation / Answer

a heterozygous child carrying one mutant (defective)allele and one wild-type allele of AP gene has 2% less chance ofsurviving into an reproductive adult compared to a homozygous childwith two wild-type alleles

fitness value ?

F + S =1

F= .98

S=.02

. A homozygous child with two mutant alleles suffers 25%decline in survival relative to a wild-type homozygote

fitness value ?

F + S =1

F= .75

S=.25

what is the probability that a wild-type AP allele willmutate to a defective allele during meiosis?

the probability that a wild-type AP allele will mutateto a defective allele during meiosis?

Is        .005 answer

OR        5 X 10-3

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